Physics Problems with Solutions

  • Uniform Acceleration Motion: Problems with Solutions

Problems on velocity and uniform acceleration are presented along with detailed solutions and tutorials can also be found in this website.

From rest, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 1

With an initial velocity of 20 km/h, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 2

A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds. a) What is the acceleration of the car in m/s 2 ? b) What is the position of the car by the time it reaches the velocity of 72 km/h? Solution to Problem 3

An object is thrown straight down from the top of a building at a speed of 20 m/s. It hits the ground with a speed of 40 m/s. a) How high is the building? b) How long was the object in the air? Solution to Problem 4

A train brakes from 40 m/s to a stop over a distance of 100 m. a) What is the acceleration of the train? b) How much time does it take the train to stop? Solution to Problem 5

A boy on a bicycle increases his velocity from 5 m/s to 20 m/s in 10 seconds. a) What is the acceleration of the bicycle? b) What distance was covered by the bicycle during the 10 seconds? Solution to Problem 6

a) How long does it take an airplane to take off if it needs to reach a speed on the ground of 350 km/h over a distance of 600 meters (assume the plane starts from rest)? b) What is the acceleration of the airplane over the 600 meters? Solution to Problem 7

Starting from a distance of 20 meters to the left of the origin and at a velocity of 10 m/s, an object accelerates to the right of the origin for 5 seconds at 4 m/s 2 . What is the position of the object at the end of the 5 seconds of acceleration? Solution to Problem 8

What is the smallest distance, in meters, needed for an airplane touching the runway with a velocity of 360 km/h and an acceleration of -10 m/s 2 to come to rest? Solution to Problem 9

Problem 10:

To approximate the height of a water well, Martha and John drop a heavy rock into the well. 8 seconds after the rock is dropped, they hear a splash caused by the impact of the rock on the water. What is the height of the well. (Speed of sound in air is 340 m/s). Solution to Problem 10

Problem 11:

A rock is thrown straight up and reaches a height of 10 m. a) How long was the rock in the air? b) What is the initial velocity of the rock? Solution to Problem 11

Problem 12:

A car accelerates from rest at 1.0 m/s 2 for 20.0 seconds along a straight road . It then moves at a constant speed for half an hour. It then decelerates uniformly to a stop in 30.0 s. Find the total distance covered by the car. Solution to Problem 12

More References and links

  • Velocity and Speed: Tutorials with Examples
  • Velocity and Speed: Problems with Solutions
  • Acceleration: Tutorials with Examples
  • Uniform Acceleration Motion: Equations with Explanations

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solving a uniform motion problem

4.8 Applications and Variation

Learning objectives.

  • Solve applications involving uniform motion (distance problems).
  • Solve work-rate applications.
  • Set up and solve applications involving direct, inverse, and joint variation.

Solving Uniform Motion Problems

Uniform motion (or distance) Described by the formula D = r t , where the distance D is given as the product of the average rate r and the time t traveled at that rate. problems involve the formula D = r t , where the distance D is given as the product of the average rate r and the time t traveled at that rate. If we divide both sides by the average rate r , then we obtain the formula

For this reason, when the unknown quantity is time, the algebraic setup for distance problems often results in a rational equation. We begin any uniform motion problem by first organizing our data with a chart. Use this information to set up an algebraic equation that models the application.

Sally traveled 15 miles on the bus and then another 72 miles on a train. The train was 18 miles per hour faster than the bus, and the total trip took 2 hours. What was the average speed of the train?

First, identify the unknown quantity and organize the data.

Let x represent the average speed (in miles per hour) of the bus.

Let x + 18 represent the average speed of the train.

solving a uniform motion problem

To avoid introducing two more variables for the time column, use the formula t = D r . The time for each leg of the trip is calculated as follows:

T i m e   s p e n t   o n   t h e   b u s : t = D r = 15 x T i m e   s p e n t   o n   t h e   t r a i n : t = D r = 72 x + 18

Use these expressions to complete the chart.

solving a uniform motion problem

The algebraic setup is defined by the time column. Add the time spent on each leg of the trip to obtain a total of 2 hours:

solving a uniform motion problem

We begin solving this equation by first multiplying both sides by the LCD, x ( x + 18 ) .

15 x + 72 x + 18 = 2 x ( x + 18 ) ⋅ ( 15 x + 72 x + 18 ) = x ( x + 18 ) ⋅ 2 x ( x + 18 ) ⋅ 15 x + x ( x + 18 ) ⋅ 72 x + 18 = x ( x + 18 ) ⋅ 2 15 ( x + 18 ) + 72 x = 2 x ( x + 18 ) 15 x + 270 + 72 x = 2 x 2 + 36 x 87 x + 270 = 2 x 2 + 36 x 0 = 2 x 2 − 51 x − 270

Solve the resulting quadratic equation by factoring.

0 = 2 x 2 − 51 x − 270 0 = ( 2 x + 9 ) ( x − 30 ) 2 x + 9 =   0 or x − 30 = 0 x = − 9 2 x = 30

Since we are looking for an average speed we will disregard the negative answer and conclude the bus averaged 30 mph. Substitute x = 30 in the expression identified as the speed of the train.

x + 18 = 30 + 18 = 48

Answer: The speed of the train was 48 mph.

A boat can average 12 miles per hour in still water. On a trip downriver the boat was able to travel 29 miles with the current. On the return trip the boat was only able to travel 19 miles in the same amount of time against the current. What was the speed of the current?

First, identify the unknown quantities and organize the data.

Let c represent the speed of the river current.

Next, organize the given data in a chart. Traveling downstream, the current will increase the speed of the boat, so it adds to the average speed of the boat. Traveling upstream, the current slows the boat, so it will subtract from the average speed of the boat.

solving a uniform motion problem

Use the formula t = D r to fill in the time column.

t r i p   d o w n r i v e r : t = D r = 29 12 + c t r i p   u p r i v e r : t = D r = 19 12 − c

solving a uniform motion problem

Because the boat traveled the same amount of time downriver as it did upriver, finish the algebraic setup by setting the expressions that represent the times equal to each other.

29 12 + c = 19 12 − c

Since there is a single algebraic fraction on each side, we can solve this equation using cross multiplication.

29 12 + c = 19 12 − c 29 ( 12 − c ) = 19 ( 12 + c ) 348 − 29 c = 228 + 19 c 120 = 48 c 120 48 = c 5 2 = c

Answer: The speed of the current was 2 1 2 miles per hour.

Try this! A jet aircraft can average 160 miles per hour in calm air. On a trip, the aircraft traveled 600 miles with a tailwind and returned the 600 miles against a headwind of the same speed. If the total round trip took 8 hours, then what was the speed of the wind?

Answer: 40 miles per hour

Solving Work-Rate Problems

The rate at which a task can be performed is called a work rate The rate at which a task can be performed. . For example, if a painter can paint a room in 6 hours, then the task is to paint the room, and we can write

1     task 6   hours          work rate

In other words, the painter can complete 1 6 of the task per hour. If he works for less than 6 hours, then he will perform a fraction of the task. If he works for more than 6 hours, then he can complete more than one task. For example,

w o r k - r a t e       ×       t i m e       =   a m o u n t   o f   t a s k   c o m p l e t e d   1   6             ×           3     h r s           =         1 2   o n e - h a l f   o f   t h e   r o o m   p a i n t e d   1   6             ×           6     h r s           =         1   o n e   w h o l e   r o o m   p a i n t e d             1   6             ×             12     h r s     =         2   t w o   w h o l e   r o o m s   p a i n t e d

Obtain the amount of the task completed by multiplying the work rate by the amount of time the painter works. Typically, work-rate problems involve people or machines working together to complete tasks. In general, if t represents the time two people work together, then we have the following work-rate formula 1 t 1 ⋅ t + 1 t 2 ⋅ t = 1 , where 1 t 1 and 1 t 2 are the individual work rates and t is the time it takes to complete the task working together. :

1 t 1 t + 1 t 2 t =   a m o u n t   o f   t a s k   c o m p l e t e d   t o g e t h e r

Here 1 t 1 and 1 t 2 are the individual work rates.

Joe can paint a typical room in 2 hours less time than Mark. If Joe and Mark can paint 5 rooms working together in a 12 hour shift, how long does it take each to paint a single room?

Let x represent the time it takes Mark to paint a typical room.

Let x − 2 represent the time it takes Joe to paint a typical room.

Therefore, Mark’s individual work-rate is 1 x rooms per hour and Joe’s is 1 x − 2 rooms per hour. Both men worked for 12 hours. We can organize the data in a chart, just as we did with distance problems.

solving a uniform motion problem

Working together, they can paint 5 total rooms in 12 hours. This leads us to the following algebraic setup:

12 x − 2 + 12 x = 5

Multiply both sides by the LCD, x ( x − 2 ) .

x ( x − 2 ) ⋅ ( 12 x − 2 + 12 x ) = x ( x − 2 ) ⋅ 5 x ( x − 2 ) ⋅ 12 x − 2 + x ( x − 2 ) ⋅ 12 x = x ( x − 2 ) ⋅ 5 12 x + 12 ( x − 2 ) = 5 x ( x − 2 ) 12 x + 12 x − 24 = 5 x 2 − 10 x 0 = 5 x 2 − 34 x + 24

0 = 5 x 2 − 34 x + 24 0 = ( 5 x − 4 ) ( x − 6 ) 5 x − 4 = 0 o r x − 6 = 0 5 x = 4 x = 6 x = 4 5

We can disregard 4 5 because back substituting into x − 2 would yield a negative time to paint a room. Take x = 6 to be the only solution and use it to find the time it takes Joe to paint a typical room.

x − 2 = 6 − 2 = 4

Answer: Joe can paint a typical room in 4 hours and Mark can paint a typical room in 6 hours. As a check we can multiply both work rates by 12 hours to see that together they can paint 5 rooms.

J o e 1     r o o m 4   h r s ⋅ 12   h r s = 3   r o o m s M a r k 1   r o o m 6   h r s ⋅ 12   h r s = 2   r o o m s   }   T o t a l   5   r o o m s    ✓

It takes Bill twice as long to lay a tile floor by himself as it does Manny. After working together with Bill for 4 hours, Manny was able to complete the job in 2 additional hours. How long would it have taken Manny working alone?

Let x represent the time it takes Manny to lay the floor alone.

Let 2 x represent the time it takes Bill to lay the floor alone.

Manny’s work rate is 1 x of the floor per hour and Bill’s work rate is 1 2 x . Bill worked on the job for 4 hours and Manny worked on the job for 6 hours.

solving a uniform motion problem

This leads us to the following algebraic setup:

1 x ⋅ 6 + 1 2 x ⋅ 4 = 1

6 x + 4 2 x = 1 x ⋅ ( 6 x + 2 x ) = x ⋅ 1 6 + 2 = x 8 = x

Answer: It would have taken Manny 8 hours to complete the floor by himself.

Consider the work-rate formula where one task is to be completed.

1 t 1 t + 1 t 2 t = 1

Factor out the time t and then divide both sides by t . This will result in equivalent specialized work-rate formulas:

      t ( 1 t 1 + 1 t 2 ) = 1 1 t 1 + 1 t 2 = 1 t

In summary, we have the following equivalent work-rate formulas:

Try this! Matt can tile a countertop in 2 hours, and his assistant can do the same job in 3 hours. If Matt starts the job and his assistant joins him 1 hour later, then how long will it take to tile the countertop?

Answer: 1 3 5 hours

Solving Problems involving Direct, Inverse, and Joint variation

Many real-world problems encountered in the sciences involve two types of functional relationships. The first type can be explored using the fact that the distance s in feet an object falls from rest, without regard to air resistance, can be approximated using the following formula:

Here t represents the time in seconds the object has been falling. For example, after 2 seconds the object will have fallen s = 16 ( 2 ) 2 = 16 ⋅ 4 = 64 feet.

In this example, we can see that the distance varies over time as the product of a constant 16 and the square of the time t . This relationship is described as direct variation Describes two quantities x and y that are constant multiples of each other: y = k x . and 16 is called the constant of variation The nonzero multiple k , when quantities vary directly or inversely. . Furthermore, if we divide both sides of s = 16 t 2 by t 2 we have

In this form, it is reasonable to say that s is proportional to t 2 , and 16 is called the constant of proportionality Used when referring to the constant of variation. . In general, we have

Used when referring to direct variation.

Here k is nonzero and is called the constant of variation or the constant of proportionality. Typically, we will be given information from which we can determine this constant.

An object’s weight on Earth varies directly to its weight on the Moon. If a man weighs 180 pounds on Earth, then he will weigh 30 pounds on the Moon. Set up an algebraic equation that expresses the weight on Earth in terms of the weight on the Moon and use it to determine the weight of a woman on the Moon if she weighs 120 pounds on Earth.

Let y represent weight on Earth.

Let x represent weight on the Moon.

We are given that the “weight on Earth varies directly to the weight on the Moon.”

To find the constant of variation k , use the given information. A 180-lb man on Earth weighs 30 pounds on the Moon, or y   =   180 when   x   =   30 .

180 = k ⋅ 30

Solve for k .

180 30 = k 6 = k

Next, set up a formula that models the given information.

This implies that a person’s weight on Earth is 6 times his weight on the Moon. To answer the question, use the woman’s weight on Earth, y = 120 lbs, and solve for x .

120 = 6 x 120 6 = x 20 = x

Answer: The woman weighs 20 pounds on the Moon.

The second functional relationship can be explored using the formula that relates the intensity of light I to the distance from its source d .

Here k represents some constant. A foot-candle is a measurement of the intensity of light. One foot-candle is defined to be equal to the amount of illumination produced by a standard candle measured one foot away. For example, a 125-Watt fluorescent growing light is advertised to produce 525 foot-candles of illumination. This means that at a distance d = 1 foot, I = 525 foot-candles and we have:

525 = k ( 1 ) 2 525 = k

Using k = 525 we can construct a formula which gives the light intensity produced by the bulb:

I = 525 d 2

Here d represents the distance the growing light is from the plants. In the following chart, we can see that the amount of illumination fades quickly as the distance from the plants increases.

This type of relationship is described as an inverse variation Describes two quantities x and y , where one variable is directly proportional to the reciprocal of the other: y = k x . . We say that I is inversely proportional Used when referring to inverse variation. to the square of the distance d , where 525 is the constant of proportionality. In general, we have

Again, k is nonzero and is called the constant of variation or the constant of proportionality.

The weight of an object varies inversely as the square of its distance from the center of Earth. If an object weighs 100 pounds on the surface of Earth (approximately 4,000 miles from the center), how much will it weigh at 1,000 miles above Earth’s surface?

Let w represent the weight of the object.

Let d represent the object’s distance from the center of Earth.

Since “ w varies inversely as the square of d ,” we can write

Use the given information to find k . An object weighs 100 pounds on the surface of Earth, approximately 4,000 miles from the center. In other words, w = 100 when d = 4,000:

100 = k ( 4,000 ) 2

( 4,000 ) 2 ⋅ 100 = ( 4,000 ) 2 ⋅ k ( 4,000 ) 2 1,600,000,000 = k 1.6 × 10 9 = k

Therefore, we can model the problem with the following formula:

w = 1.6 × 10 9 d 2

To use the formula to find the weight, we need the distance from the center of Earth. Since the object is 1,000 miles above the surface, find the distance from the center of Earth by adding 4,000 miles:

d = 4,000 + 1,000 = 5,000     miles

To answer the question, use the formula with d = 5,000.

y = 1.6 × 10 9 ( 5,000 ) 2 = 1.6 × 10 9 25,000,000 = 1.6 × 10 9 2.5 × 10 7 = 0.64 × 10 2 = 64

Answer: The object will weigh 64 pounds at a distance 1,000 miles above the surface of Earth.

Lastly, we define relationships between multiple variables, described as joint variation Describes a quantity y that varies directly as the product of two other quantities x and z : y = k x z . . In general, we have

Used when referring to joint variation.

Here k is nonzero and is called the constant of variation or the constant of proportionality.

The area of an ellipse varies jointly as a , half of the ellipse’s major axis, and b , half of the ellipse’s minor axis as pictured. If the area of an ellipse is 300 π   cm 2 , where a = 10   cm and b = 30   cm , what is the constant of proportionality? Give a formula for the area of an ellipse.

solving a uniform motion problem

If we let A represent the area of an ellipse, then we can use the statement “area varies jointly as a and b ” to write

To find the constant of variation k , use the fact that the area is 300 π when a = 10 and b = 30 .

300 π = k ( 10 ) ( 30 ) 300 π = 300 k π = k

Therefore, the formula for the area of an ellipse is

Answer: The constant of proportionality is π and the formula for the area of an ellipse is A = a b π .

Try this! Given that y varies directly as the square of x and inversely with z , where y = 2 when x = 3 and z = 27, find y when x = 2 and z = 16.

Answer: 3 2

Key Takeaways

  • When solving distance problems where the time element is unknown, use the equivalent form of the uniform motion formula, t = D r , to avoid introducing more variables.
  • When solving work-rate problems, multiply the individual work rate by the time to obtain the portion of the task completed. The sum of the portions of the task results in the total amount of work completed.
  • The setup of variation problems usually requires multiple steps. First, identify the key words to set up an equation and then use the given information to find the constant of variation k . After determining the constant of variation, write a formula that models the problem. Once a formula is found, use it to answer the question.

Topic Exercises

Part a: solving uniform motion problems.

Use algebra to solve the following applications.

Every morning Jim spends 1 hour exercising. He runs 2 miles and then he bikes 16 miles. If Jim can bike twice as fast as he can run, at what speed does he average on his bike?

Sally runs 3 times as fast as she walks. She ran for 3 4 of a mile and then walked another 3 1 2 miles. The total workout took 1 1 2 hours. What was Sally’s average walking speed?

On a business trip, an executive traveled 720 miles by jet and then another 80 miles by helicopter. If the jet averaged 3 times the speed of the helicopter, and the total trip took 4 hours, what was the average speed of the jet?

A triathlete can run 3 times as fast as she can swim and bike 6 times as fast as she can swim. The race consists of a 1 4 mile swim, 3 mile run, and a 12 mile bike race. If she can complete all of these events in 1 5 8 hour, then how fast can she swim, run and bike?

On a road trip, Marty was able to drive an average 4 miles per hour faster than George. If Marty was able to drive 39 miles in the same amount of time George drove 36 miles, what was Marty’s average speed?

The bus is 8 miles per hour faster than the trolley. If the bus travels 9 miles in the same amount of time the trolley can travel 7 miles, what is the average speed of each?

Terry decided to jog the 5 miles to town. On the return trip, she walked the 5 miles home at half of the speed that she was able to jog. If the total trip took 3 hours, what was her average jogging speed?

James drove the 24 miles to town and back in 1 hour. On the return trip, he was able to average 20 miles per hour faster than he averaged on the trip to town. What was his average speed on the trip to town?

A light aircraft was able to travel 189 miles with a 14 mile per hour tailwind in the same time it was able to travel 147 miles against it. What was the speed of the aircraft in calm air?

A jet flew 875 miles with a 30 mile per hour tailwind. On the return trip, against a 30 mile per hour headwind, it was able to cover only 725 miles in the same amount of time. How fast was the jet in calm air?

A helicopter averaged 90 miles per hour in calm air. Flying with the wind it was able to travel 250 miles in the same amount of time it took to travel 200 miles against it. What is the speed of the wind?

Mary and Joe took a road-trip on separate motorcycles. Mary’s average speed was 12 miles per hour less than Joe’s average speed. If Mary drove 115 miles in the same time it took Joe to drive 145 miles, what was Mary’s average speed?

A boat averaged 12 miles per hour in still water. On a trip downstream, with the current, the boat was able to travel 26 miles. The boat then turned around and returned upstream 33 miles. How fast was the current if the total trip took 5 hours?

If the river current flows at an average 3 miles per hour, a tour boat can make an 18-mile tour downstream with the current and back the 18 miles against the current in 4 1 2 hours. What is the average speed of the boat in still water?

Jose drove 10 miles to his grandmother’s house for dinner and back that same evening. Because of traffic, he averaged 20 miles per hour less on the return trip. If it took 1 4 hour longer to get home, what was his average speed driving to his grandmother’s house?

Jerry paddled his kayak, upstream against a 1 mph current, for 12 miles. The return trip, downstream with the 1 mph current, took one hour less time. How fast did Jerry paddle the kayak in still water?

James and Mildred left the same location in separate cars and met in Los Angeles 300 miles away. James was able to average 10 miles an hour faster than Mildred on the trip. If James arrived 1 hour earlier than Mildred, what was Mildred’s average speed?

A bus is 20 miles per hour faster than a bicycle. If Bill boards a bus at the same time and place that Mary departs on her bicycle, Bill will arrive downtown 5 miles away 1 3 hour earlier than Mary. What is the average speed of the bus?

Part B: Solving Work-Rate Problems

Mike can paint the office by himself in 4 1 2 hours. Jordan can paint the office in 6 hours. How long will it take them to paint the office working together?

Barry can lay a brick driveway by himself in 3 1 2 days. Robert does the same job in 5 days. How long will it take them to lay the brick driveway working together?

A larger pipe fills a water tank twice as fast as a smaller pipe. When both pipes are used, they fill the tank in 10 hours. If the larger pipe is left off, how long would it take the smaller pipe to fill the tank?

A newer printer can print twice as fast as an older printer. If both printers working together can print a batch of flyers in 45 minutes, then how long would it take the older printer to print the batch working alone?

Mary can assemble a bicycle for display in 2 hours. It takes Jane 3 hours to assemble a bicycle. How long will it take Mary and Jane, working together, to assemble 5 bicycles?

Working alone, James takes twice as long to assemble a computer as it takes Bill. In one 8-hour shift, working together, James and Bill can assemble 6 computers. How long would it take James to assemble a computer if he were working alone?

Working alone, it takes Harry one hour longer than Mike to install a fountain. Together they can install 10 fountains in 12 hours. How long would it take Mike to install 10 fountains by himself?

Working alone, it takes Henry 2 hours longer than Bill to paint a room. Working together they painted 2 1 2 rooms in 6 hours. How long would it have taken Henry to paint the same amount if he were working alone?

Manny, working alone, can install a custom cabinet in 3 hours less time than his assistant. Working together they can install the cabinet in 2 hours. How long would it take Manny to install the cabinet working alone?

Working alone, Garret can assemble a garden shed in 5 hours less time than his brother. Working together, they need 6 hours to build the garden shed. How long would it take Garret to build the shed working alone?

Working alone, the assistant-manager takes 2 more hours than the manager to record the inventory of the entire shop. After working together for 2 hours, it took the assistant-manager 1 additional hour to complete the inventory. How long would it have taken the manager to complete the inventory working alone?

An older printer can print a batch of sales brochures in 16 minutes. A newer printer can print the same batch in 10 minutes. After working together for some time, the newer printer was shut down and it took the older printer 3 more minutes to complete the job. How long was the newer printer operating?

Part C: Solving Variation Problems

Translate each of the following sentences into a mathematical formula.

The distance D an automobile can travel is directly proportional to the time t that it travels at a constant speed.

The extension of a hanging spring d is directly proportional to the weight w attached to it.

An automobile’s braking distance d is directly proportional to the square of the automobile’s speed v .

The volume V of a sphere varies directly as the cube of its radius r .

The volume V of a given mass of gas is inversely proportional to the pressure p exerted on it.

Every particle of matter in the universe attracts every other particle with a force F that is directly proportional to the product of the masses m 1 and m 2 of the particles, and it is inversely proportional to the square of the distance d between them.

Simple interest I is jointly proportional to the annual interest rate r and the time t in years a fixed amount of money is invested.

The time t it takes an object to fall is directly proportional to the square root of the distance d it falls.

Construct a mathematical model given the following:

y varies directly as x , and y = 30 when x = 6.

y varies directly as x , and y = 52 when x = 4.

y is directly proportional to x , and y = 12 when x = 3.

y is directly proportional to x , and y = 120 when x = 20.

y is inversely proportional to x , and y = 3 when x = 9.

y is inversely proportional to x , and y = 21 when x = 3.

y varies inversely as x , and y = 2 when x = 1 8 .

y varies inversely as x , and y = 3 2 when x = 1 9 .

y is jointly proportional to x and z , where y = 2 when x = 1 and z = 3.

y is jointly proportional to x and z , where y = 15 when x = 3 and z = 7.

y varies jointly as x and z , where y = 2 3 when x = 1 2 and z = 12.

y varies jointly as x and z , where y = 5 when x = 3 2 and z = 2 9 .

y varies directly as the square of x , where y = 45 when x = 3.

y varies directly as the square of x , where y = 3 when x = 1 2 .

y is inversely proportional to the square of x , where y = 27 when x = 1 3 .

y is inversely proportional to the square of x , where y = 9 when x = 2 3 .

y varies jointly as x and the square of z , where y = 6 when x = 1 4 and z = 2 3 .

y varies jointly as x and z and inversely as the square of w , where y = 5 when x = 1, z = 3, and w = 1 2 .

y varies directly as the square root of x and inversely as the square of z , where y = 15 when x = 25 and z = 2.

y varies directly as the square of x and inversely as z and the square of w , where y = 14 when x = 4, w = 2, and z = 2.

Solve applications involving variation.

Revenue in dollars is directly proportional to the number of branded sweatshirts sold. The revenue earned from selling 25 sweatshirts is $318.75. Determine the revenue if 30 sweatshirts are sold.

The sales tax on the purchase of a new car varies directly as the price of the car. If an $18,000 new car is purchased, then the sales tax is $1,350. How much sales tax is charged if the new car is priced at $22,000?

The price of a share of common stock in a company is directly proportional to the earnings per share (EPS) of the previous 12 months. If the price of a share of common stock in a company is $22.55, and the EPS is published to be $1.10, determine the value of the stock if the EPS increases by $0.20.

The distance traveled on a road trip varies directly with the time spent on the road. If a 126-mile trip can be made in 3 hours, then what distance can be traveled in 4 hours?

The circumference of a circle is directly proportional to its radius. The circumference of a circle with radius 7 centimeters is measured as 14 π centimeters. What is the constant of proportionality?

The area of circle varies directly as the square of its radius. The area of a circle with radius 7 centimeters is determined to be 49 π square centimeters. What is the constant of proportionality?

The surface area of a sphere varies directly as the square of its radius. When the radius of a sphere measures 2 meters, the surface area measures 16 π square meters. Find the surface area of a sphere with radius 3 meters.

The volume of a sphere varies directly as the cube of its radius. When the radius of a sphere measures 3 meters, the volume is 36 π cubic meters. Find the volume of a sphere with radius 1 meter.

With a fixed height, the volume of a cone is directly proportional to the square of the radius at the base. When the radius at the base measures 10 centimeters, the volume is 200 cubic centimeters. Determine the volume of the cone if the radius of the base is halved.

The distance d an object in free fall drops varies directly with the square of the time t that it has been falling. If an object in free fall drops 36 feet in 1.5 seconds, then how far will it have fallen in 3 seconds?

Hooke’s law suggests that the extension of a hanging spring is directly proportional to the weight attached to it. The constant of variation is called the spring constant.

solving a uniform motion problem

Robert Hooke (1635—1703)

A hanging spring is stretched 5 inches when a 20-pound weight is attached to it. Determine its spring constant.

A hanging spring is stretched 3 centimeters when a 2-kilogram weight is attached to it. Determine the spring constant.

If a hanging spring is stretched 3 inches when a 2-pound weight is attached, how far will it stretch with a 5-pound weight attached?

If a hanging spring is stretched 6 centimeters when a 4-kilogram weight is attached to it, how far will it stretch with a 2-kilogram weight attached?

The braking distance of an automobile is directly proportional to the square of its speed.

It takes 36 feet to stop a particular automobile moving at a speed of 30 miles per hour. How much breaking distance is required if the speed is 35 miles per hour?

After an accident, it was determined that it took a driver 80 feet to stop his car. In an experiment under similar conditions, it takes 45 feet to stop the car moving at a speed of 30 miles per hour. Estimate how fast the driver was moving before the accident.

solving a uniform motion problem

Robert Boyle (1627—1691)

Boyle’s law states that if the temperature remains constant, the volume V of a given mass of gas is inversely proportional to the pressure p exerted on it.

A balloon is filled to a volume of 216 cubic inches on a diving boat under 1 atmosphere of pressure. If the balloon is taken underwater approximately 33 feet, where the pressure measures 2 atmospheres, then what is the volume of the balloon?

A balloon is filled to 216 cubic inches under a pressure of 3 atmospheres at a depth of 66 feet. What would the volume be at the surface, where the pressure is 1 atmosphere?

To balance a seesaw, the distance from the fulcrum that a person must sit is inversely proportional to his weight. If a 72-pound boy is sitting 3 feet from the fulcrum, how far from the fulcrum must a 54-pound boy sit to balance the seesaw?

The current I in an electrical conductor is inversely proportional to its resistance R . If the current is 1 4 ampere when the resistance is 100 ohms, what is the current when the resistance is 150 ohms?

The amount of illumination I is inversely proportional to the square of the distance d from a light source. If 70 foot-candles of illumination is measured 2 feet away from a lamp, what level of illumination might we expect 1 2 foot away from the lamp?

The amount of illumination I is inversely proportional to the square of the distance d from a light source. If 40 foot-candles of illumination is measured 3 feet away from a lamp, at what distance can we expect 10 foot-candles of illumination?

The number of men, represented by y , needed to lay a cobblestone driveway is directly proportional to the area A of the driveway and inversely proportional to the amount of time t allowed to complete the job. Typically, 3 men can lay 1,200 square feet of cobblestone in 4 hours. How many men will be required to lay 2,400 square feet of cobblestone in 6 hours?

The volume of a right circular cylinder varies jointly as the square of its radius and its height. A right circular cylinder with a 3-centimeter radius and a height of 4 centimeters has a volume of 36 π cubic centimeters. Find a formula for the volume of a right circular cylinder in terms of its radius and height.

The period T of a pendulum is directly proportional to the square root of its length L . If the length of a pendulum is 1 meter, then the period is approximately 2 seconds. Approximate the period of a pendulum that is 0.5 meter in length.

The time t it takes an object to fall is directly proportional to the square root of the distance d it falls. An object dropped from 4 feet will take 1 2 second to hit the ground. How long will it take an object dropped from 16 feet to hit the ground?

Newton’s universal law of gravitation states that every particle of matter in the universe attracts every other particle with a force F that is directly proportional to the product of the masses m 1 and m 2 of the particles and inversely proportional to the square of the distance d between them. The constant of proportionality is called the gravitational constant.

solving a uniform motion problem

Sir Isaac Newton (1643—1727)

Source: Portrait of Isaac Newton by Sir Godfrey Kneller, from http://commons.wikimedia.org/wiki/File:GodfreyKneller-IsaacNewton-1689.

http://commons.wikimedia.org/wiki/File:Frans_Hals_-_Portret_

_van_Ren%C3%A9_Descartes.jpg.

If two objects with masses 50 kilograms and 100 kilograms are 1 2 meter apart, then they produce approximately 1.34 × 10 − 6 newtons (N) of force. Calculate the gravitational constant.

Use the gravitational constant from the previous exercise to write a formula that approximates the force F in newtons between two masses m 1 and m 2 , expressed in kilograms, given the distance d between them in meters.

Calculate the force in newtons between Earth and the Moon, given that the mass of the Moon is approximately 7.3 × 10 22 kilograms, the mass of Earth is approximately 6.0 × 10 24 kilograms, and the distance between them is on average 1.5 × 10 11 meters.

Calculate the force in newtons between Earth and the Sun, given that the mass of the Sun is approximately 2.0 × 10 30 kilograms, the mass of Earth is approximately 6.0 × 10 24 kilograms, and the distance between them is on average 3.85 × 10 8 meters.

If y varies directly as the square of x , then how does y change if x is doubled?

If y varies inversely as square of t , then how does y change if t is doubled?

If y varies directly as the square of x and inversely as the square of t , then how does y change if both x and t are doubled?

20 miles per hour

240 miles per hour

52 miles per hour

5 miles per hour

112 miles per hour

10 miles per hour

1 mile per hour

40 miles per hour

50 miles per hour

2 4 7 hours

y = 2 3 x z

y = 1 9 x z

  • y = 54 x z 2
  • y = 12 x z 2

36 π square meters

50 cubic centimeters

108 cubic inches

1,120 foot-candles

1.4 seconds

6.7 × 10 − 11   N   m 2 / kg 2

1.98 × 10 20   N

y changes by a factor of 4

y remains unchanged

Module 5: Systems of Linear Equations

5.2 – applications of systems of linear equations, learning objectives.

  • Specify what the variables in a cost/ revenue system of linear equations represent
  • Determine and apply an appropriate method for solving the system

(5.2.2) – Solve value problems with a system of linear equations

(5.2.3) – solve mixture problems with a system of linear equations, (5.2.4) – solve uniform motion problems with a system of linear equations.

A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible?

Skateboarders at a skating rink by the beach.

(credit: Thomas Sørenes)

(5.2.1) – Solve cost and revenue problems

Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[/latex], where [latex]x=[/latex] quantity and [latex]p=[/latex] price. The revenue function is shown in orange in the graph below.

The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The [latex]x[/latex] -axis represents quantity in hundreds of units. The y -axis represents either cost or revenue in hundreds of dollars.

A graph showing money in hundreds of dollars on the y axis and quantity in hundreds of units on the x axis. A line representing cost and a line representing revenue cross at the point (7,33), which is marked break-even. The shaded space between the two lines to the right of the break-even point is labeled profit.

The point at which the two lines intersect is called the break-even point . We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.

The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex]. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.

A business wants to manufacture bike frames. Before they start production, they need to make sure they can make a profit with the materials and labor force they have. Their accountant has given them a cost equation of [latex]y=0.85x+35,000[/latex] and a revenue equation of [latex]y=1.55x[/latex]:

  • Interpret x and y for the cost equation
  • Interpret x and y for the revenue equation

Cost: [latex]y=0.85x+35,000[/latex]

Revenue:[latex]y=1.55x[/latex]

The cost equation represents money leaving the company, namely how much it costs to produce a given number of bike frames. If we use the skateboard example as a model, x would represent the number of frames produced (instead of skateboards) and y would represent the amount of money it would cost to produce them (the same as the skateboard problem).

The revenue equation represents money coming into the company, so in this context x still represents the number of bike frames manufactured, and y now represents the amount of money made from selling them.  Let’s organize this information in a table:

Example: Finding the Break-Even Point and the Profit Function Using Substitution

Given the cost function [latex]C\left(x\right)=0.85x+35,000[/latex] and the revenue function [latex]R\left(x\right)=1.55x[/latex], find the break-even point and the profit function.

Write the system of equations using [latex]y[/latex] to replace function notation.

[latex]\begin{array}{l}\begin{array}{l}\\ y=0.85x+35,000\end{array}\hfill \\ y=1.55x\hfill \end{array}[/latex]

Substitute the expression [latex]0.85x+35,000[/latex] from the first equation into the second equation and solve for [latex]x[/latex].

[latex]\begin{array}{c}0.85x+35,000=1.55x\\ 35,000=0.7x\\ 50,000=x\end{array}[/latex]

Then, we substitute [latex]x=50,000[/latex] into either the cost function or the revenue function. [latex]1.55\left(50,000\right)=77,500[/latex]

The break-even point is [latex]\left(50,000,77,500\right)[/latex].

The profit function is found using the formula [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex].

[latex]\begin{array}{l}P\left(x\right)=1.55x-\left(0.85x+35,000\right)\hfill \\ \text{ }=0.7x - 35,000\hfill \end{array}[/latex]

The profit function is [latex]P\left(x\right)=0.7x - 35,000[/latex].

Analysis of the Solution

The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.

A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.

We see from the graph below that the profit function has a negative value until [latex]x=50,000[/latex], when the graph crosses the x -axis. Then, the graph emerges into positive y -values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.

A graph showing dollars profit on the y axis and quantity on the x axis. The profit line crosses the break-even point at fifty thousand, zero. The profit line's equation is P(x)=0.7x-35,000.

It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.

How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.

1) Identify unknown quantities in a problem represent them with variables.

2) Write a system of equations which models the problem’s conditions.

3) Solve the system.

4) Check proposed solution.

Now let’s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.

Example: Writing and Solving a System of Equations in Two Variables

The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?

Let c = the number of children and a = the number of adults in attendance.

The total number of people is [latex]2,000[/latex]. We can use this to write an equation for the number of people at the circus that day.

[latex]c+a=2,000[/latex]

The revenue from all children can be found by multiplying $25.00 by the number of children, [latex]25c[/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[/latex]. The total revenue is $70,000. We can use this to write an equation for the revenue.

[latex]25c+50a=70,000[/latex]

We now have a system of linear equations in two variables.

[latex]\begin{array}{c}c+a=2,000\\ 25c+50a=70,000\end{array}[/latex]

In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either [latex]c[/latex] or [latex]a[/latex]. We will solve for [latex]a[/latex].

[latex]\begin{array}{c}c+a=2,000\\ a=2,000-c\end{array}[/latex]

Substitute the expression [latex]2,000-c[/latex] in the second equation for [latex]a[/latex] and solve for [latex]c[/latex].

[latex]\begin{array}{l} 25c+50\left(2,000-c\right)=70,000\hfill \\ 25c+100,000 - 50c=70,000\hfill \\ \text{ }-25c=-30,000\hfill \\ \text{ }c=1,200\hfill \end{array}[/latex]

Substitute [latex]c=1,200[/latex] into the first equation to solve for [latex]a[/latex].

[latex]\begin{array}{l}1,200+a=2,000\hfill \\ \text{ }\text{}a=800\hfill \end{array}[/latex]

We find that [latex]1,200[/latex] children and [latex]800[/latex] adults bought tickets to the circus that day.

In this video example we show how to set up a system of linear equations that represents the total cost for admission to a museum.

Meal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets?

700 children, 950 adults

Sometimes, a system can inform a decision.  In our next example, we help answer the question, “Which truck rental company will give the best value?”

Example: Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile. [1] When will Keep on Trucking, Inc. be the better choice for Jamal?

The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.

A linear function is of the form [latex]f\left(x\right)=mx+b[/latex]. Using the rates of change and initial charges, we can write the equations

[latex]\begin{array}{l}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{array}[/latex]

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when [latex]K\left(d\right)<M\left(d\right)[/latex]. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the [latex]K\left(d\right)[/latex] function is smaller.

image

These graphs are sketched above, with K ( d ) in blue.

To find the intersection, we set the equations equal and solve:

[latex]\begin{array}{l}K\left(d\right)=M\left(d\right)\hfill \\ 0.59d+20=0.63d+16\hfill \\ 4=0.04d\hfill \\ 100=d\hfill \\ d=100\hfill \end{array}[/latex]

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that [latex]K\left(d\right)[/latex] is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is [latex]d>100[/latex].

One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution.  A solution is a mixture of two or more different substances like water and salt or vinegar and oil.  Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand.  There are many other disciplines that use solutions as well.

The concentration or strength of a liquid solution is often described  as a percentage.  This number comes from the ratio of how much mass is in a specific volume of liquid.  For example if you have 50 grams of salt in a 100mL of water you have a 50% salt solution based on the following ratio:

[latex]\frac{50\text{ grams }}{100\text{ mL }}=0.50\frac{\text{ grams }}{\text{ mL }}=50\text{ % }[/latex]

Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths.  In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.

We will use the following table to help us solve mixture problems:

To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is 120mL of a 9% solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.

First, find the total mass of solids for each solution by multiplying the volume by the concentration.

Next we add the new volumes and new masses.

Now we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution.  In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.

A chemist has 70 mL of a 50% methane solution. How much of an 80% solution must she add so the final solution is 60% methane?

Let’s use the problem solving process outlined in Module 1 to help us work through a solution to the problem.

Read and Understand:  We are looking for a new amount – in this case a volume –  based on the words “how much”.  We know two starting  concentrations and the final concentration, as well as one volume.

Define and Translate:  Solution 1 is the 70 mL of 50% methane and solution 2 is the unknown amount with 80% methane.  We can call our unknown amount x.

Write and Solve:   Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.

Multiply amount by concentration to get total, be sure to distribute on the last row: [latex]\left(70 + x\right)0.6[/latex]Add the entries in the amount column to get final amount. The concentration for this amount is 0.6 because we want the final solution to be 60% methane.

Add the total mass for solution 1 and solution 2 to get the total mass for the 60% solution. This is our equation for finding the unknown volume.

[latex]35+0.8x=42+0.6x[/latex]

[latex]\begin{array}{c}35+0.8x=42+0.6x\\\underline{-0.6x}\,\,\,\,\,\,\,\underline{-0.6x}\\35+0.2x=42\\\end{array}[/latex]

Subtract 35 from both sides

[latex]\begin{array}{c}35+0.2x=42\\\underline{-35}\,\,\,\,\,\,\,\underline{-35}\\0.2x=7\end{array}[/latex]

Divide both sides by 0.2

[latex]\begin{array}{c}0.2x=7\\\frac{0.2x}{0.2}=\frac{7}{0.2}\end{array}[/latex] [latex]x=35[/latex]

35mL must be added to the original 70 mL to gain a solution with a concentration of 60%

The above problem illustrates how we can use the mixture table to define an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.

A farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use to end up with 42 gallons of 20% butterfat?

Read and Understand:  We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is 42 gallons. There are two unknowns in this problem.

Define and Translate:  We will call the unknown volume of the  24% solution x, and the unknown volume of the 18% solution y.

Write and Solve:  Fill in the table with the information we know.

Find the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from

When you sum the amount column you get one equation: [latex]x+ y = 42[/latex] When you sum the total column you get a second equation: [latex]0.24x + 0.18y = 8.4[/latex]

Use elimination to find a value for [latex]x[/latex], and [latex]y[/latex].

Multiply the first equation by [latex]-0.18[/latex]

[latex]\begin{array}{cc}-0.18(x+y) &= (42)(-0.18) \\ -0.18x-0.18y &= -7.56 \end{array}[/latex]

Now our system of equations looks like this:

[latex]\begin{array}{cc} -0.18x-0.18y &= -7.56\\0.24x + 0.18y &= 8.4 \end{array}[/latex]

Adding the two equations together to eliminate the y terms gives this equation:

[latex]0.06x = 8.4[/latex]

Divide by 0.06 on each side:

[latex]x = 14[/latex]

Now substitute the value for x into one of the equations in order to solve for y.

[latex]\begin{array}{cc} (14) + y &= 42\\ y &= 28 \end{array}[/latex]

This can be interpreted as 14 gallons of 24% butterfat milk added to 28 gallons of 18% butterfat milk will give 42 gallons of 20% butterfat milk.

In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water [latex]b[/latex] and the speed of the river current [latex]c[/latex].

The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is [latex]b+c[/latex].

solving a uniform motion problem

Now, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is [latex]b-c[/latex].

solving a uniform motion problem

We’ll put some numbers to this situation in the next example.

Translate to a system of equations and then solve.

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Read the problem: This is a uniform motion problem and a picture will help us visualize the situation.

solving a uniform motion problem

Identify  what we are looking for: We are looking for the speed of the ship in still water and the speed of the current.

Name  what we are looking for:

Let [latex]s=[/latex] the rate of the ship in still water.

Let [latex]c=[/latex] the rate of the current.

A chart will help us organize the information. The ship goes downstream and then upstream. Going downstream, the current helps the ship and so the ship’s actual rate is [latex]s+c[/latex]. Going upstream, the current slows the ship and so the actual rate is [latex]s-c[/latex]. Downstream it takes 4 hours. Upstream it takes 5 hours. Each way the distance is 60 miles.

solving a uniform motion problem

Translate  into a system of equations. Since rate times time is distance, we can write the system of equations.

 [latex]\begin{array}{c}4(s+c)=60 \\ 5(s-c) = 60\end{array}[/latex]

Solve  the system of equations. Distribute to put both equations in standard form, then solve by elimination.

 [latex]\begin{array}{c}4s+4c=60 \\ 5s-5c = 60\end{array}[/latex]

Multiply the top equation by 5 and the bottom equation by 4. Add the equations, then solve for [latex]s[/latex].

[latex]\begin{array}{cc}20s+20c &= 300 \\ 20s-20c &= 240 \\ \hline \\ 40s &= 540 \\ s &= 13.5 \end{array}[/latex]

Substitute [latex]s=13.5[/latex] into one of the original equations.

[latex]\begin{array}{cc}4(s+c) &= 60 \\ 4(13.5+c) &= 60 \\ 54 + 4c &= 60 \\ 4c &= 6 \\ c &= 1.5 \end{array}[/latex]

Check  the answer in the problem. The downstream rate would be:

[latex]13.5+1.5 = 15[/latex] mph

In 4 hours the ship would travel:

[latex]15 \cdot 4 = 60[/latex] miles.

The upstream rate would be

[latex]13.5 - 1.5 = 12[/latex] mph.

In 5 hours the ship would travel

[latex]12\cdot 5[/latex] miles.

Answer  the question. The rate of the ship is 13.5 mph and the rate of the current is 1.5 mph.

In the next video, we present another example of a uniform motion problem which can be solved with a system of linear equations.

  • Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ ↵
  • Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
  • College Algebra. Authored by : Abramson, Jay et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
  • Solving Systems of Equations using Elimination. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/ova8GSmPV4o . License : CC BY: Attribution
  • Question ID 115164, 115120, 115110. Authored by : Shabazian, Roy. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
  • Beginning and Intermediate Algebra. Authored by : Wallace, Tyler. Located at : http://www.wallace.ccfaculty.org/book/book.html . License : CC BY: Attribution
  • Question ID 29699. Authored by : McClure, Caren. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
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  • Ex: System of Equations Application - Mixture Problem.. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning.. Located at : https://youtu.be/4s5MCqphpKo. . License : CC BY: Attribution
  • Beginning and Intermediate Algebra Textbook. . Authored by : Tyler Wallace. Located at : . License : CC BY: Attribution
  • Ex: System of Equations Application - Plane and Wind problem. Authored by : James Sousa (Mathispower4u.com). Located at : https://www.youtube.com/watch?v=OuxMYTqDhxw . License : CC BY: Attribution
  • Intermediate Algebra . Authored by : Lynn Marecek et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
  • Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution

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  • 6.2 Uniform Circular Motion
  • Introduction
  • 1.1 Physics: Definitions and Applications
  • 1.2 The Scientific Methods
  • 1.3 The Language of Physics: Physical Quantities and Units
  • Section Summary
  • Key Equations
  • Concept Items
  • Critical Thinking Items
  • Performance Task
  • Multiple Choice
  • Short Answer
  • Extended Response
  • 2.1 Relative Motion, Distance, and Displacement
  • 2.2 Speed and Velocity
  • 2.3 Position vs. Time Graphs
  • 2.4 Velocity vs. Time Graphs
  • 3.1 Acceleration
  • 3.2 Representing Acceleration with Equations and Graphs
  • 4.2 Newton's First Law of Motion: Inertia
  • 4.3 Newton's Second Law of Motion
  • 4.4 Newton's Third Law of Motion
  • 5.1 Vector Addition and Subtraction: Graphical Methods
  • 5.2 Vector Addition and Subtraction: Analytical Methods
  • 5.3 Projectile Motion
  • 5.4 Inclined Planes
  • 5.5 Simple Harmonic Motion
  • 6.1 Angle of Rotation and Angular Velocity
  • 6.3 Rotational Motion
  • 7.1 Kepler's Laws of Planetary Motion
  • 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity
  • 8.1 Linear Momentum, Force, and Impulse
  • 8.2 Conservation of Momentum
  • 8.3 Elastic and Inelastic Collisions
  • 9.1 Work, Power, and the Work–Energy Theorem
  • 9.2 Mechanical Energy and Conservation of Energy
  • 9.3 Simple Machines
  • 10.1 Postulates of Special Relativity
  • 10.2 Consequences of Special Relativity
  • 11.1 Temperature and Thermal Energy
  • 11.2 Heat, Specific Heat, and Heat Transfer
  • 11.3 Phase Change and Latent Heat
  • 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium
  • 12.2 First law of Thermodynamics: Thermal Energy and Work
  • 12.3 Second Law of Thermodynamics: Entropy
  • 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators
  • 13.1 Types of Waves
  • 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period
  • 13.3 Wave Interaction: Superposition and Interference
  • 14.1 Speed of Sound, Frequency, and Wavelength
  • 14.2 Sound Intensity and Sound Level
  • 14.3 Doppler Effect and Sonic Booms
  • 14.4 Sound Interference and Resonance
  • 15.1 The Electromagnetic Spectrum
  • 15.2 The Behavior of Electromagnetic Radiation
  • 16.1 Reflection
  • 16.2 Refraction
  • 16.3 Lenses
  • 17.1 Understanding Diffraction and Interference
  • 17.2 Applications of Diffraction, Interference, and Coherence
  • 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge
  • 18.2 Coulomb's law
  • 18.3 Electric Field
  • 18.4 Electric Potential
  • 18.5 Capacitors and Dielectrics
  • 19.1 Ohm's law
  • 19.2 Series Circuits
  • 19.3 Parallel Circuits
  • 19.4 Electric Power
  • 20.1 Magnetic Fields, Field Lines, and Force
  • 20.2 Motors, Generators, and Transformers
  • 20.3 Electromagnetic Induction
  • 21.1 Planck and Quantum Nature of Light
  • 21.2 Einstein and the Photoelectric Effect
  • 21.3 The Dual Nature of Light
  • 22.1 The Structure of the Atom
  • 22.2 Nuclear Forces and Radioactivity
  • 22.3 Half Life and Radiometric Dating
  • 22.4 Nuclear Fission and Fusion
  • 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation
  • 23.1 The Four Fundamental Forces
  • 23.2 Quarks
  • 23.3 The Unification of Forces
  • A | Reference Tables

Section Learning Objectives

By the end of this section, you will be able to do the following:

  • Describe centripetal acceleration and relate it to linear acceleration
  • Describe centripetal force and relate it to linear force
  • Solve problems involving centripetal acceleration and centripetal force

Teacher Support

The learning objectives in this section will help your students master the following standards:

  • (C) analyze and describe accelerated motion in two dimensions using equations, including projectile and circular examples.
  • (D) calculate the effect of forces on objects, including the law of inertia, the relationship between force and acceleration, and the nature of force pairs between objects.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Circular and Rotational Motion, as well as the following standards:

Section Key Terms

Centripetal acceleration.

[BL] [OL] Review uniform circular motion. Ask students to give examples of circular motion. Review linear acceleration.

In the previous section, we defined circular motion . The simplest case of circular motion is uniform circular motion , where an object travels a circular path at a constant speed . Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity , either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.

You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force . The sharper the curve and the greater your speed, the more noticeable this effect becomes.

[BL] [OL] [AL] Demonstrate circular motion by tying a weight to a string and twirling it around. Ask students what would happen if you suddenly cut the string? In which direction would the object travel? Why? What does this say about the direction of acceleration? Ask students to give examples of when they have come across centripetal acceleration.

Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine Δ s Δ s becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration a c because centripetal means center seeking .

Consider Figure 6.7 . The figure shows an object moving in a circular path at constant speed and the direction of the instantaneous velocity of two points along the path. Acceleration is in the direction of the change in velocity and points toward the center of rotation. This is strictly true only as Δ s Δ s tends to zero.

Now that we know that the direction of centripetal acceleration is toward the center of rotation, let’s discuss the magnitude of centripetal acceleration. For an object traveling at speed v in a circular path with radius r , the magnitude of centripetal acceleration is

Centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you may have noticed when driving a car, because the car actually pushes you toward the center of the turn. But it is a bit surprising that a c is proportional to the speed squared. This means, for example, that the acceleration is four times greater when you take a curve at 100 km/h than at 50 km/h.

We can also express a c in terms of the magnitude of angular velocity . Substituting v = r ω v = r ω into the equation above, we get a c = ( r ω ) 2 r = r ω 2 a c = ( r ω ) 2 r = r ω 2 . Therefore, the magnitude of centripetal acceleration in terms of the magnitude of angular velocity is

Tips For Success

The equation expressed in the form a c = rω 2 is useful for solving problems where you know the angular velocity rather than the tangential velocity.

Virtual Physics

Ladybug motion in 2d.

In this simulation, you experiment with the position, velocity, and acceleration of a ladybug in circular and elliptical motion. Switch the type of motion from linear to circular and observe the velocity and acceleration vectors. Next, try elliptical motion and notice how the velocity and acceleration vectors differ from those in circular motion.

Grasp Check

In uniform circular motion, what is the angle between the acceleration and the velocity? What type of acceleration does a body experience in the uniform circular motion?

  • The angle between acceleration and velocity is 0°, and the body experiences linear acceleration.
  • The angle between acceleration and velocity is 0°, and the body experiences centripetal acceleration.
  • The angle between acceleration and velocity is 90°, and the body experiences linear acceleration.
  • The angle between acceleration and velocity is 90°, and the body experiences centripetal acceleration.

Centripetal Force

[BL] [OL] [AL] Using the same demonstration as before, ask students to predict the relationships between the quantities of angular velocity, centripetal acceleration, mass, centripetal force. Invite students to experiment by using various lengths of string and different weights.

Because an object in uniform circular motion undergoes acceleration (by changing the direction of motion but not the speed), we know from Newton’s second law of motion that there must be a net external force acting on the object. Since the magnitude of the acceleration is constant, so is the magnitude of the net force, and since the acceleration points toward the center of the rotation, so does the net force.

Any force or combination of forces can cause a centripetal acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, the friction between a road and the tires of a car as it goes around a curve, or the normal force of a roller coaster track on the cart during a loop-the-loop.

The component of any net force that causes circular motion is called a centripetal force . When the net force is equal to the centripetal force, and its magnitude is constant, uniform circular motion results. The direction of a centripetal force is toward the center of rotation, the same as for centripetal acceleration. According to Newton’s second law of motion, a net force causes the acceleration of mass according to F net = m a . For uniform circular motion, the acceleration is centripetal acceleration: a = a c . Therefore, the magnitude of centripetal force, F c , is F c = m a c F c = m a c .

By using the two different forms of the equation for the magnitude of centripetal acceleration, a c = v 2 / r a c = v 2 / r and a c = r ω 2 a c = r ω 2 , we get two expressions involving the magnitude of the centripetal force F c F c . The first expression is in terms of tangential speed, the second is in terms of angular speed: F c = m v 2 r F c = m v 2 r and F c = m r ω 2 F c = m r ω 2 .

Both forms of the equation depend on mass, velocity, and the radius of the circular path. You may use whichever expression for centripetal force is more convenient. Newton’s second law also states that the object will accelerate in the same direction as the net force. By definition, the centripetal force is directed towards the center of rotation, so the object will also accelerate towards the center. A straight line drawn from the circular path to the center of the circle will always be perpendicular to the tangential velocity. Note that, if you solve the first expression for r , you get

From this expression, we see that, for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

Watch Physics

Centripetal force and acceleration intuition.

This video explains why centripetal force, when it is equal to the net force and has constant magnitude, creates centripetal acceleration and uniform circular motion.

Misconception Alert

Some students might be confused between centripetal force and centrifugal force. Centrifugal force is not a real force but the result of an accelerating reference frame, such as a turning car or the spinning Earth. Centrifugal force refers to a fictional center fleeing force.

  • The yoyo will fly inward in the direction of the centripetal force.
  • The yoyo will fly outward in the direction of the centripetal force.
  • The yoyo will fly to the left in the direction of the tangential velocity.
  • The yoyo will fly to the right in the direction of the tangential velocity.

Solving Centripetal Acceleration and Centripetal Force Problems

To get a feel for the typical magnitudes of centripetal acceleration, we’ll do a lab estimating the centripetal acceleration of a tennis racket and then, in our first Worked Example, compare the centripetal acceleration of a car rounding a curve to gravitational acceleration. For the second Worked Example, we’ll calculate the force required to make a car round a curve.

Estimating Centripetal Acceleration

In this activity, you will measure the swing of a golf club or tennis racket to estimate the centripetal acceleration of the end of the club or racket. You may choose to do this in slow motion. Recall that the equation for centripetal acceleration is a c = v 2 r a c = v 2 r or a c = r ω 2 a c = r ω 2 .

  • One tennis racket or golf club
  • One ruler or tape measure
  • Work with a partner. Stand a safe distance away from your partner as he or she swings the golf club or tennis racket.
  • Describe the motion of the swing—is this uniform circular motion? Why or why not?
  • Try to get the swing as close to uniform circular motion as possible. What adjustments did your partner need to make?
  • Measure the radius of curvature. What did you physically measure?
  • By using the timer, find either the linear or angular velocity, depending on which equation you decide to use.
  • What is the approximate centripetal acceleration based on these measurements? How accurate do you think they are? Why? How might you and your partner make these measurements more accurate?

The swing of the golf club or racket can be made very close to uniform circular motion. For this, the person would have to move it at a constant speed, without bending their arm. The length of the arm plus the length of the club or racket is the radius of curvature. Accuracy of measurements of angular velocity and angular acceleration will depend on resolution of the timer used and human observational error. The swing of the golf club or racket can be made very close to uniform circular motion. For this, the person would have to move it at a constant speed, without bending their arm. The length of the arm plus the length of the club or racket is the radius of curvature. Accuracy of measurements of angular velocity and angular acceleration will depend on resolution of the timer used and human observational error.

Was it more useful to use the equation a c = v 2 r a c = v 2 r or a c = r ω 2 a c = r ω 2 in this activity? Why?

  • It should be simpler to use a c = r ω 2 a c = r ω 2 because measuring angular velocity through observation would be easier.
  • It should be simpler to use a c = v 2 r a c = v 2 r because measuring tangential velocity through observation would be easier.
  • It should be simpler to use a c = r ω 2 a c = r ω 2 because measuring angular velocity through observation would be difficult.
  • It should be simpler to use a c = v 2 r a c = v 2 r because measuring tangential velocity through observation would be difficult.

Worked Example

Comparing centripetal acceleration of a car rounding a curve with acceleration due to gravity.

A car follows a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h). What is the magnitude of the car’s centripetal acceleration? Compare the centripetal acceleration for this fairly gentle curve taken at highway speed with acceleration due to gravity ( g ).

Because linear rather than angular speed is given, it is most convenient to use the expression a c = v 2 r a c = v 2 r to find the magnitude of the centripetal acceleration.

Entering the given values of v = 25.0 m/s and r = 500 m into the expression for a c gives

To compare this with the acceleration due to gravity ( g = 9.80 m/s 2 ), we take the ratio a c / g = ( 1.25  m/s 2 ) / ( 9.80 m/s 2 ) = 0.128 a c / g = ( 1.25  m/s 2 ) / ( 9.80 m/s 2 ) = 0.128 . Therefore, a c = 0.128 g a c = 0.128 g , which means that the centripetal acceleration is about one tenth the acceleration due to gravity.

Frictional Force on Car Tires Rounding a Curve

  • Calculate the centripetal force exerted on a 900 kg car that rounds a 600-m-radius curve on horizontal ground at 25.0 m/s.
  • Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
  • If the car would slip if it were to be traveling any faster, what is the coefficient of static friction between the tires and the road? Could we conclude anything about the coefficient of static friction if we did not know whether the car could round the curve any faster without slipping?

Strategy and Solution for (a)

We know that F c = m v 2 r F c = m v 2 r . Therefore,

Strategy and Solution for (b)

The image above shows the forces acting on the car while rounding the curve. In this diagram, the car is traveling into the page as shown and is turning to the left. Friction acts toward the left, accelerating the car toward the center of the curve. Because friction is the only horizontal force acting on the car, it provides all of the centripetal force in this case. Therefore, the force of friction is the centripetal force in this situation and points toward the center of the curve.

Strategy and Solution for (c)

If the car is about to slip, the static friction is at its maximum value and f = μ s N = μ s m g f = μ s N = μ s m g . Solving for μ s μ s , we get μ s = 938 900 × 9.8 = 0.11 μ s = 938 900 × 9.8 = 0.11 . Regardless of whether we know the maximum allowable speed for rounding the curve, we can conclude this is a minimum value for the coefficient.

Since we found the force of friction in part (b), we could also solve for the coefficient of friction, since f = μ s N = μ s m g f = μ s N = μ s m g . The static friction is only equal to μ s N μ s N when it is at the maximum possible value. If the car could go faster, the friction at the given speed would still be the same as we calculated, but the coefficient of static friction would be larger.

Practice Problems

Calculate the centripetal acceleration of an object following a path with a radius of a curvature of 0.2 m and at an angular velocity of 5 rad/s.

Check Your Understanding

  • Uniform circular motion is when an object accelerates on a circular path at a constantly increasing velocity.
  • Uniform circular motion is when an object travels on a circular path at a variable acceleration.
  • Uniform circular motion is when an object travels on a circular path at a constant speed.
  • Uniform circular motion is when an object travels on a circular path at a variable speed.

Which of the following is centripetal acceleration?

  • The acceleration of an object moving in a circular path and directed radially toward the center of the circular orbit
  • The acceleration of an object moving in a circular path and directed tangentially along the circular path
  • The acceleration of an object moving in a linear path and directed in the direction of motion of the object
  • The acceleration of an object moving in a linear path and directed in the direction opposite to the motion of the object
  • Yes, the object is accelerating, so a net force must be acting on it.
  • Yes, because there is no acceleration.
  • No, because there is acceleration.
  • No, because there is no acceleration.

Identify two examples of forces that can cause centripetal acceleration.

  • The force of Earth’s gravity on the moon and the normal force
  • The force of Earth’s gravity on the moon and the tension in the rope on an orbiting tetherball
  • The normal force and the force of friction acting on a moving car
  • The normal force and the tension in the rope on a tetherball

Use the Check Your Understanding questions to assess whether students master the learning objectives of this section. If students are struggling with a specific objective, the formative assessment will help identify which objective is causing the problem and direct students to the relevant content.

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Algebra Motion Problems

In these lessons, we will learn how to solve algebra word problems that involve motion.

Related Pages Rate, Time and Distance Word Problems Rate, Time and Distance More Algebra Word Problems More Algebra Lessons

What are Motion or Distance Word Problems? Motion problems are based on the formula

where d = distance, r = rate and t = time.

How to solve Motion or Distance Word Problems? Step 1: Draw a diagram to represent the relationship between the distances involved in the problem. Step 2: Set up a chart based on the formula: rate × time = distance. Step 3: Use the chart to set up one or more equations. Step 4: Solve the equations.

We will look at three types of Motion Word Problems:

  • Two objects going in opposite directions.
  • Both objects going in the same direction, but one goes further.
  • One object going and returning at different rates.

Solve Motion Word Problems: Two objects going in opposite directions

Example: John and Philip who live 14 miles apart start at noon to walk toward each other at rates of 3 mph and 4 mph respectively. After how many hours will they meet?

Solution: Let x = time walked.

3x + 4x = 14 7x = 14 x = 2

They will meet in 2 hours.

How to solve motion word problems with objects traveling in opposite directions?

Example: Two cars leave from the same place at the same time and travel in opposite direction. One car travels at 55 mph and the other at 75 mph. After how many hours will there be 520 miles apart?

Example: Two planes leave the same point at 8 AM. Plane 1 heads East at 600 mph and Plane 2 heads West at 450 mph. How long will they be 1400 miles apart? At what time will they be 1400 miles apart? How far has each plane traveled?

Solve Motion Word Problems: Two objects going in the same direction

Example: Aaron left L.A. to drive at 55 mph towards Las Vegas. Mike left L.A. an hour after Aaron (also towards Las Vegas), driving at 70 mph. How long will it take Mike to overtake Aaron?

How to solve motion word problems with objects traveling in the same direction?

Example: John left his house at 3.00 pm to drive 60 mph to drive towards Michigan. Phoebe left the same house at 5.00 pm, driving 80 mph in the same direction as John. How long will it take Phoebe to overtake John?

Solve Motion Word Problems: One object going and returning at different rates

Example: In still water, Peter’s boat goes 4 times as fast as the current in the river. He takes a 15-mile trip up the river and returns in 4 hours. Find the rate of the current.

Solution: Let x = rate of the current.

The rate of the current is 2 mph.

Example: Gordon rode his bike at 15 mph to get his car. He then drove back at 45 mph. If the entire trip took him 8 hours, how far away was his car?

Motion Word Problems This is how to set up motion problems for Algebra. Three Types of Problems

  • Both going the same direction but one going further
  • Two going in opposite directions
  • Going in one direction and then returning at a different rate.

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Mathematics LibreTexts

4.3: Solve Applications with Systems of Equations

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  • Page ID 114161

Learning Objectives

By the end of this section, you will be able to:

  • Solve direct translation applications
  • Solve geometry applications

Solve uniform motion applications

Be prepared 4.4.

Before you get started, take this readiness quiz.

The sum of twice a number and nine is 31. Find the number. If you missed this problem, review Example 2.15.

Be Prepared 4.5

Twins Jon and Ron together earned $96,000 last year. Ron earned $8000 more than three times what Jon earned. How much did each of the twins earn? If you missed this problem, review Example 2.19.

Be Prepared 4.6

An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in four hours and the local train takes five hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains. If you missed this problem, review Example 2.43.

Solve Direct Translation Applications

Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we’ll first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system.

Solve applications with systems of equations.

  • Step 1. Read the problem. Make sure all the words and ideas are understood.
  • Step 2. Identify what we are looking for.
  • Step 3. Name what we are looking for. Choose variables to represent those quantities.
  • Step 4. Translate into a system of equations.
  • Step 5. Solve the system of equations using good algebra techniques.
  • Step 6. Check the answer in the problem and make sure it makes sense.
  • Step 7. Answer the question with a complete sentence.

We solved number problems with one variable earlier. Let’s see how differently it works using two variables.

Example 4.14

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Try It 4.27

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

Try It 4.28

The sum of two numbers is −6 . −6 . One number is 10 less than the other. Find the numbers.

Example 4.15

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $ 10,000 + $ 40 $ 10,000 + $ 40 for each training session. How many training sessions would make the salary options equal?

Try It 4.29

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

Try It 4.30

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

As you solve each application, remember to analyze which method of solving the system of equations would be most convenient.

Example 4.16

Translate to a system of equations and then solve:

When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories for each minute of circuit training?

Try It 4.31

Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?

Try It 4.32

Erin spent 30 minutes on the rowing machine and 20 minutes lifting weights at the gym and burned 430 calories. During her next visit to the gym she spent 50 minutes on the rowing machine and 10 minutes lifting weights and burned 600 calories. How many calories did she burn for each minutes on the rowing machine? How many calories did she burn for each minute of weight lifting?

Solve Geometry Applications

We will now solve geometry applications using systems of linear equations. We will need to add complementary angles and supplementary angles to our list some properties of angles.

The measures of two complementary angles add to 90 degrees. The measures of two supplementary angles add to 180 degrees.

Complementary and Supplementary Angles

Two angles are complementary if the sum of the measures of their angles is 90 degrees.

Two angles are supplementary if the sum of the measures of their angles is 180 degrees.

If two angles are complementary, we say that one angle is the complement of the other.

If two angles are supplementary, we say that one angle is the supplement of the other.

Example 4.17

Translate to a system of equations and then solve.

The difference of two complementary angles is 26 degrees. Find the measures of the angles.

Try It 4.33

The difference of two complementary angles is 20 degrees. Find the measures of the angles.

Try It 4.34

The difference of two complementary angles is 80 degrees. Find the measures of the angles.

In the next example, we remember that the measures of supplementary angles add to 180.

Example 4.18

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Try It 4.35

Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.

Try It 4.36

Two angles are supplementary. The measure of the larger angle is 18 less than twice the measure of the smaller angle. Find the measures of the angles.

Recall that the angles of a triangle add up to 180 degrees. A right triangle has one angle that is 90 degrees. What does that tell us about the other two angles? In the next example we will be finding the measures of the other two angles.

Example 4.19

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

Try It 4.37

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

Try It 4.38

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

Often it is helpful when solving geometry applications to draw a picture to visualize the situation.

Example 4.20

Randall has 125 feet of fencing to enclose the part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

Try It 4.39

Mario wants to put a fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.

Try It 4.40

Alexis wants to build a rectangular dog run in her yard adjacent to her neighbor’s fence. She will use 136 feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbor’s fence will be 16 feet less than twice the width. Find the length and width of the dog run.

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was D = r t D = r t where D is the distance traveled, r is the rate, and t is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

Example 4.21

Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

A diagram is useful in helping us visualize the situation.

.

Identify and name what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart. We are looking for the length of time Kelly, k , and Joni, j , will each drive.

.

Since D = r · t D = r · t we can fill in the Distance column.

Translate into a system of equations.

To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So,

65 j = 78 k 65 j = 78 k

Also, since Kelly left later, her time will be 1 2 1 2 hour less than Joni’s time. So,

Try It 4.41

Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

Try It 4.42

Charlie left his mother’s house traveling at an average speed of 36 miles per hour. His sister Sally left 15 minutes ( 1 4 hour ) ( 1 4 hour ) later traveling the same route at an average speed of 42 miles per hour. How long before Sally catches up to Charlie?

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water b and the speed of the river current c .

The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b + c . b + c .

Figure shows a boat and two horizontal arrows, both pointing left. The one to the left of the boat is b and the one to the right is c.

Now, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is b − c . b − c .

Figure shows a boat and two horizontal arrows to its left. One, labeled b, points left and the other, labeled c, points right.

We’ll put some numbers to this situation in the next example.

Example 4.22

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Try It 4.43

A Mississippi river boat cruise sailed 120 miles upstream for 12 hours and then took 10 hours to return to the dock. Find the speed of the river boat in still water and the speed of the river current.

Try It 4.44

Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in the next example. A wind current in the same direction as the plane is flying is called a tailwind . A wind current blowing against the direction of the plane is called a headwind .

Example 4.23

A private jet can fly 1,095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Try It 4.45

A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1,035 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Try It 4.46

A commercial jet can fly 1,728 miles in 4 hours with a tailwind but only 1,536 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Access this online resource for additional instruction and practice with systems of equations.

  • Systems of Equations

Section 4.2 Exercises

Practice makes perfect.

Direct Translation Applications

In the following exercises, translate to a system of equations and solve.

The sum of two number is 15. One number is 3 less than the other. Find the numbers.

The sum of two number is 30. One number is 4 less than the other. Find the numbers.

The sum of two number is −16. One number is 20 less than the other. Find the numbers.

The sum of two number is −26 . −26 . One number is 12 less than the other. Find the numbers.

The sum of two numbers is 65. Their difference is 25. Find the numbers.

The sum of two numbers is 37. Their difference is 9. Find the numbers.

The sum of two numbers is −27 . −27 . Their difference is −59 . −59 . Find the numbers.

The sum of two numbers is −45 . −45 . Their difference is −89 . −89 . Find the numbers.

Maxim has been offered positions by two car companies. The first company pays a salary of $10,000 plus a commission of $1000 for each car sold. The second pays a salary of $20,000 plus a commission of $500 for each car sold. How many cars would need to be sold to make the total pay the same?

Jackie has been offered positions by two cable companies. The first company pays a salary of $14,000 plus a commission of $100 for each cable package sold. The second pays a salary of $20,000 plus a commission of $25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same?

Amara currently sells televisions for company A at a salary of $17,000 plus a $100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a $20 commission for each television she sells. How televisions would Amara need to sell for the options to be equal?

Mitchell currently sells stoves for company A at a salary of $12,000 plus a $150 commission for each stove he sells. Company B offers him a position with a salary of $24,000 plus a $50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal?

Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?

June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?

Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?

Drew burned 1800 calories Friday playing one hour of basketball and canoeing for two hours. Saturday he spent two hours playing basketball and three hours canoeing and burned 3200 calories. How many calories did he burn per hour when playing basketball? How many calories did he burn per hour when canoeing?

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for $116. Lisa bought two notebooks and three thumb dives for $68. Find the cost of each notebook and each thumb drive.

Nancy bought seven pounds of oranges and three pounds of bananas for $17. Her husband later bought three pounds of oranges and six pounds of bananas for $12. What was the cost per pound of the oranges and the bananas?

Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for $114 or she is able to buy 2 shirts and 4 sweaters for $164. How much does a shirt cost? How much does a sweater cost?

Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for $40 or he is able to buy 5 packages of paper and 6 staplers for $62. How much does a package of paper cost? How much does a stapler cost?

The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?

The total number of calories in 2 hot dogs and 3 cups of cottage cheese is 960 calories. The total number of calories in 5 hot dogs and 2 cups of cottage cheese is 1190 calories. How many calories are in a hot dog? How many calories are in a cup of cottage cheese?

Molly is making strawberry infused water. For each ounce of strawberry juice, she uses three times as many ounces of water as juice. How many ounces of strawberry juice and how many ounces of water does she need to make 64 ounces of strawberry infused water?

Owen is making lemonade from concentrate. The number of quarts of water he needs is 4 times the number of quarts of concentrate. How many quarts of water and how many quarts of concentrate does Owen need to make 100 quarts of lemonade?

The difference of two complementary angles is 55 degrees. Find the measures of the angles.

The difference of two complementary angles is 17 degrees. Find the measures of the angles.

Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.

Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles.

The difference of two supplementary angles is 8 degrees. Find the measures of the angles.

The difference of two supplementary angles is 88 degrees. Find the measures of the angles.

Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.

Two angles are supplementary. The measure of the larger angle is five less than four times the measure of the smaller angle. Find the measures of both angles.

The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles.

Wayne is hanging a string of lights 45 feet long around the three sides of his patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.

Darrin is hanging 200 feet of Christmas garland on the three sides of fencing that enclose his front yard. The length is five feet less than three times the width. Find the length and width of the fencing.

A frame around a family portrait has a perimeter of 90 inches. The length is fifteen less than twice the width. Find the length and width of the frame.

The perimeter of a toddler play area is 100 feet. The length is ten more than three times the width. Find the length and width of the play area.

Solve Uniform Motion Applications

Sarah left Minneapolis heading east on the interstate at a speed of 60 mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of 70 mph. How long will it take for Sarah’s sister to catch up to Sarah?

College roommates John and David were driving home to the same town for the holidays. John drove 55 mph, and David, who left an hour later, drove 60 mph. How long will it take for David to catch up to John?

At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of 40 mph. Lucy’s friend left the beach for home 30 minutes (half an hour) later, and drove 50 mph. How long did it take Lucy’s friend to catch up to Lucy?

Felecia left her home to visit her daughter driving 45 mph. Her husband waited for the dog sitter to arrive and left home twenty minutes (1/3 hour) later. He drove 55 mph to catch up to Felecia. How long before he reaches her?

The Jones family took a 12-mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.

A motor boat travels 60 miles down a river in three hours but takes five hours to return upstream. Find the rate of the boat in still water and the rate of the current.

A motor boat traveled 18 miles down a river in two hours but going back upstream, it took 4.5 4.5 hours due to the current. Find the rate of the motor boat in still water and the rate of the current.

A river cruise boat sailed 80 miles down the Mississippi River for four hours. It took five hours to return. Find the rate of the cruise boat in still water and the rate of the current.

A small jet can fly 1072 miles in 4 hours with a tailwind but only 848 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A small jet can fly 1435 miles in 5 hours with a tailwind but only 1,215 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A commercial jet can fly 1,320 miles in 3 hours with a tailwind but only 1170 miles in 3 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Writing Exercises

Write an application problem similar to Example 4.14. Then translate to a system of equations and solve it.

Write a uniform motion problem similar to Example 4.15 that relates to where you live with your friends or family members. Then translate to a system of equations and solve it.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 3 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: solve direct translation applications, solve geometry applications, solve uniform motion applications. The remaining columns are empty.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

IMAGES

  1. Problem Solving: Uniform Motion

    solving a uniform motion problem

  2. Solving Uniform Motion Problems

    solving a uniform motion problem

  3. How to Solve a Uniform Motion Problem , Intermediate Algebra , Lesson 40

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  4. 2.4.d Uniform Motion Problems

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  5. How to Solve Uniform Motion Problems

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  6. Uniformly Accelerated Motion (2/2): Problem-Solving Example

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VIDEO

  1. Explanation : Uniform Circular motion

  2. MTH100Notes8 8UniformMotione1

  3. Ch2s3E Mixture UniformSpeed Apps

  4. EXPLANATION OF UNIFORM CIRCULAR MOTION BY GROUP-XYZ

  5. Motion In 1D ( Acceleration, Uniform and Non uniform Motion and Kinematic Equations )

  6. PHYSICS TUTOR

COMMENTS

  1. How to solve uniform motion problems

    Uniform motion explains the distance of an object when it travels at a constant speed, the rate, over a period of time. To compare different rates, times, and distances you can use subscripts to keep track of which pieces go with which equation.

  2. 3.5: Solve Uniform Motion Applications

    3: Math Models 3.5: Solve Uniform Motion Applications

  3. 9.4: Uniform motion problems

    9.4: Uniform motion problems. We can recall uniform motion problems in the word problems chapter. We used the formula r ⋅ t = d r ⋅ t = d and organized the given information in a table. Now, we use the equation as. We apply the same method in this section only the equations will be rational equations.

  4. 2.5: Solve Mixture and Uniform Motion Applications

    Solve mixture word problems; Solve uniform motion applications; Be Prepared 2.10. Before you get started, take this readiness quiz. Simplify: 0.25 x + 0.10 (x + 4). 0.25 x + 0.10 (x + 4). If you missed this problem, review Example 1.52. Be Prepared 2.11. The number of adult tickets is three more than twice the number of children tickets.

  5. 8.8 Solve Uniform Motion and Work Applications

    We have solved uniform motion problems using the formula D = rt D = r t in previous chapters. We used a table like the one below to organize the information and lead us to the equation. The formula D = rt D = r t assumes we know r and t and use them to find D.

  6. How to Solve Uniform Motion Problems

    Course 27K views Uniform Motion Formula The formula for uniform motion is d = rt, meaning distance is equal to rate times time. To unlock this lesson you must be a Study.com Member. Create your...

  7. 3.5 Solve Uniform Motion Applications

    Step 1. Read the problem. Make sure all the words and ideas are understood. Draw a diagram to illustrate what it happening. Create a table to organize the information. Label the columns rate, time, distance. List the two scenarios. Write in the information you know. Step 2. Identify what we are looking for. Step 3.

  8. How to Solve Uniform Motion Problems , Example 2 ...

    This tutorial provides another example of solving a uniform motion problem.Join this channel to get access to perks:https://www.youtube.com/channel/UCn2SbZWi...

  9. Solving Uniform Motion Problems

    how to solve uniform motion problmes with examples

  10. How to Solve a Uniform Motion Problem , Intermediate Algebra ...

    392 42K views 11 years ago Intermediate Algebra ...more ...more This tutorial shows how to solve a uniform motion story problem.Join this channel to get access to...

  11. 2.4 Solve Mixture and Uniform Motion Applications

    Solve uniform motion applications Be Prepared 2.10 Before you get started, take this readiness quiz. Simplify: 0.25 x + 0.10 ( x + 4). If you missed this problem, review Example 1.52. Be Prepared 2.11 The number of adult tickets is three more than twice the number of children tickets. Let c represent the number of children tickets.

  12. 7.6: Solve Uniform Motion and Work Applications

    Solve Uniform Motion Applications. We have solved uniform motion problems using the formula D=rt in previous chapters. We used a table like the one below to organize the information and lead us to the equation. The formula D=rt assumes we know r and t and use them to find D. If we know D and r and need to find t, we would solve the equation for ...

  13. Uniform Acceleration Motion: Problems with Solutions

    Solution to Problem 1 Problem 2: With an initial velocity of 20 km/h, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 2 A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds.

  14. Applications and Variation

    Solving Uniform Motion Problems. Uniform motion (or distance) Described by the formula D = r t, where the distance D is given as the product of the average rate r and the time t traveled at that rate. problems involve the formula D = r t, where the distance D is given as the product of the average rate r and the time t traveled at that rate. If we divide both sides by the average rate r, then ...

  15. (5.2.2)

    (5.2.4) - Solve uniform motion problems with a system of linear equations. Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing ...

  16. 8.8: Solve Uniform Motion and Work Applications

    Solve Uniform Motion Applications. We have solved uniform motion problems using the formula D=rt in previous chapters. We used a table like the one below to organize the information and lead us to the equation. The formula D=rt assumes we know r and t and use them to find D. If we know D and r and need to find t, we would solve the equation for ...

  17. Solving Uniform Motion Applications

    Solve the system of equations by substitution. Substitute k = j − 1 2 k = j − 1 2 into the second equation, then solve for j. To find Kelly's time, substitute j = 3 into the first equation, then solve for k. Check the answer in the problem. Joni 3 hours (65 mph) = 195 miles. Kelly 2 1 2 2 1 2 hours (78 mph) = 195 miles.

  18. Uniform acceleration motion Problems and Solutions

    a = 15 m/s2. Problem #2. A car accelerates uniformly from 21,0 m/s to 35,0 m/s in 3,0 seconds. Determine the acceleration of the car and the distance traveled. Answer: Given: time t = 4,0 s, initial velocity vi = 21,0 m/s, and finale velocity vf = 35,0 m/s,

  19. 6.2 Uniform Circular Motion

    The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in ...

  20. Algebra Motion Problems (solutions, examples, videos)

    Step 1: Draw a diagram to represent the relationship between the distances involved in the problem. Step 2: Set up a chart based on the formula: rate × time = distance. Step 3: Use the chart to set up one or more equations. Step 4: Solve the equations. We will look at three types of Motion Word Problems: Two objects going in opposite directions.

  21. 2.5: Solve Mixture and Uniform Motion Applications

    Solve Uniform Motion Applications. When you are driving down the interstate using your cruise control, the speed of your car stays the same—it is uniform. We call a problem in which the speed of an object is constant a uniform motion application. We will use the distance, rate, and time formula, \(D=rt\), to compare two scenarios, such as two ...

  22. Archery N Motion Solves a Common Problem for Hunters

    Archery N Motion provides platforms that mimic the motion hunters and archery enthusiasts need to learn when they go hunting live animals. Stationary target shooting can be useful for many interested in shooting sports. ... Archery N Motion Solves a Common Problem for Hunters. Published: Feb 14, 2024 by Annie Pilon In Business Success Stories 0 ...

  23. 7.6: Solve Applications with Rational Equations

    Solve Uniform Motion Applications. We have solved uniform motion problems using the formula D = r t D = r t in previous chapters. We used a table like the one below to organize the information and lead us to the equation. The formula D = r t D = r t assumes we know r and t and use them to find D.

  24. 4.3: Solve Applications with Systems of Equations

    Solve uniform motion applications. We used a table to organize the information in uniform motion problems when we introduced them earlier. We'll continue using the table here. The basic equation was D = r t D = r t where D is the distance traveled, r is the rate, and t is the time.