Applied Problems: Motion

By Dr. Marcia L.Tharp

Dr. Julia Arnold

Applied Problems: Motion

Motion problems use the equation

D = RT

where D is the distance traveled.R is the rate of travel.T is the time spent traveling.

Applied Problems: MotionMotion Problem 1

Let’s suppose that you drive your car 120 miles in 2 hours than we can find the rate you traveled by using the distance equation.

D = RT

60 = R

Since 120 is a distance we let D = 120Since the time you traveled is 2 hours we let T= 2.

Placing this in the equation D=RT we have 120 = R 2

120 = R 2

2 2

So your rate is 60 miles per hour.

Motion Problem 2 Pat has been caught speeding by

the airplane patrol. He is doing 85miles per hour. A police car who is 30 miles behind him can do 100 miles per hour to catch him. How long will it take the police car to catch Pat if they continue in the same direction at this speed?

Print this page.

It is helpful to use a D = RT grid when solving motion problems as shown in the following example. The purpose of the grid is to find an algebraic name for each distance.

D=RT GridTo use this grid fill in the rate and time for each driver.We will let x = the time it takes for the police car to catch Pat. Then multiply the rate by the time to get the distance traveled.

Rate Time Distance

Pat

PoliceCar

D=RT Grid

Rate Time Distance

Pat 85

PoliceCar

100

D=RT Grid

Rate Time Distance

Pat 85 X

Police Car

100 X

D=RT GridRate Time Distance

Pat 85 X 85x

Police Car

100 X 100x

Notice that neither the policeman or Pat are represented by the 30 miles. We will use this later.

Next Draw A Picture

Picture 30 miles Pat’s

distance

Police Car’s distance

Draw A Picture

Picture 30 miles Pat’s

distance

Police Car’s distance

Use the picture to write an equation.

30 miles Pat’s distance

Police Car’s distance

Police Car’s = 30 miles + Pat’s Distance Distance

Now fill in the equation using our chart.

30 miles Pat’s distance

Police Car’s distance

Police Car’s = 30 miles + Pat’s Distance Distance

100x = 30 + 85xRate Tim

eDistance

Pat 85 X 85x

Police Car

100

X 100x

Next solve this equation.

100x = 30 + 85x

So it took the police car 2 hours to catch Pat.

100x – 85x = 30 + 85x – 85x 15x = 30

15x = 30 15 15

x = 2 hours

Subtract 85 x from both sides.

Divide by 15 on each side.

Problem 3 Juan and Amal leave DC at the

same time headed south on I-95. If Juan averages 60 mph and Amal averages 72 mph how long will it take them to be 30 miles apart?

(Now would be a good time for a guess. Write yours down and try it in this table.)

Rate Time Distance

Juan 60 x 60x

Amal 72 x 72x

72x x72Amal

60x x60Juan

DistanceTimeRate

We are looking for a time where Juan and Amal will be30 miles apart. How do we represent the distance betweenthe two men?

60x - 72x

Or is it 72x - 60x. Which of these two would be positive?

The correct equation is 72x - 60 x = 30

Sherry and Bob like to jog in the park. Sherry can jog at 5 mph, while Bob can jog at 7 mph.If Sherry starts 30 minutes ahead of Bob, how long will it take Bob to catch up to Sherry?

Draw a grid like the one below and fill in what you knowabout Sherry and Bob.

Bob

Sherry

DistanceTimeRate

When finished go to the next slide to see how you did.

Sherry and Bob like to jog in the park. Sherry can jog at 5 mph, while Bob can jog at 7 mph.If Sherry starts 30 minutes ahead of Bob, how long will it take Bob to catch up to Sherry?

Bob

Sherry

DistanceTimeRate

5

7 x

Let x = the time it takes Bob to catch Sherry

x + 1/2 5(x + 1/2)

7x

Sherry started 30 minutes or 1/2 hour before Bob, so her timemust reflect that amount. Rate is in miles per hour, time mustbe in hours so our units match.

Sherry Sherry’s jogging distance

Bob

The problem is finished when Bob catches up to Sherry.

Thus, their distances must equal each other.

5(x + 1/2) = 7x

5x + 5/2 = 7x

5/2 = 2x2[5/2] = 2 (2x)5 = 4x

x = 5/4 = 1.25 hrs

Now it’s time to see what You can do.

Practice Problems: Motion1. Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart?

2. Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute.

3. Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph?

4. Dr. John left New Orleans at 12 noon. His drummer left at 1:00 traveling 9 mph faster. If the drummer passed Dr. John at 6:00, what was the average speed of each?

Motion problem solutions:1. 3 hours2. 1/3 hour = 20 minutes3. 1.2 hours at 60 mph4. 45 mph for Dr. John and 54 mph for the drummer

For worked out solutionsclick to next slide.

1. Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Let x = time it takes for them to be 39 miles apart. Construct a table to put your information in.

Rate Time Distance

Tonya 52 x 52x

Freda 65 x 65x

Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart?

We let x stand for the time they have been driving which would be the same in this case. How far they have gone (distance) is written in the chart by using the known information and the formula R*T=D

Rate Time Distance

Tonya 52 x 52x

Freda 65 x 65x

Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart?

Tonya and Freda are headed in the same direction so we can picture their distances as this:

Tonya

Freda39 mi

52x

65x

Do you see the equation forming from our picture?

Rate Time Distance

Tonya 52 x 52x

Freda 65 x 65x

Tonya and Freda drive away from Norfolk on the same road in the same direction. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart?

Tonya

Freda39 mi

52x

65x

One way to look at it might be to say52x + 39 = 65xOr another way might be to say 65x – 52x = 39Both are correct and will give you the correct answer of 3 hrs.

2. Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute.Now they are going in opposite directions, but we will begin the same way by constructing our table.

Rate Time Distance

Tonya 52 x 52x

Freda 65 x 65x

Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute.

As you can see the table is the same, so only the picture of the event must change. Now it looks like this:

Tonya Freda52x 65x

Do you see the equation forming from this picture?

START

39 miles

Rate Time Distance

Tonya 52 x 52x

Freda 65 x 65x

Tonya and Freda drive away from Norfolk on the same road in opposite directions. If Tonya is averaging 52 mph and Freda is averaging 65 mph, how long will it take for them to be 39 miles apart? Round your answer to the nearest minute.

TonyaFreda

52x 65x

52x + 65x = 39117x=39X= 1/3 of an hour or 1/3 of 60 min = 20 minutes

START

39 miles

3. Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph? The question here deals with time. Again, letsFill in the chart.

Rate Time Distance

1st leg 60

2nd leg 48

Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph?

This time totals are given for both time traveled and distance traveled. Thus totals don’t belong in the chart. Bernadette did not travel 2.2 hours at 60 mph nor did she travel 2.2 hours at 48 mph. She traveled 2.2 hours total at both of those speeds. So how do we write this in the chart. Click to observe.

t

2.2 - t

60t

48(2.2 – t)

Distance is again computed by formula R*T = D

Rate Time Distance

1st leg 60

2nd leg 48

Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph?

What is the picture for this problem?

t

2.2 - t

60t

48(2.2 – t)

Do you see the equation from the picture?

Dist 1st leg Dist 2nd leg

60t 48(2.2 – t)

Total dist. 120 miles

60t +48(2.2 – t) = 120

Bernadette drove 120 miles. The first part of the trip she averaged 60 mph, but on the second part of the trip she ran into some congestion and averaged 48 mph. If the total driving time was 2.2 hours, how much time did she spend at 60 mph?

Dist 1st leg Dist 2nd leg

60t 48(2.2 – t)

Total dist. 120 miles

60t +48(2.2 – t) = 12060t + 105.6 – 48t = 12012t = 14.4 t = 1.2 hours at 60mph

4. Dr. John left New Orleans at 12 noon. His drummer left at 1:00 traveling 9 mph faster. If the drummer passed Dr. John at 6:00, what was the average speed of each?

Now we don’t know the speed. We do know something about time. Let’s see how we can fill in the chart.

Rate Time Distance

John x

Drummer

X + 9

Dr. John left New Orleans at 12 noon. His drummer left at 1:00 traveling 9 mph faster. If the drummer passed Dr. John at 6:00, what was the average speed of each?

The problem is over when the drummer passes Dr. John at 6 PM, so how long has Dr. John been driving?

6

How long has the drummer been driving?

5

What is our picture?

Dr. John

DrummerDrummer passing Dr. John

6x

5(x + 9)

6x

5x + 45

Rate Time Distance

John x 6x

Drummer

X + 9 5(x + 9)

Dr. John left New Orleans at 12 noon. His drummer left at 1:00 traveling 9 mph faster. If the drummer passed Dr. John at 6:00, what was the average speed of each?

6

5

Dr. John

Drummer

What equation does the picture suggest?

The two distances are equal thus:6x = 5(x + 9)6x = 5x + 45X = 45 mph for Dr. JohnX + 9 = 54 mph for the drummer

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