**Practice Problems: Motion Studies Solutions**

1. (easy) A woman walks in a straight line for 40 minutes at an average speed of 1.25 m/s. How far does she travel in this time?

(40 min)(60 s/min) = 2400 s

v_{ave} = Δx/Δt

1.25 = Δx/(2400 s)

Δx = 3000 m

2. (moderate) A car is traveling along a road. It moves according to the following equation:

x = 2.0t^{2} + 0.25t^{3} (x is in meters and t is in seconds)

a) What is the average velocity of the car between 0.0 and 2.0 seconds?

b) What is the average velocity between 2.0 and 4.0 seconds?

x(0) = 2(0) + (0.25)(0) = 0 m

x(2) = 2(4) + (0.25)(8) = 10 m

x(4) = 2(16) + (0.25)(64) = 48 m

v_{ave} = Δx/Δt

a) v = (10 - 0)/2 = 5 m/s

b) v = (48 - 10)/2 = 19 m/s

3. (moderate) Based upon the following data determine if the average acceleration of the object between 2.0 and 4.0 seconds has a bigger or smaller magnitude than the average acceleration between 4.0 and 6.0 seconds:

velocity at 2.0 seconds is 0.8 m/s

velocity at 4.0 seconds is 1.2 m/s

velocity at 6.0 seconds is 0.6 m/s

Between 2 and 4 seconds: a = Δv/Δt = (1.2 - 0.8)/2 = 0.2 m/s^{2}

Between 4 and 6 seconds: a = Δv/Δt = (0.6 - 1.2)/2 = -0.3 m/s^{2}

The average acceleration magnitude is larger between 4 and 6 seconds.

4. (moderate) Watch the video below and have a stopwatch available. Cart 1 is shown in the video. Cart 2 (not shown in video) travels 0.2 m in 4.96 seconds. If cart 1 started its motion in Pittsburgh and ended in New York City (370 miles away), how many days ahead of cart 2 would it arrive? Note that the video shows the motion of cart 1 twice. You may need to watch the video several times to make accurate measurements. The red cm markings on the track are 10 cm apart.

First find the speed of each cart:

Cart 1: v_{1} = Δx/Δt

v_{1} = 0.4/7.07 = 0.057 m/s

Cart 2: v_{2} = Δx/Δt

v_{2} = 0.2/4.96 = 0.040 m/s

Your speeds may be slightly different based on difference in timing which can be expected.

Now find the time for each cart to move 370 miles.

First convert 370 miles to meters:

370 miles(1609.344 m/mile) = 595457.3 m

Cart 1: v_{1} = Δx/Δt_{1}

Δt_{1} = 595457.3/0.057 = 10446619 s

Cart 2: v_{1} = Δx/Δt_{1}

Δt_{1} = 595457.3/0.040 = 14886433 s

Cart 2 arrives 4439814 s after cart 1

Finally, convert the time difference to days:

4439814 s (1 day/86400 s) = 51.4 days

5. (easy) Interstate 70 is a highway that runs through the heartland of the United States. If traveler A drove 100 miles along Interstate 70 at 65 mph and traveler B drove the same distance at 55 mph, how much earlier would traveler A arrive?

Time for traveler A:

v = Δx/Δt

65 = 100/Δt

Δt = 1.54 hours

Time for traveler B:

v = Δx/Δt

55 = 100/Δt

Δt = 1.82 hours

Time difference = 1.82 - 1.54 = 0.28 hours = 17 minutes

6. (easy) In a well-formed paragraph, using the concepts described in the presentation, explain why the following statement must be incorrect: "The displacement of an object during an experiment was zero, but the average velocity was not zero." Additionally, how could you change the statement (replace one word with a different word) that would allow it to be true?

The average velocity is defined as the displacement divided by the time. If there is no displacement there can be no average velocity. If the word "velocity" was replaced with "speed" the statement may be true.

7. (hard) A car is moving along the x-axis of a coordinate system accoring to the following equation:

x = 20 + 5t^{2} (x is in meters and t is in seconds)

a) Find the displacement of the car between the first and second second.

b) Find the average speed during the first and second second.

c) Find the instantaneous velocity at t = 1 second.

a) Δx = x_{2}- x_{1} = (20 + 20) - (20 + 5) = 15 m

b) v_{ave} = Δx/Δt = 15/1 = 15 m/s

c) At t = 1 second x = 25 m

At t = 2 seconds x = 40 m

At t = 1.1 seconds x = 26.05 m

At t = 1.01 seconds x = 25.1005 m

At t = 1.001 seconds x = 25.005 m

Now calculate v for smaller and smaller time intervals:

Between 1 and 2 seconds: v_{ave} = Δx/Δt = (40 - 25)/1 = 15 m/s

Between 1 and 1.1 seconds: v_{ave} = Δx/Δt = (26.05 - 25)/0.1 = 10.5 m/s

Between 1 and 1.01 seconds: v_{ave} = Δx/Δt = (25.1005 - 25)/0.01 = 10.05 m/s

Between 1 and 1.001 seconds: v_{ave} = Δx/Δt = (25.010005 - 25)/0.001 = 10.005 m/s

The speed is getting closer and closer to v = 10 m/s when the Δt is getting smaller and smaller in relation to t = 1 second. Thus, the instantaneous speed at t = 1 second is 10 m/s.

8. (hard) A student jogs from home to the store. She runs at 1 m/s for 1/3 of the time and 1.5 m/s for the remaining time. On her way back home she runs at 1 m/s for half the distance and at 1.5 m/s for the rest of the trip. Find her average speed for the round trip.

From home to the store:

Make T_{out} the time for this part of the trip.

v = Δx/Δt = [1(T_{out}/3) + 1.5(2T_{out}/3)]/T_{out}

v = 1.33 m/s

For the trip home:

x_{1} = distance traveled in v_{1 }(1 m/s)

x_{2} = distance traveled at v_{2} (1.5 m/s)

t_{1} = time traveled at v_{1}

t_{2 }= time traveled at v_{2}

Given that x_{1} = x_{2, } x_{1} = v_{1}t_{1} and x_{2} = v_{2}t_{2}

v_{1}t_{1} = v_{2}t_{2}

t_{1} = 1.5t_{2}

Thus:

v = Δx/Δt = 1(t_{1}) + 1.5(t_{2})/(t_{1} + t_{2})

v = 1(1.5t_{2}) + 1.5t_{2}/(1.5t_{2} + t_{2}) = 3t_{2}/2.5t_{2} = 1.20 m/s

Now find the average speed for the the entire trip:

Δx_{out} = Δx_{back} (magnitude only)

1.33T_{out} = 1.20T_{back}

T_{out} = 0.9T_{back}

average speed = v_{ave} = (total distance traveled)/Δt

v_{ave} = (1.33T_{out} + 1.20T_{back})/(T_{out }+ T_{back})

v_{ave} = [1.33(0.9T_{back}) + 1.20(T_{back})]/[(0.9T_{back} + T_{back})]

v_{ave} = 2.40T_{back}/1.90T_{back} = 1.26 m/s

**Click on a link below to see Scenario Solutions:**

1A

1C