**Practice Problems: Motion Studies **

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. (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

3. (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.

4. (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

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

5. (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.

6. (moderate) A particle undergoes motion according to the equation: x = 5t^{4} - 2t^{2} - 8 where x is in meters and t is in seconds. Find the average speed of the particle over the first 7 seconds and the instantaneous speed of the particle at 5.0 seconds. Also, find the instantaneous acceleration at 0.5 seconds.

At t=0 s...x = 5(0) -2(0) - 8 = -8 m

At t = 7 s...x = 5(7)^{4} - 2(7)^{2} - 8 = 11899 m

v_{ave} = Δx/Δt = [11899 - (-8)] / 7 = 1701 m/s

v = dx/dt = 20t^{3} - 4t

at t = 5.0 s: v = 20(5)^{3} - 4(5) = 2480 m/s

a = dv/dt = 60t^{2} - 4

at t = 0.5 s: a = 60(0.5)^{2} - 4 = 11 m/s^{2}

7. (moderate) A particle changes speed (in m/s) as it moves along the x axis according to the following equation: v = t^{2} - 4t + 9

Find the average acceleration of the particle between t = 5 seconds and t = 10 seconds and the instantaneous acceleration at each of those times.

at t = 10 s: v = 10^{2} - 4(10) + 9 = 69 m/s

at t = 5 s: v = 5^{2} - 4(5) + 9 = 14 m/s

a_{ave} = Δv/Δt

a_{ave} = (69 - 14)/5 = 11 m/s^{2}

Now, to find the instantaneous acclerations:

a = dv/dt = 2t - 4

at t = 10 s: a= 2(10) - 4 = 16 m/s^{2}

at t = 5 s: a = 2(5) - 4 = 6 m/s^{2}

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_{1}) + 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

9. (moderate) An object is bobbing up and down on the end of a long elastic cord according to the equation y = 5cos(3t), such that when t is in seconds, y is in in meters. Find the speed and the magnitude of the acceleration at t = 3.2 s.

v = dy/dt

v = d(5cos(3t))/dt

v = -15sin(3t)

v = -15sin(3(3.2)) = 2.6 m/s

a = dv/dt

a = d(-15sin(3t))/dt

a = -45cos(3t)

a = -45cos(3(3.2)) = 44.3 m/s^{2}