WORKSHEET 1

WORKSHEET 1: 1-Dimensional Motion

QUESTION 1: Here are three pairs of initial and final positions, respectively, along the x-axis. Which pairs give a negative displacement?
(a) –3 m, +5 m: Dx = +5 m – (-3 m) = +5 m + 3 m = +8 m
(b) –3 m, -7 m: Dx = -7 m – (-3 m) = -7 m + 3 m = -4 m
(c) +7 m, -3 m: Dx = -3 m – 7 m = -10 m
ANSWER: (b) and (c)

QUESTION 2: You drive your truck down a straight road for 5.2 mi at 43 mi/h, at which point you run out of fuel. You walk 1.2 mi farther, to the nearest gas station, in 27 min (=0.450 h). What is your average velocity from the time you started your truck to the time you arrived at the gas station?
SOLUTION:
                    To find the average velocity we must find Dx and Dt.
                    Dx = 5.2 mi + 1.2 mi = 6.4 mi
                    Dt = (time it took to go from the starting point to where you ran out of gas) + (time it took to walk to the gas station) = 5.2 mi/ (43 mi/h) + 0.450 h = 0.12 h + 0.450 h = 0.57 h
                    Therefore the average velocity is v = Dx/Dt = (6.4 mi/ 0.57 h) = 11 mi/h

QUESTION 3: Suppose that you carry the fuel back to the truck, making the return trip in 35 min. What is your average velocity for the full journey, from the start of your driving to your arrival back at the truck with fuel?
SOLUTION:
                    Dx = 5.2 mi + 1.2 mi – 1.2 mi = 5.2 mi
Note: When you walk to the gas station you have a 1.2 mi displacement. When you walk from the gas station back to your car, you have to walk in the opposite direction, therefore your displacement is –1.2 mi.
                    Dt = 0.12 h + 0.450 h + (35 min)(1 h/60min) = 0.12 h + 0.450 h + 0.58 h = 1.15 h
                    Therefore the average velocity is v = Dx/Dt = (5.2 mi/ 1.15 h) = 4.5 mi/h

QUESTION 4: In the previous problem, what is your average speed?
SOLUTION:
Recall that the average speed is equal to the total distance traveled divided by the time interval. Therefore
s = total distance / 1.15 h = (5.2 + 1.2 + 1.2) mi / (1.15 h) = 7.6 mi / 1.15 h = 6.6 mi/h

PROBLEM SOLVING TACTICS
Tactic 1: Do you understand the problem? Do you know what is being asked for in the problem?

You should be able to explain the problem in your own words.
Write down the given data and write down what is the unknown quantity.
Based on the given data and unknown, you should be able to decide which formula to use for the problem.

Tactic 2: Check for reasonable answers by checking units and order of magnitude.

QUESTION 5: The displacement versus time graph for a certain particle moving along the x axis is shown in the figure below. Find the average velocity in the following time intervals: (based on the figure below)

average velocity = displacement/time =Dx/Dt a) 0 to 2 s
                   Dx = 10 m – 0 m = 10 m
                   Dt = 2 s – 0 s = 2 s
                   average velocity = 10 m/ 2 s = 5.0 m/s
b) 0 to 4 s
                   Dx = 5 m – 0 m = 5 m
                   Dt = 4 s – 0 s = 4 s
                   average velocity = 5 m/4 s = 1.2 m/s
c) 2 s to 4 s
                Dx= 5 m – 10 m = - 5 m
                   Dt = 2 s
                    average velocity = -5 m/2 s = -2.5 m/s
d) 4 s to 7 s
                   Dx = -5 m – 5 m = -10 m
                   Dt = 7 s – 4 s = 3 s
                    average velocity = -10m/3s = -3.3 m/s
e) 0 to 8 s
                   Dx = 0 m – 0 m = 0 m
                   Dt = 8.0 s
                    average velocity = 0 m/s

QUESTION 6: A person walks first at a constant speed of 5.0 m/s along a straight line from point A to point B and then back along the line from point B to A at a constant speed of 3.0 m/s
a) What is her average speed over the entire trip?
Solution:
                    Average Speed = (total distance traveled)/(total time)
STEP 1: Find the total time assuming that the distance from A to B is d (Therefore the total distance traveled is 2d)
                    From A to B: t_ab = d/(5.0 m/s)
                    From B to A: t_ba = d/(3.0 m/s)
                    Total time T = t_ab + t_ba = d/(5.0 m/s) + d/(3.0 m/s) = d[(1/5) +(1/3)]
                    Therefore T = (8d)/15
STEP 2: Using the total distance as 2d and the total time T, we can find the average speed
                    Average speed v = 2d/[8d/15] = 30d/8d= 30/8
                   Average Speed = 3.75 m/s = 3.8 m/s

b) What is her average velocity?
Solution:
                    Average velocity =(x_f -x_i )/(t_f -t_i )
                    But she begins and ends at the same place, therefore (x_f -x_i )=0
                   Therefore her average velocity = 0

QUESTION 8: Find the instantaneous velocity of the particle described in the figure at the following times: (a) t = 1.0 s, (b) t = 3.0 s, (c) t = 4.5 s, and (d) t = 7.5 s.

(a) t = 1.0 s
                    v = slope = (10 - 0)/(2.0 - 1.0) = 5.0 m/s
(b) t = 3.0 s
                    v = slope = (5.0 - 10)/(4.0 - 2.0) = -2.5 m/s
(c) t = 4.5 s
                    the slope in this region is 0, therefore v = 0
(d) t = 7.5 s
                    v = slope = (0 - (-5))/(8 - 7) = 5.0 m/s

QUESTION 9: The velocity-time graph for an object moving along th x-axis is shown in the figure below. (a) Plot the graph of the acceleration versus time. (b) Determine the average acceleration of the object in the time intervals t = 5.00s to t = 15.0 s and t = 0 s to t = 20.0s.

(a) Acceleration Plot

(b) t = 5.00 s to t = 15.0 s
                    a = slope = (8.00 m/s -(-8.00 m/s)/(10.0s) = 1.60 m/s²
      t = 0 to 20.0 s
                    a = (16.0 m/s)/20 s = 0.80 m/s²

QUESTION 10: (a) When Kitty 0’Neil set the dragster records for the greatest speed and least elapsed time, she reached 392.54 mi/h in 3.72 s. What was her average acceleration?
SOLUTION:
Step 1: Convert mi/h into m/s: 392.54 mi/h = (392.54 mi/h)(1609 m/ mi)(1h/3600 s) = 175.44 m/s
Step 2: Find the average acceleration realizing that the initial velocity is zero at t = 0.
Average acceleration = (175.44 m/s – 0)/(3.72 s – 0) = 47.2 m/s² which is about 5g’s.

(b) What was the average acceleration when Eli Beeding Jr. reached 72.5 mi/h in 0.04 s on a rocket sled?
SOLUTION:
Step 1: Convert mi/h into m/s: 72.5 mi/h = (72.5 mi/h)(1609 m/ mi)(1h/3600 s) = 32.4 m/s
Step 2: Find the average acceleration realizing that the initial velocity is zero at t = 0.
Average acceleration = (32.4 m/s – 0)/(0.04 s – 0) = 8.1 x 10² m/s² which is about 83g’s.

PROBLEM SOLVING TACTICS: Tactic 4 - The sign of acceleration.

The sign of the acceleration is often misinterpreted the following way: if the acceleration is negative then the object is decelerating, if positive, then it is accelerating.
What if the velocity of the particle (which is a vector quantity) is initially negative and the particle has a negative acceleration is it still decelerating or slowing down? Of course not.
The best way to interpret the signs is the following way: If the signs of the velocity and acceleration of a particle are the same, the speed of the particle increases. If the signs are opposite, then the speed decreases.

QUESTION 10: A wombat moves along an x-axis. What is the sign of its acceleration if it is moving:
(a) in the positive direction with increasing speed?
positive

(b) in the positive direction with decreasing speed?
negative

(d) In the negative direction with decreasing speed?
positive