Practice 1.2D Solutions
Motion Graphs 2
Multiple Choice
At time t = 0, car X traveling with speed v0 passes car Y which is just starting to move. Both cars then travel on two
parallel lanes of the same straight road. The graphs of speed v versus time t for both cars are shown above.
1.
Which of the following is true at time t = 20 seconds?
(A) Car Y is behind car X. (B) Car Y is passing car X. (C) Car Y is in front of car X.
(D) Both cars have the same acceleration. (E) Car X is accelerating faster than car Y.
The area under the curve is the displacement. There is more area under the curve for Car X.
2.
From time t = 0 to time t = 40 seconds, the areas under both curves are equal. Therefore, which of the
following is true at time t = 40 seconds?
(A) Car Y is behind car X. (B) Car Y is passing car X. (C) Car Y is in front of car X.
(D) Both cars have the same acceleration. (E) Car X is accelerating faster than car Y.
Area under the curve is the displacement. Car Y is moving faster as they reach the same point
3.
Above is a graph of the distance vs. time for car moving along a road. According the graph, at which of the
following times would the automobile have been accelerating positively?
(A) 0, 20, 38, & 60 min. (B) 5, 12, 29, & 35 min. (C) 5, 29, & 57 min. (D) 12, 35, & 41 min.
(E) at all times from 0 to 60 min
Positive acceleration is an increasing slope (including negative slope increasing toward zero) or
upward curvature
4. Which of the following sets of graphs might be the corresponding graphs of Position, Velocity,
and Acceleration vs. Time for a moving particle?
The velocity-time graph should represent the slope of the position-time graph and the acceleration-time
graph should represent the slope of the velocity-time graph
Answer: C
5.
Consider the motion of an object given by the position vs. time graph shown. For what time(s) is the
speed of the object greatest?
(A) At all times from t = 0.0 s→ t = 2.0 s (B) At time t = 3.0 s (C) At time t = 4.0 s
(D) At all times from t = 5.0 s → t = 7.0 s (E) At time t = 8.5 s
Speed is the slope of the line.
Free Response
Problem 1
A group of students measures the motion of a cart in a long hallway. Use methods practiced in this class
(think first week of school).
time (s)
2
2.8
3.6
4.4
5.2
6
6.8
7.6
8.4
9.2
10
displacement (m)
5
9.8
16.2
24.2
33.8
45
57.8
72.2
88.2
105.8
125
a) Use excel to find the acceleration of the cart, according to the data.
Motion of the Cart
140
y = 1.25x2
R² = 1
Displacement (m)
120
100
80
60
40
20
0
0
2
4
6 (s)
Time
8
10
12
Displacement vs. Time^2
140
y = 1.25x
R² = 1
120
Displacement (m)
100
80
60
40
20
0
0
20
40
60
80
100
120
Time^2 (s^2)
Looking at the graph above, you can create the equation
Y= mx+b
In this case b appears to be 0 so
Y= mx
Y= displacement x= time^2 and the slope of the graph is 1.25 so
Displacement = 1.25 t2
Keep in mind that when an object starts from rest one of its acceleration equations is
D= (1/2) at2
So
1.25 = (1/2)a
a= 2.5 m/s/s
b) If you were a teacher and a group turned these values in to you, would you be suspicious. If you
had to confront them about cooking the data, what compelling evidence would you have?
Their R2 value is 1, which is pretty difficult to get with something as inexact as measuring a cart
with stopwatches.
Problem 2
a) True or False? The cart in the above graphs in traveling in a positive direction
True
b) True or False? the cart is experiencing a constant positive acceleration.
True
c) What is the cart's initial speed?
5m/s
d) What is the cart's initial position?
4m
e) How much did the cart's velocity change from 2 to 14 seconds?
36 m/s
f)
What is the cart's final velocity at 14 seconds?
41 m/s
g) How fast was the cart traveling at 8 seconds?
23 m/s
h) How far did the cart travel between 2 and 14 seconds?
i) 276 m
j) What is the cart's final position at 14 seconds?
280 m
The following equations describe the motion of an object. Use the power rule of derivation to find the
equation for the velocity of the object. Then, state the velocity of the object at t= 2s.
6. 𝑓(𝑡) = 5
0, 0m/s
7. 𝑓(𝑡) = −2𝑡
-2, -2m/s
8. 𝑓(𝑡) = −2𝑡 + 2
-2, -2m/s
9. 𝑓(𝑡) = −2𝑡 2 − 5
-4x, -8 m/s
10.
𝑓(𝑡) = 2𝑡 4 + 𝑡 3 − 𝑡 2 + 4
3
8t + 3t 2 -2t, 72 m/s