AP Physics 2
Preparation for Test – Chapter 2
Name ________________________
1.
A(n) ______________ quantity is completely described by its magnitude.
A.) vector
B.) scalar
C.) large
D.) small
2.
Velocity is defined as
A.) change in velocity divided by elapsed time.
B.) elapsed time divided by change in velocity.
C.) change in position divided by elapsed time.
D.) change in position times elapsed time.
E.) None of these
3.
A vector quantity has ____________.
A.) only a magnitude
B.) only a direction
C.) intensity
D.) both magnitude and direction
4.
The main difference between speed and velocity is that speed does not include
A.) acceleration.
B.) time.
C.) direction.
D.) distance.
5.
If the average velocity of an object is the same for all time intervals, then the
object moves at a(n) _____________________.
A.) constant velocity.
B.) changing speed.
C.) relative speed.
D.) instantaneous velocity
6.
The area under the curve on a velocity-time graph equals the __________________.
A.) instantaneous velocity.
B.) rise divided by the run
C.) displacement from the original position to its position at time t.
D.) relative velocity.
7.
The slope of the tangent on a position-time graph equals the __________________ .
A.) average velocity
B.) constant speed
C.) instantaneous velocity
D.) average speed
8.
The graph at the right represents the motion of a body
that is moving with
A.) increasing velocity.
B.) increasing speed.
C.) constant velocity.
D.) constant speed.
9.
The correct units for representing a velocity would be:
A.) 10 meters in the westerly direction.
B.) 10 meters/sec in the westerly direction.
C.) 10 meters/sec2 in the westerly direction.
D.) 10 meters/sec.
Distance
Time
10.
Which is constant for a freely falling body in a vacuum?
(A) displacement
(B) speed
(C) velocity
(D) acceleration
11.
Which graph best represents the motion of an object initially at its origin and
accelerating uniformly?
(A)
(B)
d
d
t
(C)
(D)
d
d
t
t
t
12.
On a velocity - time graph, the slope of the tangent to the curve at a given point is
the ___________.
(A) average velocity
(B) average acceleration
(C) instantaneous velocity
(D) instantaneous acceleration
13.
Constant acceleration on a velocity - time graph produces a curve that is _________.
A.) a straight line
B.) parallel to the horizontal axis
C.) half a parabola
D.) parallel to the vertical axis
14.
Which graph represents the relationship between velocity and time for an object
accelerating at a constant positive rate?
(A)
v
(B)
v
t
(C)
v
t
(D)
v
t
15.
Ignoring air resistance, if a 10 kg ball and a 200 kg crate were both dropped
from the top of a building, the acceleration of the crate would be _____________ the
acceleration of the ball.
(A) less than
(B) ten times
(C) equal to
(D) twenty times
16.
The magnitude of the acceleration of a baseball thrown straight up in the air at the
moment it reaches the top of its trajectory is ___________________.
(A) zero
(B) 9.8 m/s
2
(C) 9.8 m/s
(D) dependant on the mass of the bullet
17.
A rocket is thrown straight up into the air. When the rocket reaches its
maximum height, its velocity in the vertical direction is ______________.
(A) at its maximum
(B) equal to its displacement multiplied by time
(C) equal to its displacement divided by time
(D) zero
t
18.
19.
A baseball is thrown straight up into the air. The total displacement upon
landing is _______________. Assume there is no wind and no air resistance.
(A) zero
(B) equal to the distance it traveled up multiplied by two
(C) equal to the distance it traveled up
(D) equal to the difference between the final and initial velocities divided by the
acceleration due to gravity.
Given:
position (m)
0
time(s)
Draw:
(A) the velocity vs. time graph.
20.
(B) the acceleration vs. time graph
A train travels due east 95 km/h for 0.52 h, 50 km/h for the next 0.24 h, and then 100
km/h for the next 0.71 h. What is its average velocity?
𝑣𝑎𝑣𝑔 =
𝑘𝑚
𝑘𝑚
𝑘𝑚
(.52 ℎ𝑟)+ 50
(.24 ℎ𝑟)+100 (.71 ℎ𝑟)
ℎ𝑟
ℎ𝑟
ℎ𝑟
95
.52 ℎ𝑟+ .24 ℎ𝑟+ .71 ℎ𝑟
𝑘𝑚
= 90.1 ℎ𝑟 East
[must convert the km/hr to displacement before you can determine average velocity]
21.
This question concerns the motion of a car on a straight track; the car’s velocity as a
function of time is plotted below.
v(m/s)
20
10
0
Time(s)
- 10
- 20
A. Describe what happens to the car at time t = 1 s.
It decelerated from positive 20 m/s while still traveling in the positive direction
B. How does the car’s average velocity between time t = 0 and t = 1 s compare to its
average velocity between times t = 1 s and t = 5 s.
𝑣𝑎𝑣𝑔 =
𝑣0−1
𝑣1−5
𝑣𝑓 + 𝑣𝑖
2
𝑚
𝑚
20 𝑠 + 0 𝑠
𝑚
=
= 10
2
𝑠
𝑚
𝑚
0 𝑠 + 20 𝑠
𝑚
=
= 10
2
𝑠
1
𝑚
1
𝑠
𝑚
Or, using displacement under the curve, the displacement from 0-1 s is 2 20
10 𝑚 divided by 1 s yields v = 10 m/s. The displacement from 1-5 s is 2 20
40 𝑚 divided by 4 s yields v = 10 m/s.
𝑠
𝑥1𝑠 =
𝑥4𝑠 =
C. Plot the car’s acceleration during this interval as a function of time.
[NOTE. MY PREVIOUS GRAPH WAS WRONG. Here, in red, is the right graph. Use the V-t
diagram to figure out the area under the curve (displacement) and plot the displacement as dots
on the graph. Then freehand the curve to show acceleration as below. My free hand is terrible
but I hope it communicates the basic look.
a(m/s/s)
20
10
0
Time(s)
- 10
- 20
D. Plot the object’s position during this interval as a function of time. Assume that the
car begins at x = 0.
d(m)
20
10
0
- 10
- 20
Time(s)