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WorksheetsPhysics Midterm Review
Total questions: 150
Worksheet time: 1hrs 15mins
A frame of reference is best defined as:
a force diagram of an object
a coordinate system used to measure motion
the mass of an object
the acceleration due to gravity
Motion is described relative to:
the object only
a reference point or observer
gravity only
mass only
A book on a desk is at rest relative to the desk but moving relative to the Sun. This shows:
time is relative
mass depends on speed
motion depends on reference frame
friction causes motion
You walk 1 m/s forward inside a bus moving 10 m/s forward. Your speed relative to the ground is:
9 m/s
10 m/s
11 m/s
1 m/s
Same situation as the previous, but you walk 1 m/s backward (toward the rear). Ground speed:
9 m/s
10 m/s
11 m/s
1 m/s
Two cars move east: Car A at 25 m/s, Car B at 18 m/s. Velocity of A relative to B is:
43 m/s E
7 m/s E
7 m/s W
0 m/s
Car A 25 m/s east, Car B 18 m/s west. A relative to B:
7 m/s E
43 m/s E
43 m/s W
7 m/s W
An inertial frame is one that:
accelerates with the object
is rotating
moves at constant velocity (no acceleration)
always stays on Earth
Newton’s laws apply most directly in:
any rotating frame
inertial frames
only frames attached to Earth
only frames with friction
A person drops a ball inside a train moving at constant speed. To a person on the train, the ball falls:
backward
straight down
forward
upward
To an observer standing outside watching the same drop, the ball’s path is best described as:
straight down
straight forward
a forward-moving curve (parabola-like)
a circle
A frame moving at constant velocity relative to Earth will measure:
different accelerations for the same object
the same acceleration for the same object (in classical physics)
different masses
different times always
A student says, “Velocity is absolute.” Best response:
True, velocity is always the same
False, velocity depends on reference frame
True, because speed is absolute
False, because mass changes
In most classroom labs, Earth is treated as:
a non-inertial frame because it rotates
an inertial frame approximation
a frame where forces don’t act
a frame with zero gravity
Which frame is most convenient for analyzing a cart on a track in the lab?
the Sun-centered frame
the cart’s accelerating frame
the lab/track frame
a rotating frame attached to the wheels
If you stand on a moving walkway and do not walk, you are:
at rest in all frames
moving relative to the ground
accelerating relative to the walkway
moving relative to the walkway but not ground
Two observers disagree on whether a skateboard is speeding up. Most likely reason:
they used different meters
they used different reference frames or sign conventions
gravity changed
mass changed
“At rest” means:
not moving in any frame
not moving in the chosen frame
no forces act
zero mass
A rider on a carousel feels “pulled outward.” In the rotating frame, this is explained by:
friction
a fictitious (inertial) force
increased gravity
tension disappears
A frame is non-inertial if it is:
moving steadily in a straight line
accelerating or rotating
far from Earth
measuring distance
A plane flies 250 m/s east. Wind is 40 m/s west. Plane’s speed relative to air is 250 m/s; speed relative ground is:
210 m/s E
250 m/s E
290 m/s E
210 m/s W
Same as the previous but wind 40 m/s east:
210 m/s E
250 m/s E
290 m/s E
290 m/s W
If you choose a different origin on the x-axis, displacement values:
become wrong
change, but physics conclusions do not
never change
make velocity negative always
If you reverse the positive direction, a velocity of +5 m/s becomes:
+5 m/s
−5 m/s
0 m/s
undefined
Which quantity is most clearly reference-frame dependent?
mass
temperature
velocity
charge
In classical mechanics, acceleration measured in two frames moving at constant relative velocity is:
always different
always the same
opposite
zero
A person walking on a bus tosses keys straight up. To the person on the bus, the keys come down:
behind them
in front of them
back to their hand
never return
To a stationary observer outside, the keys follow:
a straight line up/down
a forward curve
a circle
a backward curve only
Relative velocity is found by:
multiplying velocities
subtracting velocities (vector subtraction)
dividing velocities
adding masses
You must design a demo proving “motion can be different in different frames.” Best choice:
drop a ball while standing still
measure gravity with a spring scale
roll a ball on a moving cart and film from two locations
weigh the ball twice
A student claims “If my velocity is 0, no forces act.” Best counterexample:
a book sliding at constant speed
a book resting on a desk
a falling ball at the top (v=0 but a≠0)
a cart at constant speed
Which is a good “reference object” for describing motion in class?
“the universe”
the classroom wall
“anything moving”
a feeling
Two people in different frames can disagree on:
whether the object is moving
the object’s mass
the value of g
the object’s charge
They will agree most reliably on:
velocity
displacement
acceleration in inertial frames (classical)
position value
A skateboarder coasts at constant speed. In skateboarder’s frame, the ground moves:
forward
backward
not at all
upward
In the ground frame, the skateboarder moves:
forward
backward
not at all
upward
If an observer is accelerating, they may introduce:
scalar forces
fictitious forces to apply Newton’s laws in that frame
fewer forces
more mass
You’re analyzing a cart in an accelerating elevator. The simplest accurate frame is:
elevator frame with fictitious force included
Earth frame only
cart frame always
a rotating frame
If two observers use different frames, they must still agree on:
the chosen positive direction
the same measured velocities
physical events occurring (what happens)
exact coordinate values
A GPS system must account for:
only friction
only distance
motion and time measurements in different frames
color of satellites
Best scientific question to investigate frames in a lab:
“Is physics real?”
“How does measured velocity change when the observer moves?”
“Do heavier objects move faster?”
“What color is motion?”
Best data collection method for the previous question:
one measurement only
measure from two observer locations and compare
guess the result
use only one stopwatch without distances
A boat aims north at 6 m/s in still water; river flows east at 4 m/s. Boat’s ground speed is:
2 m/s
6 m/s
10 m/s
62+42 m/s
Direction of boat’s ground velocity in the previous scenario is:
due north
northeast
due east
northwest
If you change frames, which is most likely to change sign?
mass
velocity component
time
charge
If you say “the car is fast,” what is missing scientifically?
its mass
a reference frame and direction
its color
its engine type
Which statement is best?
Object moves at 3 m/s.
Object moves at 3 m/s east relative to the lab floor.
Object moves.
Object is fast.
If the reference frame changes from ground to train moving east, a thrown ball’s measured horizontal velocity:
stays identical
changes by subtracting train speed
doubles
becomes zero always
A drone flies 12 m/s north relative to air; wind 5 m/s east. Ground velocity magnitude is:
7 m/s
12 m/s
17 m/s
122+52 m/s
Best conclusion from the drone scenario is:
wind has no effect
wind changes measured motion between frames
gravity caused it
mass changed
Displacement is:
total path length
change in position (with direction)
(DOK1) Speed is:
displacement/time
distance/time
acceleration/time
force/mass
(DOK2) A runner goes 50 m east then 50 m west. Displacement is:
0 m
100 m
50 m
50 m W
(DOK2) Same trip: distance is:
0 m
100 m
50 m
25 m
(DOK1) Average velocity equals:
distance/time
displacement/time
acceleration/time
slope of v–t graph
(DOK2) A car’s position changes from x=−10 m to x=40 m in 5 s. Average velocity:
6 m/s
10 m/s
50 m/s
−10 m/s
(DOK2) If velocity is constant and positive, position-time graph is:
flat line
straight line with positive slope
curve upward
straight line with negative slope
(DOK2) If acceleration is constant and positive, velocity-time graph is:
flat
straight line with positive slope
curve
straight line with negative slope
(DOK1) Acceleration is:
change in position/time
change in velocity/time
distance/time
force/time
(DOK3) A car moves east but slows down. Acceleration is:
east
west
zero
upward
(DOK3) A car moves west and slows down. Acceleration is:
west
east
zero
downward
(DOK3) A car moves west and speeds up. Acceleration is:
west
east
zero
upward
(DOK2) Which is possible?
v=0 and a=0 only
v=0 and a≠0
v≠0 and a≠0 never
a≠0 implies v always positive
(DOK3) At the instant an object changes direction in 1D, its velocity is:
maximum
zero
negative always
constant
(DOK2) The kinematic equations require:
constant speed
constant acceleration
constant force
no time
(DOK2) If a=0, then velocity is:
increasing
decreasing
constant
undefined
(DOK2) If a is opposite v, speed:
increases
decreases
stays constant
becomes zero instantly
(DOK3) A cart has v=+2 m/s and a=−0.5 m/s². After 4 s, v =
0 m/s
+4 m/s
−0.5 m/s
+0.5 m/s
(DOK3) Using #18, displacement in 4 s:
8 m
4 m
0 m
2 m
(DOK2) If an object’s position does not change, its velocity is:
constant positive
constant negative
zero
increasing
(DOK3) A car starts from rest and accelerates 3.0 m/s² for 5.0 s. Final speed:
15 m/s
8 m/s
3 m/s
1.5 m/s
(DOK3) Same as #21: displacement:
75 m
37.5 m
15 m
7.5 m
(DOK3) A bike slows from 12 m/s to 4 m/s in 4 s. Acceleration:
+2 m/s²
−2 m/s²
−4 m/s²
+4 m/s²
(DOK3) Same as #23: displacement during 4 s:
32 m
24 m
16 m
8 m
(DOK2) The slope of an x–t graph gives:
acceleration
velocity
displacement
time
(DOK2) The area under a v–t graph gives:
acceleration
displacement
force
speed
(DOK3) A v–t graph is a horizontal line at −6 m/s for 10 s. Displacement:
−60 m
+60 m
0 m
6 m
(DOK3) A v–t graph increases linearly from 0 to 20 m/s in 4 s. Acceleration:
5 m/s²
20 m/s²
80 m/s²
0 m/s²
(DOK3) Displacement for #28:
80 m
40 m
20 m
10 m
(DOK2) A negative velocity means:
slowing down
moving opposite the positive direction
speeding up
acceleration is negative
(DOK3) A negative acceleration means:
slowing down always
speeding up always
acceleration points in the negative direction
velocity is negative
(DOK3) If v is negative and a is negative, the object:
speeds up in negative direction
slows down in negative direction
turns around instantly
stops immediately
(DOK3) A car goes from x=5 m to x=−15 m in 4 s. Average velocity:
+5 m/s
−5 m/s
+10 m/s
−10 m/s
(DOK2) Which is a scalar?
velocity
displacement
speed
acceleration
(DOK3) A cart moving +3 m/s experiences a=+1 m/s² for 6 s. Final velocity:
9 m/s
6 m/s
3 m/s
−3 m/s
(DOK3) Displacement for #35:
18 m
36 m
54 m
27 m
(DOK4) A student’s data show constant velocity but changing position. Best explanation:
impossible
constant velocity means position changes linearly
constant velocity means position is constant
acceleration must be changing
(DOK4) Two objects start at same position. One has constant v=4 m/s. Other starts at rest with a=1 m/s². When does second catch first?
4 s
8 s
2 s
never
(DOK4) If an object has v=0 at t=2 s and v=0 at t=6 s with constant a, then:
it never moved
it changed direction between 2 s and 6 s
acceleration was zero
velocity was constant
(DOK4) Which representation best shows turning around?
x–t straight line
v–t crossing 0
v–t horizontal above 0
x–t flat line
(DOK3) A car accelerates at 2 m/s² from 5 m/s to 25 m/s. Time:
5 s
10 s
20 s
40 s
(DOK3) Same as #41: displacement:
150 m
75 m
50 m
100 m
(DOK2) If you double acceleration (same time), change in velocity:
halves
doubles
stays same
becomes zero
(DOK4) A student uses distance instead of displacement in average velocity. This will:
always increase the magnitude of velocity
always decrease it
sometimes change it, especially if direction reverses
never matter
(DOK5) Best experiment to test constant acceleration on a cart:
measure mass once
measure position vs time and check if v–t is linear
measure color of cart
measure only final position
(DOK3) A cart has v=−8 m/s and a=+2 m/s². After 3 s, v:
−14 m/s
−2 m/s
+2 m/s
+14 m/s
(DOK3) How far does it go in 3 s (from #46)?
−15 m
−9 m
+9 m
+15 m
(DOK4) A v–t graph has equal positive and negative areas over a time interval. Net displacement is:
maximum
zero
negative
positive
(DOK4) A car’s x–t graph is getting steeper with time. This means:
slowing down
speeding up
moving backward
stopped
(DOK4) A car’s x–t graph is a curve concave down while moving positive. This implies acceleration is:
positive
negative
zero
undefined
Slope of a position-time graph is:
acceleration
velocity
force
time
Slope of a velocity-time graph is:
acceleration
displacement
mass
speed only
A flat (horizontal) x–t line means:
constant acceleration
constant velocity
at rest
moving backward
A straight x–t line with constant positive slope means:
speeding up
constant positive velocity
constant negative velocity
stopped
A curved x–t graph means:
constant velocity
changing velocity
zero displacement
negative time
Area under v–t graph represents:
acceleration
displacement
speed
force
A v–t line above the axis indicates:
negative displacement
positive velocity
zero acceleration always
stopped
A v–t line below the axis indicates:
negative velocity
positive acceleration
speed is zero
position is constant
If v–t crosses from positive to negative, the object:
speeds up always
turns around (changes direction)
stops forever
accelerates upward
If v–t is a horizontal line at 8 m/s, acceleration is:
8 m/s^2
0 m/s2
−8 m/s^2
cannot tell
If v–t is a line increasing from 2 to 10 m/s in 4 s, acceleration is:
2m/s2
4 m/s2
8 m/s^2
12 m/s^2
Displacement for #11:
24 m
16 m
48 m
32 m
A v–t graph is a triangle from 0 to 12 m/s over 6 s. Displacement:
72 m
36 m
12 m
6 m
A v–t graph is −5 m/s for 10 s. Displacement:
50 m
−50 m
5 m
−5 m
Distance-time graphs cannot slope downward because:
time can’t be negative
distance is always nonnegative total path length
velocity can’t be negative
acceleration is constant
Position-time graphs CAN slope downward because:
position can decrease with time
distance decreases
time decreases
speed is negative
A steeper slope on x–t means:
less velocity
more velocity magnitude
more acceleration
less displacement
A x–t graph that becomes less steep over time while moving positive indicates:
speeding up
slowing down
turning around
constant speed
A v–t graph with positive slope means:
velocity is negative
acceleration is positive
displacement is negative
object is stopped
A v–t graph with negative slope means:
acceleration is negative
object must be moving negative
displacement is zero
time is negative
If v is negative and becoming more negative, then on v–t graph:
below axis with positive slope
below axis with negative slope
above axis with negative slope
above axis with positive slope
If v is positive but decreasing toward zero, v–t graph:
above axis with negative slope
below axis with positive slope
below axis with negative slope
flat at zero
A v–t graph shows equal positive and negative areas. Net displacement is:
maximum
zero
negative
positive
If two v–t graphs have the same area but different shapes, they have:
same acceleration
same displacement
same final velocity
same slope
Units of slope on x–t graph:
m/s2
m/s
s/m
N
Units of area under v–t graph:
m/s2
m/s
m
N
A v–t graph is a rectangle: v=6 m/s for 8 s. Displacement:
48 m
14 m
6 m
8 m
A v–t graph increases from 0 to 18 m/s in 3 s. a=?
6
9
18
54 (m/s^2)
Displacement for #28:
27 m
54 m
18 m
9 m
A student says “negative area under v–t means negative distance.” Correct response:
true
false; it means negative displacement
false; it means zero distance
true only when speeding up
Best evidence of “turning around” on graphs is:
x–t slope constant
v–t crosses zero
x–t is flat
v–t is flat
A position-time graph that is concave up (steepening) indicates:
negative acceleration
positive acceleration (if moving positive)
constant velocity
zero displacement
If x–t is concave down while moving positive, acceleration is:
positive
negative
zero
cannot tell
Which graph directly shows acceleration as slope?
x–t
v–t
a–t
distance–t
If a–t is constant positive, v–t is:
flat
straight line increasing
curve
random
If a student calculates slope of x–t but labels it “speed,” error is that:
slope gives acceleration
slope gives velocity (can be negative), not speed
slope gives time
slope gives force
A v–t graph is a line from −4 to +8 m/s over 6 s. Acceleration:
2m/s2
−2m/s2
12 m/s^2
−12m/s2
Displacement for #37: average v = (−4+8)/2 = 2 m/s, so displacement =
12 m
6 m
2 m
−12 m
If x–t slope is zero at a moment, velocity at that moment is:
maximum
zero
negative
constant
Best way to reduce errors when reading a graph:
estimate randomly
use two clear points and include units
ignore axis labels
use only one point
A v–t graph shows v=0 for 2 s then v=5 m/s for 4 s. Total displacement:
10 m
20 m
5 m
0 m
Average velocity for #41 over 6 s:
5 m/s
3.33 m/s
2.5 m/s
0.83 m/s
A v–t graph shows v=+6 for 3 s then v=−6 for 3 s. Net displacement:
36 m
18 m
0 m
−18 m
Total distance for #43:
0 m
18 m
36 m
12 m
The slope of a v–t line is 0.5 m/s^2. After 10 s, change in velocity is:
0.05 m/s
5 m/s
10 m/s
50 m/s
If the area under v–t is 120 m, that represents:
distance only
displacement (could be signed)
acceleration
force
A student confuses distance–t with position–t. Key difference is:
distance can be negative
distance always increases or stays constant; position can increase/decrease
position is scalar
both must slope downward sometimes
A v–t line is steep. That indicates:
large displacement
large acceleration magnitude
large position
large time
If two motions have same final velocity but different slopes on v–t, they had:
same acceleration
different accelerations
same displacement
same time always
Best conclusion when x–t is linear but v–t is also linear:
impossible
both cannot be linear
