7 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 7
A student is asked for the average velocity of a 200 m long train that travels 150 km along a straight track between two cities. She decides to model the train as an object. Which statement about her choice of model is correct?
Answer and reasoning
AIt is unsuitable, since an object as large as a train cannot be treated as a point. A student who thinks only small objects can be modeled as points picks this. What matters is whether size affects the quantity asked for; here it does not, so even a train can be treated as an object.
BIt is suitable, provided the train is given no mass, since a point has no size. A student who thinks a point has no mass picks this. The object model ignores size and shape but keeps extensive properties such as mass: the point has the whole mass of the train.
CIt is suitable, since the train's size and shape do not affect the answer.Correct In the object model, size, shape and internal configuration are ignored. The train's 200 m length is small compared with the 150 km trip, and every part of the train has the same displacement, so treating it as a single point does not change its average velocity.
DIt is unsuitable, since the train's front and back are at different positions. A student who thinks an extended object has no single position picks this. The front and back of the train are at different places, but they have the same displacement, so one point represents the train's motion.
Swimmer A swims one 50 m length of a straight pool and stops at the far end. Swimmer B swims two lengths, to the far end and back, and stops where she started. Which statement correctly compares the two swimmers?
Answer and reasoning
AB's displacement is greater, since B swam twice as far as A. A student who treats displacement as the distance traveled picks this. B swam 100 m, but she ended where she started, so her displacement is zero.
BTheir displacements are equal, as both got 50 m from the start. A student who takes displacement to be the farthest distance reached picks this. Both reached the far end, but B came back: displacement uses only the final position, so B's is zero.
CB's displacement is negative, as B was swimming back at the end. A student who gives the displacement the direction of the final motion picks this. B was moving back toward the start at the end, but she finished exactly where she started, so her displacement is zero, neither positive nor negative.
DA's displacement is greater, but B swam the greater distance.Correct Displacement depends only on the start and finish: A's is 50 m toward the far end, and B's is zero, since she ends where she started. Distance is the path length: 50 m for A and 100 m for B.
Cars 1 and 2 both leave town P at 1:00 p.m. and arrive at town Q, 60 km away along a straight road, at 2:00 p.m. Car 1 travels at a steady speed. Car 2 stops for 10 minutes on the way and drives faster than car 1 for the rest of the trip. Which statement about the cars' average velocities for the trip is correct?
Answer and reasoning
AThey are equal: cars 1 and 2 have the same displacement in the same time.Correct Average velocity is Δx/Δt, which depends only on the initial and final states. Both cars have a displacement of 60 km toward Q in the same 1 h interval, so both have an average velocity of 60 km/h toward Q.
BCar 1's is greater, since car 2 was not moving for part of the trip. A student who treats average velocity as the typical velocity while moving picks this. Car 2 makes up for the stop by driving faster, and it still has the same displacement in the same time interval.
CCar 2's is greater, since car 2 reached a greater top speed than car 1 did. A student who thinks the top speed sets the average picks this. Car 2's higher speed only makes up for its stop; its displacement and time interval are the same as car 1's.
DThey cannot be compared without car 2's velocity at every instant. A student who thinks an average needs every value picks this. An average velocity needs only the displacement and the time interval, and both are given.
Working Both cars: Δx = 60 km toward Q and Δt = 1 h (1:00 p.m. to 2:00 p.m.), so vavg = Δx/Δt = 60 km/h toward Q for each car.
The table shows the position x of a cart moving along a straight track, recorded at 1 s intervals. The cart turns around at t = 2 s. What is the cart's average velocity from t = 0 to t = 4 s?
Answer and reasoning
A+1.5 m/s A student who divides the final position by the clock time, 6 m ÷ 4 s, picks this. The cart started at x = 4 m, not at the origin, so its displacement is 2 m, not 6 m.
B+5.0 m/s A student who uses the distance traveled, 11 m out and 9 m back, picks this: 20 m ÷ 4 s is the average speed. Average velocity uses the displacement, +2 m.
C+0.5 m/sCorrect Average velocity uses only the initial and final positions: Δx = 6 m − 4 m = +2 m over Δt = 4 s, so vavg = +0.5 m/s. The trip out to x = 15 m and back does not enter.
D+2.0 m/s A student who divides the time interval by the displacement, 4 s ÷ 2 m, picks this. That has units s/m, not m/s; the displacement goes on top.
Working Δx = x(4 s) − x(0) = 6 m − 4 m = +2 m; Δt = 4 s. vavg = (+2 m)/(4 s) = +0.5 m/s. (Distance traveled = 11 m out + 9 m back = 20 m, so the average speed is 5.0 m/s.)
A cart on a level track collides with a spring bumper and rebounds along the same line. The diagram shows the cart's velocity just before and just after the collision, which lasts 0.20 s. Taking the direction toward the bumper as positive, what is the cart's average acceleration during the collision?
Answer and reasoning
A−1.0 m/s² A student who uses the change in speed, 0.40 m/s − 0.60 m/s, picks this. The velocity also reverses direction, so the change in velocity is −1.00 m/s, not −0.20 m/s.
B−5.0 m/s²Correct Taking toward the bumper as positive, v0 = +0.60 m/s and v = −0.40 m/s, so Δv = −0.40 m/s − (+0.60 m/s) = −1.00 m/s. aavg = Δv/Δt = (−1.00 m/s) ÷ 0.20 s = −5.0 m/s², directed away from the bumper.
C+5.0 m/s² A student who subtracts final from initial velocity, 0.60 m/s − (−0.40 m/s), picks this. A change is final minus initial, −1.00 m/s; the bumper turns the cart back, so the acceleration is directed away from it and is negative.
D+2.0 m/s² A student who divides the final velocity by the time, reading 0.40 m/s from the diagram, picks this: 0.40 m/s ÷ 0.20 s = 2.0 m/s². Average acceleration uses the change in velocity, and the cart did not start from rest: with the direction toward the bumper positive, Δv = (−0.40 m/s) − (+0.60 m/s) = −1.00 m/s.
Working v0 = +0.60 m/s, v = −0.40 m/s. Δv = v − v0 = −0.40 m/s − 0.60 m/s = −1.00 m/s. aavg = Δv/Δt = (−1.00 m/s)/(0.20 s) = −5.0 m/s², directed away from the bumper.
A ball moving at 5.0 m/s toward a wall bounces straight back at 5.0 m/s. A student claims that the ball does not accelerate during the bounce. Which statement correctly evaluates the claim?
Answer and reasoning
ACorrect: the ball's speed after the bounce is the same as its speed before it. A student who judges acceleration by the change in speed alone picks this. Velocity includes direction, and reversing it is a change in velocity, so the ball accelerates.
BCorrect: the ball comes to rest during the bounce, and a ball at rest has no acceleration. A student who thinks an object at rest cannot be accelerating picks this. At the instant the ball is at rest its velocity is changing from toward the wall to away from it, so its acceleration is not zero, and the ball accelerates throughout the bounce.
CIncorrect: it accelerates as it speeds back up, but not while it slows down. A student who thinks 'accelerating' means only speeding up picks this. While the ball slows down against the wall its velocity is changing too, so it is accelerating then as well.
DIncorrect: the direction of the ball's velocity changes, so the ball accelerates.Correct An object accelerates if the magnitude or the direction of its velocity changes. The speed is the same before and after, but the velocity reverses, from 5.0 m/s toward the wall to 5.0 m/s away from it: a change of 10 m/s directed away from the wall.
The table shows the position x of an object moving along a straight line at four instants. Which value is the best estimate of the object's instantaneous velocity at t = 2.00 s?
Answer and reasoning
A25 m/s A student who uses the longest interval, (55.0000 m − 30.0000 m) ÷ 1.00 s, picks this. The velocity changes during that second; averages over shorter intervals, 20.5 m/s and then 20.05 m/s, come closer to the value at 2.00 s.
B15 m/s A student who divides the position by the clock time, 30.0000 m ÷ 2.00 s, picks this. That would be an average velocity only if the object had been at x = 0 at t = 0, and even then it would be an average over 2.00 s, not the velocity at 2.00 s.
C20 m/sCorrect Average velocities over intervals starting at t = 2.00 s are 25 m/s (to 3.00 s), 20.5 m/s (to 2.10 s) and 20.05 m/s (to 2.01 s). As the interval shrinks the average approaches 20 m/s, so the average over the shortest interval gives the best estimate.
D30 m/s A student who reads the position at t = 2.00 s as the velocity picks this. 30.0000 m says where the object is; a velocity needs a change in position divided by a change in time.
Working Average velocities over intervals starting at t = 2.00 s: to 3.00 s, (55.0000 − 30.0000) m ÷ 1.00 s = 25 m/s; to 2.10 s, 2.0500 m ÷ 0.10 s = 20.5 m/s; to 2.01 s, 0.2005 m ÷ 0.01 s = 20.05 m/s. As the interval shrinks the averages approach 20 m/s.
In preparation: 0 of 7 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
1.2.A.1 Object model Fix
Object model
A model in which an object's size, shape and internal configuration are ignored and the object is treated as a single point. It applies whenever those features do not affect the quantity being found, however large the object is.
Extensive property
A property whose value for a whole object is the total for all its parts, such as mass (kg) and charge. An object treated as a point keeps its full mass and charge.
Students often think Only objects that are physically small, such as a ball or a bead, can be treated as a point; a large object such as a train or a planet cannot. In fact No. Whether an object can be modeled as a single point depends on the question, not on its size. A 200 m train traveling 150 km between cities can be treated as an object when finding its average velocity, because its size and shape do not affect the answer.
Students often think A point has no size, so an object modeled as a point also has no mass, or its mass can be set to zero. In fact No. The object model ignores size, shape and internal configuration, but the point keeps the object's extensive properties, such as its whole mass and its whole charge.
1.2.A.2 Position, x Fix
Position, x
The location of an object along an axis, measured from a chosen origin, with a sign that gives the side of the origin. SI unit: m.
Displacement, Δx
The change in an object's position: Δx = x − x0, final position minus initial position. It is a vector: its sign gives its direction, and it depends only on the start and end positions, not on the path between them. SI unit: m.
Distance traveled
The total length of the path followed, a scalar. It equals the magnitude of the displacement only if the object never reverses direction. SI unit: m.
Students often think An object's displacement is its position at the end of the motion, read directly from the axis, whatever its starting position. In fact Only if the object starts at the origin. Displacement is the change in position, Δx = x − x0. An object that moves from x0 = +10 m to x = −15 m has final position −15 m but displacement −25 m.
Students often think A change in a quantity can be found as the initial value minus the final value, so Δx = x0 − x and Δv = v0 − v. In fact No. A change is final minus initial: Δx = x − x0 and Δv = v − v0. Reversing the subtraction reverses the sign, which in one dimension reverses the direction.
1.2.B.1 Time interval, Δt Fix
Time interval, Δt
The duration between the initial and final instants of the part of the motion being described: Δt = t − t0. SI unit: s.
Average over an interval
An average velocity or average acceleration is found from the object's states at the start and at the end of a time interval and the length of that interval. What happens in between does not enter the calculation.
Students often think Average velocity is the typical velocity while moving, so a trip with a stop has a smaller average velocity than a steady trip, even if both have the same displacement and take the same time. In fact No. Average velocity depends only on the initial and final positions and the time interval. Two trips with the same displacement in the same time interval have the same average velocity, whatever happens in between.
Students often think The object that reaches the higher top speed during a trip has the greater average velocity. In fact No. A higher top speed matters only through its effect on the displacement and the time interval. If two objects have the same displacement in the same time interval, their average velocities are equal, whatever their top speeds.
1.2.B.2 Average velocity Fix
Average velocity
The displacement divided by the time interval in which it occurs: v⃗avg = Δx⃗/Δt. In one dimension it is a signed component whose sign gives its direction. SI unit: m/s.
Average speed
The distance traveled divided by the time interval, a scalar. It differs from the magnitude of the average velocity whenever the object reverses direction. SI unit: m/s.
Students often think An object's velocity, over an interval or at an instant, is its position divided by the time on the clock, x/t. In fact No. Average velocity is the displacement divided by the time interval, Δx/Δt. x/t equals it only if the object was at x = 0 at t = 0, and it is not the velocity at an instant when the velocity is changing.
Students often think A rate such as average velocity or average acceleration can be found by dividing the time interval by the change (Δt/Δx or Δt/Δv), as if the order of division did not matter. In fact No. Average velocity is Δx/Δt and average acceleration is Δv/Δt: the change goes on top. Δt/Δx has units s/m, which is not a velocity.
1.2.B.3 Change in velocity, Δv Fix
Change in velocity, Δv
Final velocity minus initial velocity, Δv = v − v0, found with signed components. An object that reverses from +3 m/s to −3 m/s has Δv = −6 m/s even though its speed is unchanged. SI unit: m/s.
Average acceleration
The change in velocity divided by the time interval in which that change occurs: a⃗avg = Δv⃗/Δt. SI unit: m/s², read as (m/s) per second.
Students often think An object's average acceleration is its final velocity divided by the time taken, a = v/t, whatever its initial velocity. In fact Only if the object started from rest. Average acceleration is the CHANGE in velocity divided by the time interval: aavg = (v − v0)/Δt.
Students often think Acceleration is the change in velocity, so the object whose velocity changes by more has the greater acceleration, whatever time the change takes. In fact No. Acceleration is the change in velocity divided by the time interval in which it happens. A change of 6 m/s in 3 s is an average acceleration of 2 m/s²; the same change in 1 s is 6 m/s².
1.2.B.4 Accelerating Fix
Accelerating
An object is accelerating whenever its velocity is changing in magnitude, in direction, or in both: speeding up, slowing down and reversing are all accelerations.
Students often think Acceleration depends only on the change in speed, so an object whose speed is the same at the start and at the end, even after reversing, has zero acceleration, and the largest change in speed means the largest accelera… In fact No. Acceleration is the change in VELOCITY divided by the time interval. A change of direction is a change in velocity, so an object that reverses at the same speed is accelerating.
Students often think An object is accelerating only while it speeds up; slowing down, or changing direction at the same speed, is not accelerating. In fact No. An object is accelerating whenever the magnitude or the direction of its velocity is changing: speeding up, slowing down or reversing.
1.2.B.5 Instantaneous velocity Fix
Instantaneous velocity
The velocity at one instant. Averages of velocity calculated over shorter and shorter time intervals that include the instant come closer and closer to it, so an average over a very small interval is a good estimate of it. SI unit: m/s.
Instantaneous acceleration
The acceleration at one instant, closely approximated by the average acceleration over a very small time interval that includes the instant. SI unit: m/s².
Students often think An average over any interval, and preferably a long one, gives the velocity at the instant the interval begins, because a longer interval smooths out errors. In fact No. When the velocity is changing, the average over an interval differs from the velocity at its start; the shorter the interval, the closer the average is to the instantaneous value.
Students often think The velocity at an instant is the position value recorded at that instant, so a position of 30 m at t = 2.00 s means a velocity of 30 m/s. In fact No. Position (in m) tells where the object is; velocity (in m/s) tells how quickly, and in which direction, the position is changing. They are different quantities with different units.
6 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 6
The diagram shows the motion of a toy car along a straight track. The car starts at the point marked start, moves to the right, turns around, and stops at the point marked finish. What is the car's displacement from start to finish?
Answer and reasoning
A+25 m A student who subtracts the final position from the initial one, x0 − x = 10 m − (−15 m), picks this. A change is final minus initial; the car ends to the left of where it started, so Δx is negative.
B−25 mCorrect Displacement is the change in position: Δx = x − x0 = (−15 m) − (+10 m) = −25 m. The trip out to x = 20 m and back does not matter; only the start and finish positions do.
C−15 m A student who takes the final position as the displacement picks this. −15 m is where the car ends up, measured from the origin; the car started at +10 m, so its displacement is (−15 m) − (+10 m).
D+45 m A student who gives the distance traveled picks this: 10 m out to the turning point plus 35 m back. Displacement is a vector that depends only on the start and finish positions.
Working x0 = +10 m and x = −15 m. Δx = x − x0 = (−15 m) − (+10 m) = −25 m. (Distance traveled = 10 m out to x = 20 m + 35 m back = 45 m.)
A runner on a straight track runs a distance D in the +x direction in a time T. She then turns and sprints a distance D/2 in the −x direction in a time T/3. Which expression gives her average velocity for the whole run?
Answer and reasoning
A9D/(8T) A student who uses the total distance, D + D/2 = 3D/2, picks this: (3D/2) ÷ (4T/3) = 9D/(8T) is the average speed. Average velocity uses the displacement, D/2.
B8T/(3D) A student who divides the time interval by the displacement, (4T/3) ÷ (D/2), picks this. That has units s/m; average velocity is Δx/Δt.
C−D/(4T) A student who averages the velocities of the two parts, [D/T + (−3D/(2T))]/2, picks this. The first part lasts three times as long as the second, so the two velocities cannot simply be averaged; the runner ends on the +x side of her start, so her average velocity is positive.
D3D/(8T)Correct Displacement: +D and then −D/2, so Δx = +D/2. Time interval: T + T/3 = 4T/3. vavg = Δx/Δt = (D/2) ÷ (4T/3) = 3D/(8T), in the +x direction.
Working Δx = (+D) + (−D/2) = +D/2. Δt = T + T/3 = 4T/3. vavg = Δx/Δt = (D/2) ÷ (4T/3) = 3D/(8T). (Distance traveled = 3D/2, so the average speed is (3D/2) ÷ (4T/3) = 9D/(8T). The velocities of the two parts are +D/T and −3D/(2T), whose mean is −D/(4T).)
Cart A speeds up from 2.0 m/s to 8.0 m/s in 3.0 s. Cart B speeds up from 24.0 m/s to 28.0 m/s in 2.0 s. Both carts move in the +x direction along straight tracks. Which statement correctly compares the magnitudes of their average accelerations?
Answer and reasoning
AThey are equal, since each cart's velocity changes by 2.0 m/s per second on average.Correct aavg = Δv/Δt. Cart A: 6.0 m/s ÷ 3.0 s = 2.0 m/s². Cart B: 4.0 m/s ÷ 2.0 s = 2.0 m/s². The average accelerations are equal.
BB's is greater, since cart B is moving much faster than cart A throughout. A student who links a greater speed with a greater acceleration picks this. Acceleration depends on how quickly the velocity changes, not on how large it is; both average accelerations are 2.0 m/s².
CA's is greater, since A's velocity changes by 6.0 m/s but B's by only 4.0 m/s. A student who equates acceleration with the change in velocity picks this. A's larger change takes longer: 6.0 m/s in 3.0 s and 4.0 m/s in 2.0 s are both 2.0 m/s².
DB's is greater, since B's final velocity divided by its time is the larger value. A student who divides the final velocity by the time picks this (28.0 ÷ 2.0 for B, 8.0 ÷ 3.0 for A). Neither cart started from rest, so the change in velocity must be used.
Working aA = (8.0 − 2.0) m/s ÷ 3.0 s = 2.0 m/s². aB = (28.0 − 24.0) m/s ÷ 2.0 s = 2.0 m/s². They are equal. (Final velocity ÷ time would give 2.7 m/s² for A and 14 m/s² for B.)
The diagram shows the velocities of four carts, P, Q, R and S, moving along straight tracks, at t = 0 and at t = 2.0 s. The positive direction is to the right, and each arrow is labeled with its speed. Which cart has the greatest magnitude of average acceleration between t = 0 and t = 2.0 s?
Answer and reasoning
ACart PCorrect Average acceleration is Δv/Δt, using signed velocities. This cart goes from +3 m/s to −3 m/s, so Δv = −6 m/s and the magnitude of its average acceleration is 6 m/s ÷ 2.0 s = 3.0 m/s², the largest of the four (the others are 2.0, 2.5 and 0 m/s²).
BCart Q A student who thinks only speeding up counts as accelerating picks this, the one cart that speeds up. Its velocity changes by 4 m/s, an average acceleration of 2.0 m/s²; the cart that reverses has a larger change of velocity, 6 m/s.
CCart R A student who uses the change in speed picks this: its speed falls by 5 m/s, more than any other cart's. But the cart that reverses changes its velocity by 6 m/s, from +3 m/s to −3 m/s, even though its speed ends where it began.
DCart S A student who links a greater speed with a greater acceleration picks this, the fastest cart. Its velocity is the same at both times, so its average acceleration is zero.
Working Δt = 2.0 s. P: +3 m/s → −3 m/s, Δv = −6 m/s, |aavg| = 3.0 m/s²; Q: +2 m/s → +6 m/s, Δv = +4 m/s, |aavg| = 2.0 m/s²; R: +6 m/s → +1 m/s, Δv = −5 m/s, |aavg| = 2.5 m/s²; S: +7 m/s → +7 m/s, Δv = 0 m/s, |aavg| = 0.0 m/s². Greatest magnitude: cart P.
The velocity of cart 1 changes by Δv during a time interval Δt. The velocity of cart 2 changes by 2Δv during a time interval Δt/2. The magnitude of cart 2's average acceleration is how many times that of cart 1?
Answer and reasoning
A×1 A student who thinks doubling one quantity and halving another always cancel picks this. They cancel in a product; here the time interval divides, so halving it doubles the acceleration: 2 × 2 = 4.
B×4Correct aavg = Δv/Δt. For cart 2: (2Δv) ÷ (Δt/2) = 4Δv/Δt. Doubling the change in velocity doubles the average acceleration, and halving the time interval doubles it again: 2 × 2 = 4.
C×2 A student who takes acceleration to be the change in velocity picks this, doubling Δv and ignoring the time interval. The shorter time interval doubles the acceleration again.
D×¼ A student who divides the time interval by the change in velocity picks this: (Δt/2) ÷ (2Δv) is ¼ of Δt/Δv. Acceleration is Δv/Δt, the reciprocal, so the factor is 4, not ¼.
A cyclist rides in a straight line in the +x direction. She covers the first half of the total distance D at a constant speed v and the second half at a constant speed 2v. Which expression gives the magnitude of her average velocity for the whole ride?
Answer and reasoning
A(3/2)v A student who averages the two speeds, (v + 2v)/2, picks this. The halves do not last equal times (the slower half takes twice as long), so the simple mean gives the faster half too much weight; divide the total displacement by the total time instead.
B3/(4v) A student who divides the time interval by the displacement, (3D/(4v)) ÷ D, picks this. It has units s/m, not m/s; average velocity is Δx/Δt.
C(2/3)v A student who divides the displacement of one half, D/2, by the time for the whole ride, 3D/(4v), picks this. The displacement and the time interval must describe the same interval: over the whole ride, Δx = D.
D(4/3)vCorrect The two halves take different times: D/(2v) at speed v and D/(4v) at speed 2v, so Δt = 3D/(4v). The average velocity uses the initial and final states of the whole interval: Δx/Δt = D ÷ (3D/(4v)) = (4/3)v. It is nearer v than 2v because the slower half lasts twice as long.
Working First half: t1 = (D/2)/v = D/(2v). Second half: t2 = (D/2)/(2v) = D/(4v). Whole ride: Δt = t1 + t2 = 3D/(4v). She rides in one direction only, so Δx = +D. vavg = Δx/Δt = D ÷ (3D/(4v)) = (4/3)v. (Mean of the two speeds: (v + 2v)/2 = (3/2)v. Time divided by displacement: (3D/(4v)) ÷ D = 3/(4v), units s/m. Displacement of one half over the whole time: (D/2) ÷ (3D/(4v)) = (2/3)v.) The slower half lasts twice as long as the faster half, so the average lies nearer v than 2v.
Compiled from the AP Physics 1 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
