9 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 9
A student uses the object model to describe a 1200 kg car on a 30 km highway trip. Which statement about this model is correct?
Answer and reasoning
AIt is invalid: a car is much too large an object to be treated as a point. A student who thinks only small objects can be modeled as points picks this. What matters is whether the size affects the question; a few meters is negligible on a 30 km trip.
BIt is valid: the car becomes a single point, and a point has no mass. A student who thinks a point object has no mass picks this. The object model keeps the object's extensive properties: the point has the car's mass, 1200 kg.
CIt is valid: the car becomes a single point that has all of the car's mass.Correct In the object model the car's size, shape and internal configuration are ignored, and the car is treated as a single point that keeps its extensive properties, including its 1200 kg mass. For a 30 km trip the car's size does not matter.
DIt is invalid: the car's wheels turn, so it cannot be treated as a point. A student who thinks any internal motion rules out the object model picks this. The model ignores internal configuration, including the turning of the wheels.
Working Object model: size, shape and internal configuration ignored; the car becomes one point with its mass, 1200 kg. A car's size is negligible compared with a 30 km trip.
The figure shows the path of a bead that slides along a straight wire lying along an x-axis. The bead starts at P, moves to Q, and then moves to R, where it stops. What is the bead's displacement Δx for the whole trip?
Answer and reasoning
A13 m A student who takes displacement to be the distance traveled picks this: 4 m from P to Q plus 9 m from Q to R. Displacement uses only the start and end positions.
B+5 m A student who subtracts the final position from the initial one picks this: (+2 m) − (−3 m) = +5 m. Δx = x − x0, so the bead's net change is −5 m, toward −x.
C−3 m A student who takes the final position as the displacement picks this. The bead started at x = +2 m, not at the origin, so Δx = −3 m − 2 m = −5 m.
D−5 mCorrect Displacement depends only on the start and end: Δx = x − x0 = (−3 m) − (+2 m) = −5 m. The trip out to Q and back does not affect it.
Working x0 = +2 m (P), x = −3 m (R). Δx = x − x0 = −3 m − (+2 m) = −5 m. (Distance traveled: 4 m + 9 m = 13 m.)
A car starts from rest in a parking space, is driven 12 km through town, and 30 min later is again at rest in the same space. For the 30 min interval, which statement is correct?
Answer and reasoning
AIts average velocity and its average acceleration are both zero.Correct Averages depend only on the initial and final states. The car ends where it started, so its displacement and average velocity are zero; it starts and ends at rest, so its change in velocity and average acceleration are zero.
BIts average velocity is not zero; it was driven 12 km in all. A student who uses the distance driven in place of the displacement picks this. The 12 km is a distance; the displacement from the space back to the same space is zero.
CIts average acceleration is not zero, as it moved fast for a while. A student who links acceleration with how fast the car moves picks this. Average acceleration depends on the change in velocity, and the car's velocity is zero at both the start and the end.
DNeither can be found without its velocity at every instant. A student who thinks an average needs the whole history of the motion picks this. Average velocity and average acceleration need only the initial and final states and the time interval.
Working Initial and final positions equal: Δx⃗ = 0, so v⃗avg = 0. Initial and final velocities both zero: Δv⃗ = 0, so a⃗avg = 0.
The graph shows the position x of a cart moving along a straight track as a function of time t. What is the cart's average velocity from t = 1.0 s to t = 5.0 s?
Answer and reasoning
A+4.0 m/s A student who divides the distance traveled by the time picks this: 4 m out to x = 10 m plus 12 m back, 16 m in 4.0 s. That is the average speed; the average velocity uses the displacement, −8 m.
B−0.4 m/s A student who divides the final position by the clock time picks this: (−2 m)/(5.0 s) = −0.4 m/s. Average velocity is a change in position divided by the time interval, here from t = 1.0 s to 5.0 s.
C+2.0 m/s A student who subtracts the final position from the initial one picks this: (6 m − (−2 m))/(4.0 s) = +2.0 m/s. Δx = x − x0 = −8 m, so the average velocity is negative.
D−2.0 m/sCorrect Read the positions at the two instants: 6 m at t = 1.0 s and −2 m at t = 5.0 s. vavg,x = Δx/Δt = (−2 m − 6 m)/(4.0 s) = −2.0 m/s. The turnaround at t = 2 s does not enter.
Working From the graph: x = 6 m at t = 1.0 s and x = −2 m at t = 5.0 s. vavg,x = Δx/Δt = (−2 m − 6 m)/(5.0 s − 1.0 s) = −8 m/4.0 s = −2.0 m/s.
The motion diagram shows a car's positions at equal time intervals as it moves along a straight road; the dots are numbered in time order. What is the direction of the car's average acceleration between dots 1 and 5?
Answer and reasoning
A−x, which is the same as the direction of the car's motion A student who thinks acceleration points along the motion picks this. The car is slowing, so its velocity change, and its acceleration, points opposite to its motion.
BNo direction: its acceleration is zero while it slows A student who thinks acceleration means only speeding up picks this. Slowing down is a change in velocity, so the car has a nonzero acceleration, in the +x direction.
C+x, opposite to the direction of the car's motionCorrect The car moves in the −x direction and covers less distance in each interval, so it is slowing down. Its velocity change, and so its average acceleration, points opposite to its motion: in the +x direction.
DNo direction: acceleration is a scalar, with only a size A student who treats acceleration as a scalar picks this. Acceleration is a vector; here it points in the +x direction, opposite to the car's velocity.
Working The car moves in the −x direction (dots 1 to 5 run right to left) and the gaps shrink, so it is slowing. Δv⃗ is opposite to v⃗, toward +x.
Which object is NOT accelerating at the instant described?
Answer and reasoning
AA car going around a curve in the road at a steady 15 m/s A student who thinks constant speed means no acceleration picks this. On a curve the direction of the velocity changes, so the car is accelerating.
BA cart moving along a straight track at a steady 3 m/sCorrect A straight path and a steady speed mean neither the magnitude nor the direction of the velocity is changing, so the cart's acceleration is zero.
CA ball tossed straight up, at its highest point A student who thinks zero velocity means zero acceleration picks this. At the top the ball's velocity is passing from upward to downward, so it is changing: the ball is accelerating.
DA train slowing at a steady rate as it approaches a station A student who thinks only speeding up counts as acceleration picks this. A decreasing speed is a change in velocity, so the train is accelerating.
Working Accelerating ⇔ velocity changing in magnitude or direction. Only the cart on a straight track at a steady 3 m/s has constant velocity.
A particle moves along the x-axis with position x(t) = bt³, where b is a positive constant. What is the particle's average velocity vavg,x over the interval from t = T − h to t = T + h, where 0 < h < T?
Answer and reasoning
A3bT² A student who takes the average velocity over an interval containing t = T to equal the instantaneous velocity there, dx/dt = 3bT², picks this. That holds only in the limit as h approaches zero; over this finite interval the average is larger by bh².
Bb(T + h)² A student who takes the velocity to be the position divided by the time picks this: x(T + h)/(T + h) = b(T + h)³/(T + h) = b(T + h)². The average velocity is the displacement over the interval, x(T + h) − x(T − h), divided by the interval's duration, 2h.
Cb(3T² + h²)Correct Average velocity is the displacement divided by the time interval: [b(T + h)³ − b(T − h)³]/(2h) = b(6T²h + 2h³)/(2h) = b(3T² + h²). It exceeds the instantaneous velocity at t = T, 3bT², by bh², a difference that shrinks toward zero as the interval is made smaller.
D3b(T + h)² A student who takes the average velocity to be the instantaneous velocity at the end of the interval, t = T + h, picks this. The average velocity is the displacement divided by the 2h interval, not the final value of dx/dt.
Working Δx = b(T + h)³ − b(T − h)³ = b(6T²h + 2h³); Δt = 2h. vavg,x = Δx/Δt = b(3T² + h²). For comparison, vx = dx/dt = 3bt²: at t = T it is 3bT²; at t = T + h it is 3b(T + h)²; position over time at the end, x(T + h)/(T + h) = b(T + h)². As h → 0, vavg,x → 3bT², the instantaneous velocity at T. Check (b = 1.3, T = 2.0, h = 0.5): key 15.925; 3bT² = 15.6; 3b(T + h)² = 24.375; b(T + h)² = 8.125.
A cart moves along a straight track with a velocity that changes continuously. Which statement describes the cart's instantaneous velocity at t = 3.0 s?
Answer and reasoning
AThe value its average velocity approaches as an interval around 3.0 s shrinks to zeroCorrect The instantaneous velocity is the limit of the average velocity as the time interval containing t = 3.0 s approaches zero; for a smooth motion this limit is dx/dt at t = 3.0 s.
BIts average velocity over any interval that starts at 3.0 s, whatever the interval's length A student who thinks any interval gives the instantaneous value picks this. The velocity changes continuously, so averages over long intervals differ from the value at t = 3.0 s.
CIts position at t = 3.0 s divided by the 3.0 s that have passed since t = 0 A student who equates velocity with position divided by clock time picks this. That gives at most an average over 0 to 3.0 s, and only if the cart started at the origin.
DThe mean of its average velocities over several intervals that include t = 3.0 s A student who takes 'instantaneous' to be a mean of averages picks this. The instantaneous value is the limit the averages approach as the interval shrinks, not their mean.
Working Instantaneous value = limit of the average value as the time interval around the instant approaches zero.
An object starts from rest at t = 0 and moves along the x-axis with acceleration ax(t) = kt, where k is a positive constant. What is the ratio vx(2T)/vx(T) of its velocities at t = 2T and t = T, where T > 0?
Answer and reasoning
A2 A student who assumes a constant acceleration, so that velocity grows in proportion to time, picks this. Here the acceleration itself grows with t, so vx ∝ t².
B8 A student who lets the velocity change with time as the position does, as t³, picks this. The position from rest is kt³/6, but the velocity is kt²/2.
C1 A student who differentiates ax instead of integrating it picks this: dax/dt = k is a constant, so the 'velocity' seems the same at both times. Velocity is found from acceleration by integration.
D4Correct Integrate from rest: vx(t) = ∫₀ᵗ kt′ dt′ = kt²/2, which is proportional to t². Doubling the time multiplies the velocity by 2² = 4.
Working vx(t) = 0 + ∫₀ᵗ kt′ dt′ = kt²/2, proportional to t². vx(2T)/vx(T) = (2T)²/T² = 4.
In preparation: 0 of 9 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, so the object is treated as a single point that keeps its extensive properties, such as mass.
Students often think Only objects that are physically small, such as a pebble, can be treated as points; a large object such as a car cannot. In fact No. Whether an object can be treated as a point depends on the question asked, not on the object's absolute size: a car, or even a planet, can be modeled as a point when its size, shape and internal motion do not affect the motion being described.
Students often think A point has no size, so an object modeled as a point has no mass either. In fact No. The object model ignores size, shape and internal configuration, but the point keeps the object's extensive properties, such as its mass (and its charge).
1.2.A.2 Position, x Fix
Position, x
The location of an object along an axis, measured from a chosen origin; its sign shows on which side of the origin the object is. SI unit: m.
Displacement, Δx = x − x0
The change in an object's position: final position minus initial position. It is a vector and depends only on the initial and final positions, not on the path between them. SI unit: m.
Distance traveled
The total length of the path an object follows, a scalar that is never negative. It equals the magnitude of the displacement only for motion in one direction along a straight line. SI unit: m.
Students often think Displacement is the total distance traveled, so the average velocity of a trip is the distance traveled divided by the time. In fact No. Displacement is final position minus initial position and can be positive, negative or zero; distance is the total path length and is never negative. Average velocity uses displacement; average speed uses distance.
Students often think A change is the initial value minus the final value (or the order of subtraction does not matter). In fact No. Every change in this course is final minus initial: Δx = x − x0 and Δv⃗ = v⃗ − v⃗0. Reversing the order reverses the sign (and the direction) of the result.
1.2.B.1 Time interval, Δt Fix
Time interval, Δt
The elapsed time between an initial and a final instant, Δt = t − t0. SI unit: s.
Students often think An average velocity or acceleration is a mean of the values at every instant, so it cannot be found without the full details of the motion. In fact No. Average velocity and average acceleration depend only on the initial and final states (position or velocity) and the time interval, whatever happened in between.
1.2.B.2 Average velocity, v⃗avg = Δx⃗/Δt Fix
Average velocity, v⃗avg = Δx⃗/Δt
The displacement divided by the time interval in which it occurs. It depends only on the initial and final positions and the interval, not on how the object moved in between. SI unit: m/s.
Average speed
The distance traveled divided by the time interval, a scalar. It is greater than the magnitude of the average velocity whenever the path is not a straight line in one direction. SI unit: m/s.
Students often think An object's velocity, average or instantaneous, is its position divided by the time, x/t. In fact No. Average velocity is a change in position divided by the time interval for that change, Δx/Δt. Position divided by clock time, x/t, gives the average velocity only over an interval from t = 0 for an object that starts at the origin, and in general it is not the instantaneous velocity.
Students often think The average velocity over an interval is the instantaneous velocity at the end of the interval (average and instantaneous velocity are the same thing). In fact No. Average velocity is the displacement divided by the time interval. It equals the instantaneous velocity at the end only if the velocity is constant over the interval.
1.2.B.3 Average acceleration, a⃗avg = Δv⃗/Δt Fix
Average acceleration, a⃗avg = Δv⃗/Δt
The change in velocity, final minus initial (a vector subtraction), divided by the time interval. SI unit: m/s².
Students often think The change in velocity is the change in speed, so an object whose speed is the same at the start and end of an interval has zero change in velocity (and zero average acceleration). In fact No. Velocity is a vector, so Δv⃗ = v⃗ − v⃗0 must account for direction. A ball that hits a wall at 8 m/s and rebounds at 8 m/s has the same speed before and after, but Δv⃗ has magnitude 16 m/s.
Students often think The magnitude of a velocity change is the sum of the initial and final speeds, whatever the directions. In fact Only when the final velocity is exactly opposite to the initial velocity. For velocities in other directions, subtract the vectors and then take the magnitude: for perpendicular velocities |Δv⃗| = √(v₁² + v₂²).
1.2.B.4 Accelerating Fix
Accelerating
An object is accelerating whenever its velocity changes, in magnitude, in direction or in both: speeding up, slowing down and turning all count.
Students often think Acceleration means speeding up, so an object that is slowing down (or keeping its speed) has zero acceleration. In fact No. In physics, acceleration is any change in velocity: speeding up, slowing down or changing direction.
Students often think The sign of an object's position gives its direction of motion, so an object passing from negative x to positive x has changed the direction of its velocity. In fact No. The sign of x tells on which side of the origin the object is; the sign of vx tells which way it is moving. An object can cross the origin without any change in its velocity.
1.2.B.5 Estimate from a small interval Fix
Estimate from a small interval
An average velocity or acceleration calculated over a very small time interval that contains an instant is very close to the instantaneous value at that instant.
Students often think The average velocity over any interval that contains (or starts at) an instant equals the instantaneous velocity at that instant, whatever the size of the interval. In fact Only if the velocity is constant. When the velocity changes, the average over an interval approaches the instantaneous value only as the interval is made very small; a wide interval can give a quite different value.
1.2.C.1 Instantaneous value Fix
Instantaneous value
The limit that the average value of a quantity approaches as the time interval used to calculate it approaches zero.
Instantaneous velocity, v⃗ = dr⃗/dt (vx = dx/dt)
The rate of change of position: the derivative of position with respect to time. Its magnitude is the instantaneous speed. SI unit: m/s.
The rate of change of velocity: the derivative of velocity with respect to time. It is zero only when the velocity is not changing at that instant, whatever the value of the velocity. SI unit: m/s².
Students often think Velocity and acceleration are not distinguished: a large velocity means a large acceleration, a velocity value (or an average velocity) can serve as the acceleration or stand in for the change in velocity, and accelerat… In fact No. Acceleration is the rate of change of velocity, a different quantity from velocity itself. A fast object can have zero acceleration, and an object with a large acceleration can be at rest for an instant.
Students often think Position and velocity are not distinguished: an object's velocity is read as its position value, objects at the same position have the same velocity, and the velocity changes with time in the same way as the position. In fact No. Velocity is the rate of change of position. Two objects at the same position can have very different velocities, and the velocity generally changes with time differently from the position: for x ∝ t³, vx ∝ t².
1.2.C.2 Integration with initial conditions Fix
Integration with initial conditions
Integrating ax(t) gives vx(t) up to a constant, fixed by a known velocity at one instant; integrating vx(t) gives x(t) up to a constant, fixed by a known position at one instant.
Students often think Differentiation and integration are interchanged: velocity is found by integrating position, or position by differentiating velocity. In fact No. Going from position to velocity to acceleration is differentiation (vx = dx/dt, ax = dvx/dt); going back from acceleration to velocity to position is integration.
Students often think An object's acceleration is constant during its motion, so it does not change between two instants (and velocity grows in direct proportion to time). In fact No. The acceleration is constant only if ax(t) does not depend on t. For ax = kt, or for vx = kt², the acceleration grows with time.
14 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 14
A particle moves from position r⃗0 = −3aî + 2aĵ to position r⃗ = aî + aĵ, where a is a positive constant. What is the particle's displacement?
Answer and reasoning
A−4aî + aĵ A student who subtracts the final position from the initial position picks this: r⃗0 − r⃗ = −4aî + aĵ, the reverse of the displacement.
B4aî − aĵCorrect Displacement is final position minus initial position, component by component: x: a − (−3a) = 4a; y: a − 2a = −a. So Δr⃗ = 4aî − aĵ.
Caî + aĵ A student who takes the final position as the displacement picks this. The particle did not start at the origin, so its initial position must be subtracted.
D−2aî + 3aĵ A student who adds the two position vectors picks this: r⃗ + r⃗0 = −2aî + 3aĵ. Displacement is the difference r⃗ − r⃗0.
Working Δr⃗ = r⃗ − r⃗0 = (a − (−3a))î + (a − 2a)ĵ = 4aî − aĵ.
A particle moves along the x-axis with position x(t) = A − Bt², where A and B are positive constants. What is the particle's average velocity vavg,x from t = 0 to t = T?
Answer and reasoning
A−BTCorrect Average velocity is the displacement divided by the time interval: x(T) − x(0) = (A − BT²) − A = −BT², and dividing by T gives −BT.
BA/T − BT A student who divides the final position by the time picks this: (A − BT²)/T. The particle starts at x = A, not at the origin, so A must be subtracted first.
C−2BT A student who takes the instantaneous velocity at the end of the interval, dx/dt = −2Bt at t = T, as the average picks this. The velocity changes during the interval, so the average is the displacement over the time, −BT.
D2A/T − BT A student who adds the initial and final positions instead of subtracting them picks this: (x(T) + x(0))/T = (2A − BT²)/T = 2A/T − BT. Displacement is the difference x(T) − x(0) = −BT², so the average velocity is −BT.
Working x(0) = A; x(T) = A − BT². vavg,x = (x(T) − x(0))/(T − 0) = (−BT²)/T = −BT.
The figure shows the velocity of a puck sliding on a horizontal surface at two instants. What is the magnitude of the puck's average acceleration between those instants?
Answer and reasoning
A1.0 m/s² A student who uses the change in speed picks this: (8.0 − 6.0) m/s ÷ 2.0 s. The direction of the velocity also changed, so Δv⃗ must be found by vector subtraction.
B5.0 m/s²Correct Subtract the velocity vectors: Δv⃗ = (8.0 m/s north) − (6.0 m/s east) = (−6.0î + 8.0ĵ) m/s, of magnitude 10 m/s. Dividing by the 2.0 s interval gives 5.0 m/s².
C7.0 m/s² A student who adds the two speeds to get the change in velocity picks this: (6.0 + 8.0) m/s ÷ 2.0 s. The speeds add only for opposite velocities; for perpendicular ones |Δv⃗| = √(6.0² + 8.0²) m/s.
D4.0 m/s² A student who divides the final velocity by the time, as if a = v/t, picks this: 8.0 m/s ÷ 2.0 s. Average acceleration is the change in velocity divided by the time, and the puck did not start from rest.
Working With x east and y north: v⃗0 = (6.0 m/s)î, v⃗ = (8.0 m/s)ĵ. Δv⃗ = −6.0î + 8.0ĵ m/s, |Δv⃗| = √(6.0² + 8.0²) = 10 m/s. |a⃗avg| = 10 m/s ÷ 2.0 s = 5.0 m/s².
The table shown gives the initial and final velocity components of three carts, X, Y and Z, moving along the same straight track, and the time interval for each change. Which ranking of the magnitudes of the carts' average accelerations is correct?
Answer and reasoning
AZ > X > Y A student who uses changes in speed picks this: X's speed changes only from 4.0 to 10 m/s, giving 1.5 m/s². X reverses direction, so its velocity changes by 14 m/s, not 6 m/s.
BY > X > Z A student who divides each final velocity by the time, as if a = v/t, picks this: 2.5, 4.0 and 0.5 m/s² for X, Y and Z. Average acceleration uses the change in velocity, and no cart starts from rest.
CY > Z > X A student who ranks the signed values +1.0, −2.5 and −3.5 m/s² as numbers picks this. The question asks for magnitudes: 3.5 > 2.5 > 1.0 m/s².
DX > Z > YCorrect aavg,x = Δvx/Δt: X, (−10 − 4.0)/4.0 = −3.5 m/s²; Y, (8.0 − 6.0)/2.0 = +1.0 m/s²; Z, (2.0 − 12)/4.0 = −2.5 m/s². The magnitudes are 3.5, 1.0 and 2.5 m/s², so X > Z > Y.
Working aavg,x = Δvx/Δt. X: (−10 − 4.0)/4.0 = −3.5 m/s². Y: (8.0 − 6.0)/2.0 = +1.0 m/s². Z: (2.0 − 12)/4.0 = −2.5 m/s². Magnitudes 3.5, 1.0, 2.5: X > Z > Y.
A ball moving horizontally at speed v hits a wall head-on and rebounds along its original line with speed 0.40v. The ball is in contact with the wall for a time Δt. What is the magnitude of the ball's average acceleration during the contact?
Answer and reasoning
A1.4v/ΔtCorrect With +x toward the wall, v0x = +v and vx = −0.40v, so Δvx = −0.40v − v = −1.4v. The magnitude of the average acceleration is 1.4v/Δt: reversing direction adds the two speeds.
B0.6v/Δt A student who uses the change in speed, v − 0.40v = 0.60v, picks this. The ball reverses, so its velocity changes by v + 0.40v = 1.4v.
C1.0v/Δt A student who divides the impact velocity by the contact time, as if a = v/t, picks this. Average acceleration is the change in velocity, 1.4v, divided by Δt.
D0.3v/Δt A student who does not distinguish a velocity from a change in velocity uses the ball's average velocity during contact, (v + (−0.40v))/2 = 0.30v, in place of Δv and divides by Δt, picking this. Average acceleration uses the change in velocity, of magnitude 1.4v.
Working Take +x toward the wall. v0x = +v, vx = −0.40v. Δvx = −0.40v − v = −1.40v. |aavg| = 1.40v/Δt ≈ 1.4v/Δt.
Positions on a flat parking lot are measured from an origin with x- and y-axes. A car's speedometer reads 20 m/s throughout a 10 s interval. A student claims that the car accelerated during the interval. Which observation, if true, would support the student's claim?
Answer and reasoning
AThe car passed the origin, going from negative x to positive x. A student who reads the sign of the position as the direction of motion picks this. Crossing the origin changes the sign of x, not the velocity; a car can do it at constant velocity.
BThe car moved in the −x direction for the whole interval. A student who reads a negative velocity as slowing down picks this. Moving in the −x direction says only which way the car moves; at a steady 20 m/s in a fixed direction it is not accelerating.
CThe car followed a curved path during the interval.Correct With a constant speed of 20 m/s, the velocity can change only in direction. A curved path means the direction of motion changed, so the velocity changed and the car accelerated.
DThe car's odometer reading rose by 20 m in each second. A student who thinks any moving object is accelerating picks this. Covering 20 m every second is consistent with a constant velocity and zero acceleration.
Working Constant speed: acceleration only if the direction of the velocity changes. A curved path changes the direction.
The table shown gives the position x of a cart moving along a straight track at five instants. Which value is the best estimate of the cart's instantaneous velocity at t = 2.00 s?
Answer and reasoning
A4.00 m/s A student who divides the position at t = 2.00 s by the time picks this: 8.000 m/2.00 s. Velocity is a change in position over a time interval, here a small one around t = 2.00 s.
B13.0 m/s A student who thinks any interval around t = 2.00 s will do picks this, using the widest one, 1.00 s to 3.00 s: (27.00 − 1.000) m/(2.00 s) = 13.0 m/s. The cart's velocity changes, so a wide interval gives a poor estimate.
C8.00 m/s A student who reads the position value at t = 2.00 s as the velocity picks this. 8.000 m tells where the cart is, not how fast its position is changing.
D12.0 m/sCorrect An average velocity over a very small interval around t = 2.00 s is very close to the instantaneous velocity. Using 1.99 s to 2.01 s: (8.121 m − 7.881 m)/(0.02 s) = 12.0 m/s; the one-sided 0.01 s intervals give 12.1 and 11.9 m/s, all close to 12.0 m/s.
Working Use a very small interval around t = 2.00 s: 1.99 s to 2.01 s gives (8.121 m − 7.881 m)/(0.02 s) = 12.0 m/s (the one-sided intervals give 12.1 and 11.9 m/s).
A particle moves along the x-axis with position x(t) = At³ + C, where A and C are positive constants. What is the particle's velocity vx at time t = T?
Answer and reasoning
AAT² + C/T A student who divides the position by the time picks this: (AT³ + C)/T. Instantaneous velocity is the derivative dx/dt, not x/t.
BAT⁴/4 + CT A student who integrates the position function instead of differentiating it picks this. Velocity is the rate of change of position, dx/dt.
C3AT²Correct Differentiate: vx = dx/dt = 3At², since the derivative of the constant C is zero. At t = T, vx = 3AT².
D6AT A student who differentiates twice picks this: 6AT is the second derivative, the acceleration at t = T. The velocity is the first derivative, 3AT².
Working vx = dx/dt = d(At³ + C)/dt = 3At² (the derivative of the constant C is zero). At t = T: vx = 3AT².
A drone moving in a horizontal plane has position r⃗(t) = (1.5 m/s²)t² î + (2.0 m/s)t ĵ. What is the drone's speed at t = 2.0 s?
Answer and reasoning
A3.6 m/s A student who divides the position by the time picks this: r⃗(2.0 s)/(2.0 s) = (3.0î + 2.0ĵ) m/s, magnitude 3.6 m/s. The instantaneous velocity is the derivative dr⃗/dt.
B6.3 m/sCorrect Differentiate each component: v⃗ = dr⃗/dt = (3.0 m/s²)t î + (2.0 m/s)ĵ, which is (6.0î + 2.0ĵ) m/s at t = 2.0 s. The speed is √(6.0² + 2.0²) m/s ≈ 6.3 m/s.
C8.0 m/s A student who finds v⃗ = (6.0î + 2.0ĵ) m/s correctly but adds the components picks this. Perpendicular components combine as √(6.0² + 2.0²) m/s ≈ 6.3 m/s.
D7.2 m/s A student who reports the magnitude of the position, |r⃗(2.0 s)| = |(6.0î + 4.0ĵ) m| ≈ 7.2 m, as the speed picks this. Speed is the magnitude of dr⃗/dt.
Working v⃗ = dr⃗/dt = (3.0 m/s²)t î + (2.0 m/s)ĵ. At t = 2.0 s: v⃗ = (6.0î + 2.0ĵ) m/s. Speed = √(6.0² + 2.0²) m/s = √40 m/s ≈ 6.3 m/s.
An object moves along the x-axis with velocity vx(t) = kt², where k is a positive constant. What is the ratio ax(2t₁)/ax(t₁) of its accelerations at t = 2t₁ and t = t₁, where t₁ > 0?
Answer and reasoning
A2Correct ax = dvx/dt = 2kt, which is directly proportional to t. Doubling the time doubles the acceleration, so the ratio is 2.
B4 A student who lets the acceleration change with time as the velocity does, as t², picks this. The acceleration is the derivative of vx, 2kt, which grows only in proportion to t.
C8 A student who integrates vx instead of differentiating it picks this: kt³/3 gives a ratio of 2³ = 8. Acceleration is the derivative of velocity.
D1 A student who assumes the acceleration is constant picks this. vx = kt² is not linear in t, so ax = 2kt changes with time.
Working ax = dvx/dt = 2kt, which is proportional to t. ax(2t₁)/ax(t₁) = (2k·2t₁)/(2k·t₁) = 2.
An object moves along the x-axis with velocity vx(t) = (2.0 m/s³)t² − (2.0 m/s²)t. What is its acceleration ax at t = 2.0 s?
Answer and reasoning
A2.0 m/s² A student who divides the velocity at t = 2.0 s, 4.0 m/s, by the time, as if a = v/t, picks this. The acceleration is the derivative of vx at that instant.
B4.0 m/s² A student who takes the velocity value at t = 2.0 s, 4.0 m/s, as the acceleration picks this. Acceleration is the rate of change of velocity, dvx/dt.
D1.3 m/s² A student who integrates vx instead of differentiating it picks this: (2.0/3)(2.0)³ − (2.0/2)(2.0)² ≈ 1.3. Acceleration is the derivative of velocity.
Working ax = dvx/dt = (4.0 m/s³)t − 2.0 m/s². At t = 2.0 s: ax = 8.0 − 2.0 = 6.0 m/s².
An object moves along the x-axis with acceleration ax(t) = kt, where k is a positive constant. At t = 0 it is at x = 0 with velocity v0. What is its position x(t)?
Answer and reasoning
Akt³/6 A student who integrates without the constant from the initial velocity picks this: vx = kt²/2 and x = kt³/6. The object starts with velocity v0, which adds v0t to the position.
Bv₀t + kt³/6Correct Integrate twice, using the initial conditions: vx = v0 + ∫₀ᵗ kt′ dt′ = v0 + kt²/2, then x = 0 + ∫₀ᵗ (v0 + kt′²/2) dt′ = v0t + kt³/6.
Cv₀ + kt²/2 A student who stops at the velocity and reports it as the position picks this. v0 + kt²/2 is vx(t); it must be integrated once more, and it does not even have the unit of a position.
Dkt³ A student who uses v = at and x = vt with a = kt picks this: v = kt·t = kt² and x = kt²·t = kt³. Those products hold only for constant quantities starting from rest; here the acceleration changes and the object starts with velocity v0, so both steps must be integrated with initial conditions, giving v0t + kt³/6.
An object's velocity along the x-axis is known as a function of time, vx(t), for t ≥ 0. A student claims that this function alone is enough to determine the object's position x at t = 3.0 s. Which statement correctly evaluates the claim?
Answer and reasoning
AIt is right: integrating vx(t) from t = 0 to t = 3.0 s gives the object's position at t = 3.0 s. A student who leaves out the initial condition picks this. The integral gives the displacement; the position also depends on where the object was at t = 0.
BIt is wrong: integrating vx(t) gives x(t) only up to an added constant, so x0 is needed.Correct ∫₀3.0 s vx dt gives the displacement from t = 0 to t = 3.0 s. To get the position, a known position at one instant (for example x0 at t = 0) must be added: x = x0 + ∫vx dt.
CIt is wrong: x(t) is found by differentiating vx(t), and that also needs the object's acceleration. A student who reverses the calculus picks this. Differentiating vx(t) gives the acceleration; position comes from integrating vx(t).
DIt is right: the position at t = 3.0 s is vx at t = 3.0 s multiplied by the 3.0 s of motion. A student who uses x = vt with a single velocity picks this. The velocity changes with time, and the starting position is unknown, so neither the product nor vx(t) alone gives x.
Working x(t) = x0 + ∫₀ᵗ vx dt′. The integral gives only the displacement; x0 is not determined by vx(t).
A particle moves in the xy plane with position r⃗(t) = bt² î + ct⁴ ĵ for t ≥ 0, where b and c are positive constants. What is the particle's acceleration a⃗ at the instant t > 0 when the x- and y-components of its velocity are equal?
Answer and reasoning
Ab(2î + 6ĵ)Correct Differentiate each component: v⃗ = 2bt î + 4ct³ ĵ. The components are equal when 2bt = 4ct³, so t² = b/(2c). Differentiate again: a⃗ = 2b î + 12ct² ĵ, and at that instant 12ct² = 6b, so a⃗ = b(2î + 6ĵ).
Bb(2î + 12ĵ) A student who does not distinguish position from velocity sets the position components equal, bt² = ct⁴, so t² = b/c, and picks this: a⃗ = 2b î + 12ct² ĵ = b(2î + 12ĵ). The condition is on the velocity components, 2bt = 4ct³.
Cb(î + ĵ) A student who takes velocity as position divided by time and acceleration as velocity divided by time, v⃗ = r⃗/t and a⃗ = r⃗/t², picks this: the 'velocity' components bt and ct³ are equal at t² = b/c, where r⃗/t² = b î + ct² ĵ = b(î + ĵ). Velocity and acceleration are derivatives: v⃗ = dr⃗/dt and a⃗ = dv⃗/dt.
Db(î + 3ĵ) A student who takes the velocity to be the position divided by the time, v⃗ = r⃗/t = bt î + ct³ ĵ, and then differentiates it picks this: those components are equal at t² = b/c, and d(r⃗/t)/dt = b î + 3ct² ĵ = b(î + 3ĵ). The velocity is dr⃗/dt, not r⃗/t.
Working v⃗ = dr⃗/dt = 2bt î + 4ct³ ĵ. vx = vy: 2bt = 4ct³, so t² = b/(2c). a⃗ = dv⃗/dt = 2b î + 12ct² ĵ = 2b î + 12c(b/(2c)) ĵ = 2b î + 6b ĵ = b(2î + 6ĵ). Distractors: equal position components, bt² = ct⁴ → t² = b/c → a⃗ = b(2î + 12ĵ); v⃗ = r⃗/t equal at t² = b/c, with a⃗ = r⃗/t² = b(î + ĵ) or a⃗ = d(r⃗/t)/dt = b î + 3ct² ĵ = b(î + 3ĵ). Check (b = 1, c = 1): a⃗ = 2î + 6ĵ; distractors 2î + 12ĵ, î + ĵ, î + 3ĵ.
Compiled from the AP Physics C: Mechanics 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
