Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.

FRQ: Average Velocity with a Reversal of Direction — The Warehouse Inventory Robot

Assessments aligned to 2026 AP Physics 1 standards

Translation Between Representations (TBR)  |  MID-LEVEL  |  12 points  |  30 min

▤ Scenario

A warehouse inventory robot moves along a straight, level aisle marked with a position scale. Beginning at the origin of that scale, x = 0, the robot travels in the +x direction at constant velocity and reaches the far scanning station at x = +D after a time interval t_0.

At the scanning station the robot reverses direction in a negligibly short time and returns along the same aisle at a new constant velocity, arriving at the docking bay at x = +D/4 after an additional time interval 2t_0. Figure 1 shows the aisle, with the origin, the docking bay and the scanning station marked.

Figure 1 — The aisle, with the origin, the docking bay and the scanning station marked.

All motion is one-dimensional along the x-axis, with +x pointing from the origin toward the scanning station. Positions are measured in the reference frame of the aisle floor. Treat the robot as a point object and each change of velocity as instantaneous.

✎ Free Response Questions

Part A

On Figure 1, draw an arrow for each leg of the robot’s trip to represent the robot’s velocity on that leg.

•   Draw one arrow for the outbound leg, with its tail at x = 0.

•   Draw one arrow for the return leg, with its tail at x = +D.

•   Draw the length of each arrow to represent the magnitude of that leg’s velocity, relative to the other arrow.

•   Label each arrow with the direction of that leg along the x-axis.

Part B

Derive an expression for the average velocity v_avg,1 of the robot for the entire trip, from the instant it leaves x = 0 until it arrives at the docking bay at the end of its return. Express your final answer in terms of D, t_0, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Part C

On the axes shown in Figure 2, sketch a graph of the robot’s position x as a function of time t for the entire trip, from the instant it leaves x = 0 until it arrives at the docking bay.

Figure 2 — Axes for part C, with positions in fractions of D and times in multiples of t_0.

Part D

A second robot starts at x = 0 and reaches the scanning station at x = +D in the same time interval t_0, but takes a time T, where T > 2t_0, to return to the docking bay at x = +D/4. Let v_avg,2 be the second robot’s average velocity for its entire trip, defined as in part B.

Indicate whether |⁠v_avg,2| is greater than, less than, or equal to |⁠v_avg,1| by writing one of the following.

•   |⁠v_avg,2| > |⁠v_avg,1|

•   |⁠v_avg,2| < |⁠v_avg,1|

•   |⁠v_avg,2| = |⁠v_avg,1|

Justify how your response is consistent with the expression you derived in part B.

❖ Answer Key & Scoring Guide

▸ earns credit  ⚠︎ common error, partial credit  ✗ common error, no credit  

Points are independent and there are no deductions. ▸ criteria follow College Board’s published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.

Part A — Model Answer

Each arrow represents the robot’s velocity on one leg, not its displacement: the outbound arrow has its tail at x = 0 and points in the +x direction, and the return arrow has its tail at x = +D and points in the −x direction, each labelled with its direction.

Figure 3 — The two velocity arrows. Lengths show relative speed, not distance travelled.

The outbound leg covers D in a time t_0, a speed of D/t_0; the return leg covers 3D/4 in 2t_0, a speed of 3D/(8t_0). Less ground in twice the time makes the return the slower leg, so its arrow is drawn three-eighths as long as the outbound arrow. A velocity arrow’s length shows speed, not a distance on the position scale. Its tail marks where that leg begins, and only its direction and its length relative to the other arrow carry the velocity information.

Scoring (3 points):

▸ A1 — 1 point: For an arrow with its tail at x = 0, pointing in the +x direction, labelled with that direction or as the outbound leg’s velocity.

▸ A2 — 1 point: For an arrow with its tail at x = +D, pointing in the −x direction, labelled with that direction or as the return leg’s velocity.

▸ A3 — 1 point: For a return arrow more than one-quarter and less than one-half as long as the outbound arrow.

Scoring Note: The speeds are in the ratio 3 : 8, and no numerical value is required. Ratios of 3 : 4 (the leg lengths) or 1 : 8 (the docking-bay position taken as the return distance) do not meet A3. A1, A2 and A3 are scored independently of one another.

⚠︎ Common error (partial credit): Both arrows drawn the same length — earns 2 of 3 points, A1 and A2. The relative lengths are what carry the comparison.

⚠︎ Common error (partial credit): Draws each arrow from where its leg begins to where it ends, x = 0 to x = +D and x = +D to x = +D/4, so the lengths are in the ratio of the distances, 3 : 4, rather than of the speeds, 3 : 8 — earns 2 of 3 points, A1 and A2. These are displacement arrows, not velocity arrows.

⚠︎ Common error (partial credit): Both arrows drawn with their tails at the origin, the return arrow starting at x = 0 rather than at x = +D, with the relative lengths correct — earns 2 of 3 points, A1 and A3.

✗ Common error (no credit): A single arrow drawn from x = 0 to x = +D/4, labelled only as the robot’s average velocity, with no direction — one whole-trip arrow replaces the two leg arrows: with no direction label A1 is not met, with no arrow at x = +D A2 is not met, and with no second arrow there is no return arrow for A3, so no criterion is earned.

Part B — Model Answer

Average velocity is the displacement divided by the time interval over which that displacement occurs:

v_avg = Δx / Δt

The displacement is fixed by the initial and final positions alone, so the stretch from x = +D/4 out to x = +D and back cancels out of it:

Δx = x_f − x_i = D/4 − 0 = +D/4
Δt = t_0 + 2t_0 = 3t_0
v_avg,1 = (D/4) / (3t_0) = D / (12t_0)

The average velocity for the entire trip is D/(12t_0), in the +x direction. The return leg undoes three-quarters of the outbound displacement, so the average velocity is far smaller than the speed on either leg.

Scoring (4 points):

▸ B1 — 1 point: For a multistep derivation that includes the definition of average velocity, v_avg = Δx/Δt, or an equivalent relationship — Δx = v_avg Δt, x = x_0 + v_avg t, or the time-weighted mean of the leg velocities. This point is earned even if the steps that follow are incorrect.

▸ B2 — 1 point: For the displacement for the whole trip, Δx = +D/4, from the initial and final positions or as the sum of the leg displacements, (+D) + (−3D/4).

▸ B3 — 1 point: For the total time interval, Δt = 3t_0.

▸ B4 — 1 point: For a correct expression for v_avg,1, such as D/(12t_0).

Scoring Note: A correct, isolated, final expression for v_avg,1 earns points B2, B3 and B4. B1–B4 do not depend on the response in part A.

⚠︎ Common error (partial credit): Writes v_avg = Δx/Δt with Δt = 3t_0 but substitutes the path length, 7D/4, for Δx, giving 7D/(12t_0) — earns 2 of 4 points, B1 and B3. That is the average speed; average velocity uses the displacement.

⚠︎ Common error (partial credit): Writes v_avg = Δx/Δt with Δt = 3t_0 but takes the displacement as zero, as though the robot had returned to x = 0, and gives v_avg,1 = 0 — earns 2 of 4 points, B1 and B3. The robot ends at x = +D/4, not at its starting point.

⚠︎ Common error (partial credit): Finds each leg’s velocity from v = Δx/Δt, then takes their plain mean, ½[D/t_0 + (−3D/(8t_0))] = 5D/(16t_0) — earns 1 of 4 points, B1. The mean of the leg velocities equals the average velocity only when the legs last equally long; here the return leg lasts twice as long.

Part C — Model Answer

Constant velocity appears on a position–time graph as a straight line whose slope is that velocity, so the trip is two straight segments meeting at t = t_0, where the robot reverses.

Figure 4 — The completed position–time graph for the whole trip.

outbound slope = (D − 0) / (t_0 − 0) = +D / t_0
return slope = (D/4 − D) / (3t_0 − t_0) = −3D / (8t_0)

The return slope is negative and smaller in magnitude, so that segment is less steep: three-quarters of the ground in twice the outbound interval. The graph ends at (3t_0, D/4), above the time axis, because the robot finishes at x = +D/4, a quarter of the way from the origin to the scanning station, not back at x = 0.

A sharp corner on a position–time graph is an idealised instantaneous change of velocity, not a pause. A pause would show as a flat segment. In the real turnaround the velocity passes through zero, so the robot is momentarily at rest at x = +D.

Scoring (3 points):

▸ C1 — 1 point: For a straight line from the origin to the point (t_0, D).

▸ C2 — 1 point: For a straight line with negative slope continuing from the end of the first line, less steep than the first line.

Scoring Note: A sharp corner at t = t_0, or a turn rounded within about one-tenth of a grid division either side of it, is acceptable for C1 and C2.

▸ C3 — 1 point: For ending the second line at (3t_0, D/4), above the time axis.

Scoring Note: C1–C3 do not depend on the responses in parts A and B.

⚠︎ Common error (partial credit): Draws the return line down to the time axis at t = 3t_0, as though the robot came back to x = 0 — earns 2 of 3 points, C1 and C2. The robot ends at x = +D/4.

⚠︎ Common error (partial credit): Draws both legs as curves through the correct corner and end points — earns 1 of 3 points, C3. Curvature during a leg asserts an acceleration that the scenario’s constant velocities exclude.

✗ Common error (no credit): Sketches velocity as a function of time, a positive constant step followed by a negative constant step — the representation asked for was not produced, so no criterion is earned.

Part D — Model Answer

|⁠v_avg,2| < |⁠v_avg,1|

Both robots start at x = 0 and finish at x = +D/4, so both have the same displacement, Δx = +D/4. Only the total time interval differs. Applying the relationship used in part B to the second robot,

v_avg,2 = Δx / Δt = (D/4) / (t_0 + T)

and because T > 2t_0, the total time satisfies t_0 + T > 3t_0. The numerator is unchanged while the denominator is larger, and at fixed displacement the average velocity varies inversely with the time interval, so |⁠v_avg,2| is less than D/(12t_0). How the second robot makes its return — steadily, with a pause, or at varying speed — does not enter; only the total time does. The second robot’s longer return changes nothing about where it ends up, only how long the trip takes.

Scoring (2 points):

▸ D1 — 1 point: For indicating the relation that follows from the response’s expression in part B applied to the second robot: |⁠v_avg,2| < |⁠v_avg,1| if that expression decreases as the total time interval increases, as (D/4)/Δt does; = if it does not depend on the time; > if it increases; and |⁠v_avg,2| < |⁠v_avg,1| if the response gives no expression in part B.

▸ D2 — 1 point: For a justification, consistent with the response’s expression in part B (or with v_avg = Δx/Δt if part B gives none), that correctly relates each robot’s displacement and total time interval to the magnitude of its average velocity — for the correct expression, equal displacements and a longer total time, t_0 + T > 3t_0, give a smaller average velocity; for v_avg,1 = 0, equal zero displacements give equal average velocities for any times.

Scoring Note: A path length used in place of the displacement, or a plain mean of the leg velocities, does not relate them correctly.

⚠︎ Common error (partial credit): Having derived D/(12t_0) in part B, indicates |⁠v_avg,2| < |⁠v_avg,1| and justifies it only by “it takes longer, so it is slower”, never saying that the displacement is unchanged — earns 1 of 2 points, D1.

⚠︎ Common error (partial credit): Having taken the plain mean of the leg velocities in part B, indicates |⁠v_avg,2| > |⁠v_avg,1| because the second robot’s return velocity is smaller in magnitude — earns 1 of 2 points, D1. The plain mean ignores how long each leg lasts.

✗ Common error (no credit): Having derived D/(12t_0) in part B, indicates |⁠v_avg,2| = |⁠v_avg,1| because both robots start and finish at the same positions — the displacement is fixed but the time interval is not, so neither criterion is earned.

Keep going:

Equations & Traps deck · Mind Maps bundle