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FRQ: Average Velocity over a Round Trip — 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.

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

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

On Figure 1, draw two arrows to represent the robot’s velocity on each leg of its trip.

•   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.

•   The relative lengths of the two arrows should indicate the relative magnitudes of the two constant velocities. Arrow lengths are not measured against the position scale on the figure, and no numerical value is required.

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

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

Part B

Derive an expression for the average velocity of the robot for the entire trip, from x = 0 to the docking bay. 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 t = 0 to t = 3t_0. On your graph, label the position of the robot at t = t_0 and at t = 3t_0 with algebraic expressions, and label the times t_0 and 3t_0 on the time axis.

Figure 2 — Axes for part C. Position is measured from the origin of the aisle scale.

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.

Indicate whether the magnitude of the second robot’s average velocity for the entire trip is greater than, less than, or equal to that of the first robot 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 follows from 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.

The outbound leg covers D in a time t_0; the return leg covers 3D/4 in 2t_0. Less ground in twice the time makes the return the slower leg, so its arrow is drawn about three-eighths as long, 3D/(8t_0) against D/t_0. A velocity arrow’s length on a position axis says nothing about how far the robot went; its tail marks where that leg begins, and only its direction and its length relative to the other arrow carry the velocity information.

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

Scoring (3 points):

▸ A1 — 1 point: 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: 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: The return arrow drawn clearly shorter than the outbound arrow, no more than about two-thirds of its length. The return speed is three-eighths of the outbound speed; no numerical value is required.

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

⚠︎ 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 A1 and A3, 2 of 3 points.

✗ Common error (no credit): A single arrow drawn from x = 0 to x = +D/4 and labelled as the robot’s average velocity — A1’s label condition is not met, no arrow has its tail at x = +D, and there is no second arrow to compare against, 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; the excursion out to x = +D does not enter it:

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

The average velocity for the entire trip is D/(12t_0), in the +x direction. The robot’s average velocity is one twelfth of D per t_0 even though its speed was never less than 3D/(8t_0) on either leg — the reversal cancels most of the ground it covered.

Scoring (4 points):

▸ B1 — 1 point: Begins from the definition of average velocity, or an equivalent starting relationship from the reference information. Awarded for the correct starting relationship alone, before any substitution. Accept v_avg = Δx/Δt, Δx = v_avg Δt, or x = x_0 + v_avg t.

▸ B2 — 1 point: Correct displacement, Δx = +D/4, identified from the initial and final positions rather than from the path travelled.

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

▸ B4 — 1 point: Correct final expression, v_avg = D/(12t_0).

Scoring Note: A correct, isolated, final expression v_avg = D/(12t_0) earns points B2, B3 and B4.

⚠︎ Common error (partial credit): Uses the path length 7D/4 in place of the displacement, giving 7D/(12t_0) — earns B1 and B3, 2 of 4 points.

✗ Common error (no credit): Averages the two leg velocities, ½[D/t_0 + (−3D/(8t_0))] = 5D/(16t_0) — no criterion is earned. The mean of the leg velocities equals the average velocity only when the legs occupy equal time intervals, and here the return leg takes twice as long as the outbound leg.

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 a corner at t = t_0.

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 therefore negative and smaller in magnitude, so that segment is visibly shallower: three-quarters of the ground in twice the outbound interval. The graph finishes at x = +D/4 and never reaches the time axis; label D at t = t_0 and D/4 at t = 3t_0. A corner on a position–time graph is a change of velocity, not a stop: the robot has no single velocity at t = t_0.

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

Scoring (3 points):

▸ C1 — 1 point: A straight segment from the origin to the point (t_0, D), with constant positive slope.

▸ C2 — 1 point: A straight segment from (t_0, D) to (3t_0, D/4), with constant negative slope, terminating above the time axis and spanning twice the horizontal extent of the outbound segment.

▸ C3 — 1 point: The positions at t = t_0 and at t = 3t_0 labelled on the graph with the algebraic expressions D and D/4, and the two times labelled t_0 and 3t_0 on the time axis.

⚠︎ Common error (partial credit): Return segment drawn down to the time axis at t = 3t_0 — earns C1, 1 of 3 points. The final position is wrong, so C2 is not earned and the position at t = 3t_0 cannot be labelled D/4 on the curve drawn.

⚠︎ Common error (partial credit): Both positions labelled correctly but the two legs drawn as curves through them — earns C3, 1 of 3 points. Curvature on a position–time graph asserts an acceleration the scenario excludes.

✗ 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 and neither position label applies to it, 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. Carrying the same relationship forward from part B,

v_avg = 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. The second robot’s slower return changes nothing about where it ends up, only about how long it took to get there — which is the whole of the difference.

Scoring (2 points):

▸ D1 — 1 point: Indicates |v_avg,2| < |v_avg,1|.

▸ D2 — 1 point: Justification that identifies the displacement as unchanged at +D/4 and the total time interval as increased to t_0 + T > 3t_0, and applies the inverse dependence of v_avg on Δt at fixed Δx. Consistent with the expression derived in part B.

⚠︎ Common error (partial credit): Correct indication justified only by “it takes longer, so it is slower”, with the displacement never identified as unchanged — earns D1, 1 of 2 points.

✗ Common error (no credit): Indicates the magnitudes are equal 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.