FRQ: Constant Acceleration — The Two-Phase Rocket Sled

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Qualitative/Quantitative Translation (QQT)  |  MID-LEVEL  |  8 points

▤ Scenario

A rocket sled is tested on a straight, level track. The sled starts from rest at the origin at time t = 0, and its motion has two phases.

Phase 1 (0 ≤ t ≤ 4.0 s): the rocket engine fires and gives the sled a constant acceleration a_1 = +5.0 m/s².

Phase 2 (t > 4.0 s): the engine shuts off and the brakes are applied, giving the sled a constant acceleration a_2 = −2.5 m/s² until the sled comes to rest. The brakes then hold the sled in place, so it stays at rest and does not move again.

All motion is one-dimensional along the track. Take the sled’s direction of travel as the positive x-direction, with the origin at the sled’s starting point and the track as the reference frame. In one dimension the sign of a component completely describes its direction along that axis.

✎ Free Response Questions

(a) Starting from a relationship on the AP Physics 1 equation sheet, derive an expression for the velocity of the sled at the end of Phase 1 in terms of a_1 and the Phase 1 duration t_1. Then calculate its numerical value.

(b) Calculate the displacement of the sled from its starting position to the point where it comes to rest.

(c) On the axes provided below, sketch a velocity–time graph for the sled’s entire journey, from t = 0 until it comes to rest. Label the maximum velocity with its value and unit, the time at which that maximum occurs, and the time at which the sled stops.

Figure 1 — Axes for part (c). Time runs from 0 to 14 s; velocity runs from 0 to 25 m/s.

(d) A student says: “Since the braking acceleration has half the magnitude of the engine acceleration, the sled must travel twice as far while braking as it does while speeding up.”

In a clear, coherent paragraph-length response, evaluate the student’s claim. State whether the conclusion is correct, support your answer with quantitative evidence from your calculations or from the graph in part (c), and state the condition on the sled’s motion that the student’s reasoning depends on.

❖ Answer Key & Scoring Guide

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

Part (a) — Model Answer

The sled starts from rest, so v_x0 = 0. The equation-sheet relationship for velocity under constant acceleration is

v_x = v_x0 + a_x t

Applying it to Phase 1, with v_x0 = 0, a_x = a_1 and t = t_1:

v_x = 0 + a_1 t_1 = a_1 t_1

Substituting the given values:

v_x = (+5.0 m/s²)(4.0 s) = +20 m/s

At the end of Phase 1 the sled is moving at 20 m/s in the positive x-direction.

Scoring (2 points):

▸ 1 point: Starts from the equation-sheet relationship v_x = v_x0 + a_x t and, using v_x0 = 0, reaches the symbolic result v_x = a_1 t_1. This point is awarded for the starting relationship together with the symbolic expression, before any numbers are substituted.

▸ 1 point: Substitutes correctly to obtain v_x = +20 m/s, with the unit and the direction (or sign) stated.

⚠︎ Common error (partial credit): correct work throughout but the answer is reported as a bare speed, “20 m/s”, with no sign or direction — earns 1 of 2 points. Velocity is a vector, and in one dimension the sign is what carries its direction.

✗ Common error (no credit): substitutes into v_x² = v_x0² + 2a_xΔx using Δx = 4.0, treating the 4.0 s duration as a displacement. Time and displacement are different quantities and cannot be interchanged inside a kinematic relationship.

Part (b) — Model Answer

Phase 1 starts from rest at the origin, so use x = x_0 + v_x0 t + ½ a_x t²:

Δx_1 = v_x0 t_1 + ½ a_1 t_1² = 0 + ½(+5.0 m/s²)(4.0 s)² = +40 m

Phase 2 begins at v_x0 = +20 m/s and ends at v_x = 0, and the displacement is wanted rather than the time, so use v_x² = v_x0² + 2a_xΔx:

0 = (+20 m/s)² + 2(−2.5 m/s²)Δx_2

Δx_2 = −(400 m²/s²) ÷ (−5.0 m/s²) = +80 m

Both phases carry the sled in the same direction, so the two displacements add:

Δx = Δx_1 + Δx_2 = +40 m + 80 m = +120 m

The sled comes to rest 120 m from its starting position, in the positive x-direction.

Scoring (2 points):

▸ 1 point: Obtains both Δx_1 = +40 m (Phase 1, from v_x0 = 0) and Δx_2 = +80 m (Phase 2, carrying a_2 as a negative component), each from a correct kinematic relationship.

▸ 1 point: Adds the two signed displacements to Δx = +120 m and states the unit and the direction.

⚠︎ Common error (partial credit): obtains +40 m and +80 m correctly but reports the total as 40 m − 80 m = −40 m, subtracting because a_2 is negative — earns 1 of 2 points (the phase-calculation point only). A negative acceleration does not make the displacement negative; the sled keeps moving in the +x direction throughout Phase 2.

⚠︎ Common error (partial credit): correct method for one phase with an arithmetic slip in the other — the phase-calculation point is lost, but the addition point is still earned if the total is consistent with the student’s own (incorrect) phase values.

✗ Common error (no credit): uses Δx_2 = ½ a_2 t_2² with v_x0 = 0 for Phase 2, treating the braking phase as though it also started from rest. Phase 2 begins at +20 m/s, and dropping that term discards the whole of the sled’s motion into the phase — no credit for the phase-calculation point.

Part (c) — Model Answer

The graph is two straight segments, because the acceleration is constant within each phase and the slope of a velocity–time graph is the acceleration. From t = 0 to t = 4.0 s the velocity rises in a straight line from 0 to +20 m/s. From t = 4.0 s it falls in a straight line back to 0, at half the slope magnitude, so this segment is shallower and twice as long. The Phase 2 duration follows from v_x = v_x0 + a_x t:

t_2 = (0 − 20 m/s) ÷ (−2.5 m/s²) = 8.0 s

so the sled stops at t = 4.0 s + 8.0 s = 12.0 s.

Figure 2 — The completed velocity–time graph. The peak is +20 m/s at t = 4.0 s and the sled reaches v_x = 0 at t = 12.0 s.

Scoring (2 points):

▸ 1 point: Two connected straight segments — a steeper positive slope from the origin up to the peak, then a shallower negative slope back down to the time axis — with both axes labelled with quantity and unit.

▸ 1 point: Key values marked: maximum velocity +20 m/s at t = 4.0 s, and v_x = 0 at t = 12.0 s.

⚠︎ Common error (partial credit): correct shape with the peak in the right place but the braking segment ending at t = 8.0 s — earns 1 of 2 points. 8.0 s is the duration of Phase 2, not the clock time at which it ends.

✗ Common error (no credit): draws either phase as a curve, or continues the line below the time axis after t = 12.0 s. Constant acceleration gives a straight line on a velocity–time graph, and the brakes hold the sled at rest once it stops.

Part (d) — Model Answer

The student’s conclusion is correct, and the calculations confirm it. The sled travels +40 m in Phase 1 and +80 m in Phase 2, which is exactly twice as far. The same result is readable as area on the velocity–time graph: the accelerating triangle has area ½(4.0 s)(20 m/s) = 40 m and the braking triangle ½(8.0 s)(20 m/s) = 80 m.

The reason is not that displacement is inversely proportional to acceleration in general. Both phases change the sled’s speed between the same two values, 0 and 20 m/s, so both satisfy v_x² = v_x0² + 2a_xΔx with the same magnitude of change in v_x². Rearranging gives Δx = v² ÷ (2|a|), so with v² fixed, halving |a| doubles Δx. On the graph the two triangles have the same height, and the braking triangle needs twice the base because it removes the same 20 m/s at half the rate.

Figure 3 — The two areas that give the displacements: Δx_1 = +40 m and Δx_2 = +80 m. Both triangles have the same height, 20 m/s; the braking triangle has twice the base.

That shared speed change is the condition the student’s reasoning depends on. It holds here only because Phase 1 starts from rest and Phase 2 ends at rest. Had the sled entered Phase 1 already moving, or had the brakes been released before it stopped, the two phases would not span the same change in v_x², and the displacements would not stand in the inverse ratio of the acceleration magnitudes. The conclusion is right for this journey, but it is not a general rule about accelerations.

Scoring (2 points):

▸ 1 point: States that the conclusion is correct, supports it with quantitative evidence — the displacements +40 m and +80 m from part (b), or the two triangular areas from the graph in part (c) — and gives the physical reason: both phases change the speed between the same two values, so Δx = v² ÷ (2|a|) with v² fixed, and halving |a| doubles Δx (the equivalent graph argument — same triangle height, so half the slope needs twice the base — also earns this point).

▸ 1 point: States the condition the reasoning depends on: Phase 1 starts from rest and Phase 2 ends at rest, so the two phases span the same change in speed.

⚠︎ Common error (partial credit): states the conclusion is correct and quotes 40 m and 80 m but gives no physical reason connecting them — bare verdict-plus-arithmetic does not earn the first point. A paragraph-length argument has to connect the quantitative result to a physical relationship.

⚠︎ Common error (partial credit): correct conclusion and physical reason (first point earned), but never names the shared speed change, leaving “half the acceleration gives twice the distance” standing as a general rule — the second point is not earned.

✗ Common error (no credit): rejects the claim on the grounds that acceleration and displacement are “not proportional”, without testing it against the calculated values. Evaluating a claim means checking it against the evidence, and here the evidence supports the student.