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FRQ: Constant Acceleration — The Two-Phase Rocket Sled

Assessments aligned to 2026 AP Physics 1 standards

Mathematical Routines (MR)  |  MID-LEVEL  |  10 points  |  25 min

▤ 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.

Figure 1 — The sled’s two phases. Δx_1 and Δx_2 are the Phase 1 and Phase 2 displacements.

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

Part A

i.    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, and then state its numerical value and direction for the values given in the scenario. Express your final answer in terms of a_1, t_1, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

ii.   Calculate the sled’s Phase 1 displacement, its Phase 2 displacement, and the displacement of the sled from its starting position to the point where it comes to rest.

iii.  Figure 2 shows axes for a velocity–time graph of the sled’s motion. On these axes, sketch the velocity of the sled from t = 0 to t = 14 s. Label the maximum velocity with its value and unit, the time at which that maximum occurs, and the time at which the sled comes to rest.

Figure 2 — Axes for part A (iii). Time runs from 0 to 14 s and velocity from 0 to 25 m/s.

Part B

The first sled is removed from the track, and a second rocket sled is tested on the same track. This second sled is already moving at v_0 = +10 m/s when its engine fires. Its engine then gives it the same constant acceleration a_1 = +5.0 m/s² for the same Phase 1 duration t_1 = 4.0 s, after which its brakes give it the same constant acceleration a_2 = −2.5 m/s² until it comes to rest.

A student claims that this sled travels exactly twice as far while braking as it does while speeding up, because the braking acceleration has half the magnitude of the engine acceleration.

Let Δx_1′ and Δx_2′ be the second sled’s Phase 1 and Phase 2 displacements. Calculate Δx_1′ and Δx_2′.

Then indicate whether Δx_2′ is greater than, less than, or equal to twice Δx_1′ by writing one of the following.

•   Δx_2′ > 2Δx_1′

•   Δx_2′ < 2Δx_1′

•   Δx_2′ = 2Δx_1′

Justify your answer. In your justification, include qualitative reasoning beyond mathematical derivations or expressions.

❖ 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 (i) — Model Answer

The sled starts from rest, so v_x0 = 0. The relationship for velocity under constant acceleration from the reference information 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, and then substituting the given values:

v_x = 0 + a_1 t_1 = a_1 t_1
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. Because the sled starts from rest, the velocity the engine builds is the whole of the velocity the sled reaches.

Scoring (3 points):

▸ A1 — 1 point: For beginning the derivation from a fundamental physics principle or a relationship from the reference information — here v_x = v_x0 + a_x t. This point is awarded for the starting relationship alone, before any algebra is checked.

▸ A2 — 1 point: For reaching the symbolic result v_x = a_1 t_1, using v_x0 = 0. This point does not depend on any earlier result.

Scoring Note: A correct, isolated, final expression v_x = a_1 t_1 earns point A2, even where the starting relationship is not written out separately.

▸ A3 — 1 point: For evaluating that expression to v_x = +20 m/s, with the unit and the direction or sign stated. This point may be earned with a value consistent with a correct expression for the end-of-Phase-1 velocity obtained in A2.

⚠︎ Common error (partial credit): correct work throughout but the answer reported as a bare speed, “20 m/s”, with no sign or direction — earns 2 of 3 points (A1 and A2). Velocity is a vector, and in one dimension the sign is what carries its direction.

⚠︎ Common error (partial credit): writes v_x² = v_x0² + 2a_xΔx and substitutes Δx = 4.0, treating the 4.0 s duration as a displacement — earns 1 of 3 points (A1 only). The starting relationship is from the reference information, so A1 stands; but it relates velocity to displacement, not to time, so no expression in a_1 and t_1 follows and A2 and A3 are lost.

✗ Common error (no credit): starts from a relationship that is neither a fundamental principle nor on the reference information — for example v_x = a_1 ÷ t_1, giving 1.25 m/s. A1 is lost because the starting relationship must be one the reference information supplies, A2 because no expression of the correct form follows, and A3 because there is no correct expression for the value to be consistent with.

Part A (ii) — Model Answer

Phase 1 starts from rest at the origin, so use x = x_0 + v_x0 t + ½ a_x t². Phase 2 begins at +20 m/s and ends at rest, and the displacement is wanted rather than the time, so use v_x² = v_x0² + 2a_xΔx:

Δx_1 = v_x0 t_1 + ½ a_1 t_1² = 0 + ½(+5.0 m/s²)(4.0 s)² = +40 m
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 = (+40 m) + (+80 m) = +120 m

The sled comes to rest 120 m from its starting position, in the positive x-direction. A negative acceleration component does not reverse the displacement — it only means the velocity is being reduced.

Scoring (2 points):

▸ A4 — 1 point: For obtaining both phase displacements, Δx_1 = +40 m and Δx_2 = +80 m, by any correct route — including reading the two areas from a velocity–time graph, or working from the magnitude of a_2 and the distance needed to stop. The Phase 2 value may be consistent with the velocity the student found in part A (i).

▸ A5 — 1 point: For combining the two phase displacements into a total of 120 m in the +x direction, with the unit and the direction or sign stated. This point may be earned with a total consistent with the student’s own Phase 1 and Phase 2 values.

⚠︎ 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 (A4 only). The sled keeps moving in the +x direction throughout Phase 2.

⚠︎ Common error (partial credit): uses v_x² = v_x0² + 2a_xΔx for Phase 2 but takes v_x0 = 0, giving Δx_2 = 0 and a total of +40 m — earns 1 of 2 points (A5 only). Phase 2 begins at the velocity Phase 1 ended with; dropping that term discards the whole of the sled’s motion into the phase. A5 stands because the student’s own two values are added consistently.

✗ Common error (no credit): applies Δx = ½(v_x0 + v_x)t across the whole journey, giving ½(0 + 0)(12.0 s) = 0. That relationship holds only while the acceleration is constant, and the acceleration changes at t = 4.0 s. A4 is lost because neither phase displacement is obtained, and A5 because there is nothing to add.

Part A (iii) — 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 zero at half the slope magnitude, so that 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

Figure 3 — The completed velocity–time graph: peak +20 m/s at t = 4.0 s, at rest from t = 12.0 s.

So the sled comes to rest at t = 4.0 s + 8.0 s = 12.0 s, and stays at rest thereafter. Halving the magnitude of the acceleration halves the slope, which is why the braking segment takes twice as long as the accelerating one.

Scoring (2 points):

▸ A6 — 1 point: For two connected straight segments — a steeper positive slope from the origin up to the peak, which is marked at t = 4.0 s, then a shallower negative slope down to the time axis — and the graph held on the axis after the sled stops. This point does not depend on any earlier result.

▸ A7 — 1 point: For labelling the maximum velocity with its value and unit, +20 m/s, and the time at which the sled comes to rest, t = 12.0 s. The maximum may be consistent with the velocity found in part A (i), and the stopping time consistent with that maximum.

⚠︎ Common error (partial credit): draws the braking segment as a curve flattening toward the axis, on the intuition that the sled eases to a stop — earns 1 of 2 points (A7 only, where both labelled values are correct). Constant acceleration is a constant slope, so the segment is straight.

✗ Common error (no credit): brings the braking segment down to the time axis at t = 8.0 s, reading 8.0 s as a clock time rather than as the duration of Phase 2. A6 is lost because that segment falls at 5.0 m/s² rather than 2.5 m/s², so it is not the shallower slope; and A7 is lost because the stopping time is 12.0 s, not 8.0 s.

✗ Common error (no credit): sketches position against time instead — a curve starting flat, steepening through Phase 1, then bending over to horizontal at t = 12.0 s — and labels its maximum as a distance in metres. A6 is lost because neither segment is a straight line of the required slope, and A7 because the labelled maximum is a position, not a velocity.

Part B — Model Answer

The second sled does not start from rest, so its two phases do not span the same change in v_x², and the claim fails.

v_x = +10 m/s + (+5.0 m/s²)(4.0 s) = +30 m/s
Δx_1′ = (+10 m/s)(4.0 s) + ½(+5.0 m/s²)(4.0 s)² = (+40 m) + (+40 m) = +80 m
0 = (+30 m/s)² + 2(−2.5 m/s²)Δx_2′, so Δx_2′ = +180 m

Twice the Phase 1 displacement is 2Δx_1′ = +160 m, and 180 m is greater than that, so Δx_2′ > 2Δx_1′. The claim is not correct for this sled.

Rearranging v_x² = v_x0² + 2a_xΔx gives Δx = Δ(v_x²) ÷ (2a_x), so halving the magnitude of the acceleration doubles the displacement only if Δ(v_x²) is the same in both phases. Here Phase 1 raises v_x² from 100 m²/s² to 900 m²/s², a change of 800 m²/s², while Phase 2 has to remove the whole 900 m²/s². Phase 2 has more speed to shed than Phase 1 built, so it takes more than twice the distance.

The same argument reads off the graph. For the first sled both phases are triangles of the same height, so the braking one at half the slope needs exactly twice the base. For the second sled the accelerating region is a trapezoid standing on a 4.0 s base, while the braking region is still a full triangle dropping from the peak.

Figure 4 — Both sleds. Two triangles of equal height give the 1 : 2 ratio; a trapezoid does not.

The inverse relation between displacement and acceleration holds only between phases spanning the same change in v_x², which is why the first sled — starting from rest and ending at rest — obeyed it exactly and this one does not.

Scoring (3 points):

▸ B1 — 1 point: For obtaining the second sled’s two displacements, Δx_1′ = +80 m and Δx_2′ = +180 m, each from a correct kinematic relationship that uses the entry speed v_0. Any correct route earns this point, including reading the two areas off a correctly drawn velocity–time graph. This point does not depend on the responses in Part A.

▸ B2 — 1 point: For indicating Δx_2′ > 2Δx_1′. This point may be earned with an indication consistent with the student’s own values from B1.

▸ B3 — 1 point: For the qualitative reason — the two phases do not span the same change in v_x², because the sled enters Phase 1 already moving, so the inverse relation between displacement and acceleration magnitude does not apply. An equivalent graphical argument, that the accelerating region is a trapezoid rather than a triangle of the same height as the braking one, also earns this point. This point may be earned independently of B1 and B2.

⚠︎ Common error (partial credit): obtains both displacements correctly and indicates Δx_2′ > 2Δx_1′, but justifies it only by observing that 180 m is more than 160 m — earns 2 of 3 points (B1 and B2). The part asks for reasoning beyond the arithmetic, and B3 is where it sits.

⚠︎ Common error (partial credit): obtains Δx_1′ = +80 m correctly but computes Δx_2′ from the entry speed rather than the peak speed, giving +20 m, and so indicates Δx_2′ < 2Δx_1′ — earns 2 of 3 points (B2 and B3) where the reasoning is otherwise correct. B1 requires both displacements; Phase 2 begins at the velocity Phase 1 ended with.

⚠︎ Common error (partial credit): ignores the entry speed and takes v_x0 = 0 in Phase 1, obtaining Δx_1′ = +40 m and Δx_2′ = +80 m, and so indicates Δx_2′ = 2Δx_1′ — earns 1 of 3 points (B2 only). B1 is lost because the entry speed is part of Phase 1’s motion, but the indication is consistent with the student’s own values, so B2 stands.

✗ Common error (no credit): indicates Δx_2′ = 2Δx_1′ by applying the ratio of the acceleration magnitudes directly, without computing either displacement, and supports it by quoting the first sled’s result. B1 is lost because no displacement is obtained, B2 because the indication itself is wrong and is not consistent with any values the student has produced, and B3 because treating the ratio as a general rule is the very error the part tests.

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