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

FRQ: Acceleration and Displacement from a Velocity–Time Graph — The Braking Truck

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

A delivery truck moves along a straight, level road. At time t = 0, while the truck is travelling at 18 m/s, the driver applies the brakes. The truck slows at a constant rate and comes to rest at t = 6.0 s.

The truck then stays at rest for 4.0 s. At t = 10.0 s the driver reverses, and the truck speeds up at a constant rate while moving in the reverse direction, reaching a speed of 6.0 m/s at t = 13.0 s.

All motion is one-dimensional along the road. Take the truck’s original direction of travel as the positive x-direction, and use the road as the reference frame. Treat the truck as a point object.

Where symbols are required, write the truck’s initial speed as v_0, the instants t = 6.0 s, t = 10.0 s and t = 13.0 s as t_1, t_2 and t_3, and the truck’s speed at t = 13.0 s as v_f.

Figure 1 shows the axes for part A (i).

Figure 1 — Axes for part A (i). Grid lines are 1 s apart in time and 2 m/s apart in velocity.

✎ Free Response Questions

Part A

i.    On the axes shown in Figure 1, sketch a graph of the x-component v_x of the truck’s velocity as a function of time t from t = 0 to t = 13.0 s.

ii.   Calculate the x-component of the truck’s average acceleration over each of the three phases of the motion: the braking phase from t = 0 to t = 6.0 s, the phase at rest from t = 6.0 s to t = 10.0 s, and the reversing phase from t = 10.0 s to t = 13.0 s.

iii.  Derive an expression for the truck’s displacement Δx over the whole run, from t = 0 to t = t_3. Express your final answer in terms of v_0, v_f, t_1, t_2, t_3, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Part B

A student makes the following claim: “A negative acceleration always means that an object is slowing down.”

Indicate whether the x-component of the truck’s velocity increases, decreases, or remains constant during the reversing phase, from t = 10.0 s to t = 13.0 s.

______ Increases

______ Decreases

______ Remains constant

Justify your response.

Identify the phase of the truck’s motion, if any, that shows the student’s claim is incorrect. Justify your response.

❖ 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 graph is three straight segments, because the acceleration is constant within each phase, and a constant acceleration changes the velocity by equal amounts in equal times.

From t = 0 to t = 6.0 s the velocity falls in a straight line from v_x = +18 m/s to 0. From t = 6.0 s to t = 10.0 s the truck is at rest, so the graph lies along the time axis. After t = 10.0 s the truck moves in the −x direction with increasing speed, so v_x is negative and becoming more negative: the graph falls in a straight line from 0 to v_x = −6.0 m/s, below the axis. The reversing segment lies below the time axis, because the axis carries the velocity component v_x and not the speed.

Figure 2 — The completed graph: three straight segments, the end of each phase marked.

Scoring (2 points):

▸ A1 — 1 point: For sketching a straight line from +18 m/s at t = 0 to zero at t = 6.0 s, the graph then staying at zero from t = 6.0 s to t = 10.0 s, and no later.

▸ A2 — 1 point: For sketching a straight line that leaves the time axis at zero velocity and lies below the axis, ending at −6.0 m/s at t = 13.0 s.

Scoring Note: Judge values against the grid in Figure 1; a value within half a grid square is accepted, and the lines need not be ruled. The time at which the graph leaves the axis is scored in A1 only. A1 and A2 are scored independently of each other.

⚠︎ Common error (partial credit): Sketches the braking and rest phases correctly but draws the reversing phase above the time axis, rising from 0 to +6.0 m/s at t = 13.0 s — earns 1 of 2 points, A1. The vertical axis carries v_x, which is negative whenever the truck moves in the −x direction; the quantity that rises in that phase is the speed.

✗ Common error (no credit): Draws the braking and reversing phases as curves — a constant acceleration changes the velocity by equal amounts in equal times, which is a straight line on a velocity–time graph, so neither A1 nor A2 is earned.

Part A (ii) — Model Answer

Average acceleration is the change in velocity divided by the time interval over which that change occurs:

a_avg = Δv_x / Δt

Read each velocity as a signed component along the x-axis: the truck starts at v_x = +18 m/s, is at rest at v_x = 0 from t = 6.0 s to t = 10.0 s, and finishes at v_x = −6.0 m/s, negative because it is then moving in the −x direction.

braking: a_avg = (0 − 18 m/s) / (6.0 s − 0) = −3.0 m/s²
at rest: a_avg = (0 − 0) / (10.0 s − 6.0 s) = 0
reversing: a_avg = (−6.0 m/s − 0) / (13.0 s − 10.0 s) = −2.0 m/s²

The braking acceleration is 3.0 m/s² in the −x direction, the acceleration at rest is zero, and the reversing acceleration is 2.0 m/s², also in the −x direction. Average acceleration depends only on the change in v_x, sign included, and on how long the interval lasts.

Scoring (2 points):

▸ A3 — 1 point: For the average accelerations of the first two phases: −3.0 m/s², or 3.0 m/s² in the −x direction, for the braking phase, and zero for the phase at rest.

▸ A4 — 1 point: For the average acceleration of the reversing phase: −2.0 m/s², or 2.0 m/s² in the −x direction.

Scoring Note: Units are required on each nonzero value; the relation a_avg = Δv_x/Δt need not be written. A3 and A4 do not depend on the response in part A (i).

⚠︎ Common error (partial credit): Gives +2.0 m/s² for the reversing phase, treating “speeding up” as a positive acceleration, with the other two phases correct — earns 1 of 2 points, A3. The change in v_x over that phase is −6.0 m/s, so its average acceleration is negative; part B turns on this.

⚠︎ Common error (partial credit): Obtains −3.0 m/s² and −2.0 m/s² for the two moving phases, but says the acceleration during the phase at rest is undefined “because the truck is not moving” — earns 1 of 2 points, A4. A truck at rest has a velocity of zero that does not change, so Δv_x = 0 and a_avg = 0.

✗ Common error (no credit): Gives 3.0 m/s², 0 and 2.0 m/s², calling the first a “deceleration”, with no sign or direction on either nonzero value — the prompt asks for x-components, whose sign is their direction, and “deceleration” names slowing down, not a direction, so neither A3 nor A4 is earned.

✗ Common error (no credit): Divides each phase’s final velocity by the clock reading at the end of that phase — 0 for the braking phase and for the phase at rest, and −6.0 m/s ÷ 13.0 s ≈ −0.46 m/s² for the reversing phase — an average acceleration is a change in velocity divided by the interval over which it occurs, not a velocity divided by a clock reading, so neither A3 nor A4 is earned.

Part A (iii) — Model Answer

Treat the run as three phases and find the displacement in each. For each moving phase use the constant-acceleration relationship from the reference information,

x = x_0 + v_x0 t + ½a_x t²

with t measured from the start of that phase, so that a phase lasting a time t has Δx = v_x0 t + ½a_x t².

Braking phase. The truck enters at v_x0 = +v_0 and comes to rest in a time t_1, so its constant acceleration is a_x = −v_0/t_1 and

Δx_1 = v_0t_1 + ½(−v_0/t_1)t_1² = ½v_0t_1

Phase at rest. The velocity is zero throughout, so Δx_2 = 0.

Reversing phase. The truck starts from rest, moves in the −x direction and reaches a speed v_f, so v_x changes from 0 to −v_f in a time t_3 − t_2. Its acceleration is a_x = −v_f/(t_3 − t_2), and

Δx_3 = ½(−v_f/(t_3 − t_2))⁠(t_3 − t_2)² = −½v_f(t_3 − t_2)

Displacements add, so over the whole run

Δx = Δx_1 + Δx_2 + Δx_3 = ½v_0t_1 − ½v_f(t_3 − t_2)

An equally complete route reads each displacement as a signed area between the graph and the time axis: ½v_0t_1 above the axis, nothing while the truck is at rest, and ½v_f(t_3 − t_2) below it, counted negative.

Figure 3 — Displacement as signed area: braking counts positive, reversing negative.

The reversing term is negative because the truck moves in the −x direction during that phase. With the Scenario’s values, Δx = ½(18 m/s)⁠(6.0 s) − ½(6.0 m/s)⁠(3.0 s) = +54 m − 9.0 m = +45 m. Adding the two magnitudes instead gives 63 m, the distance travelled, a different quantity. Displacements add with their signs, so the truck ends 45 m from its start, in the +x direction, after travelling 63 m.

Scoring (3 points):

▸ A5 — 1 point: For a multistep derivation that includes a relationship giving the displacement of the braking phase or of the reversing phase, applied to that phase.

Scoring Note: The minimum requirement for A5 is a displacement worked out for the braking phase or the reversing phase: an equation from the reference information with that phase’s quantities substituted, the phase’s average velocity multiplied by its duration, or the area of that phase’s region of the graph, in words or in symbols. An equation from the reference information written but not applied, or a final expression on its own, falls short of it.

▸ A6 — 1 point: For an expression in the named symbols for the reversing phase’s displacement that is negative and has the units of a length — for example −½v_f(t_3 − t_2), or ½a_x(t_3 − t_2)² or v_f²/(2a_x) with a_x = −v_f/(t_3 − t_2). Its magnitude is scored in A7.

▸ A7 — 1 point: For a correct expression for the displacement over the whole run, Δx = ½v_0t_1 − ½v_f(t_3 − t_2), or any algebraic equivalent in the named symbols.

Scoring Note: A correct, isolated, final expression for Δx earns points A6 and A7. A5, A6 and A7 do not depend on parts A (i) and A (ii).

⚠︎ Common error (partial credit): Finds the braking displacement correctly but takes the reversing phase’s displacement as positive, writing Δx = ½v_0t_1 + ½v_f(t_3 − t_2) — earns 1 of 3 points, A5. That sum is the distance travelled, 63 m for the Scenario’s values; the reversing phase carries the truck back toward its start, so the displacement is smaller than the distance.

⚠︎ Common error (partial credit): In an otherwise complete derivation, takes t_3 as the duration of the reversing phase, writing Δx = ½v_0t_1 − ½v_f t_3 — earns 2 of 3 points, A5 and A6. t_3 is the clock reading at the end of the run; the reversing phase lasts t_3 − t_2.

⚠︎ Common error (partial credit): Works throughout with the Scenario’s numbers, reaching Δx = +54 m − 9.0 m = +45 m, with no expression in v_0, v_f, t_1, t_2 and t_3 — earns 1 of 3 points, A5. The prompt asks for the expression; +45 m is its value for this one run.

✗ Common error (no credit): Takes each phase’s displacement as the slope of its segment on the velocity–time graph, writing Δx = −v_0/t_1 − v_f/(t_3 − t_2) — on a velocity–time graph the slope is the acceleration and the area is the displacement; no relationship giving a displacement is used and no term is a displacement, so no criterion is earned.

Part B — Model Answer

Decreases

Over the reversing phase the truck’s velocity component runs from v_x = 0 at t = 10.0 s to v_x = −6.0 m/s at t = 13.0 s, so v_x decreases: its acceleration over the phase is negative, −2.0 m/s², and the reversing segment of the graph slopes down. Over the same phase the truck’s speed runs from 0 to 6.0 m/s, so the speed increases. Both are true at once, because the speed is the magnitude of the velocity and v_x also carries a sign.

The reversing phase shows the student’s claim is incorrect. Its acceleration is negative, −2.0 m/s², and yet the truck speeds up, because after t = 10.0 s its velocity and its acceleration both point in the −x direction. The braking phase does not contradict the claim: its acceleration, −3.0 m/s², is negative too, and there the truck slows, because before t = 6.0 s its velocity points in the +x direction and its acceleration in the −x direction. Accelerations of the same sign, opposite outcomes: the sign of an acceleration component gives its direction along the chosen axis and nothing more. An object slows down when its velocity and its acceleration point in opposite directions and speeds up when they point the same way.

Scoring (3 points):

▸ B1 — 1 point: For indicating “Decreases”.

▸ B2 — 1 point: For a justification that indicates one of the following: v_x changes from zero to a negative value, for example from 0 to −6.0 m/s; the truck moves in the −x direction with increasing speed, so v_x becomes more negative; the acceleration during the phase is negative, or in the −x direction, so the change in v_x is negative; or the reversing segment of the velocity–time graph has a negative slope.

Scoring Note: A justification based only on the rising speed, or only on the truck moving in the −x direction, falls short of B2: the speed rises in this phase, and motion in the −x direction alone does not fix whether v_x rises or falls.

▸ B3 — 1 point: For identifying the reversing phase, with a justification that indicates one of the following: its acceleration is negative, yet the truck’s speed increases, for example from 0 to 6.0 m/s; its velocity and its acceleration both point in the −x direction, so the truck speeds up even though its acceleration is negative; or the two moving phases have accelerations of the same sign, yet the truck slows in the braking phase and speeds up in the reversing phase.

Scoring Note: Stating that the claim fails in the braking phase, alone or alongside the reversing phase, falls short of B3: in the braking phase the truck does slow down. Identifying no phase falls short of B3. B2 and B3 may each be earned from anywhere in the response to part B. B1, B2 and B3 are scored independently of one another and of part A.

⚠︎ Common error (partial credit): Indicates “Decreases”, justified by the negative acceleration, but identifies no phase that contradicts the claim, reading a falling v_x as a falling speed — earns 2 of 3 points, B1 and B2. From 0 to −6.0 m/s the component falls while the speed rises.

⚠︎ Common error (partial credit): Indicates “Increases”, reading the question as asking about the speed, but identifies the reversing phase, where the truck speeds up while its acceleration, −2.0 m/s², is negative — earns 1 of 3 points, B3.

⚠︎ Common error (partial credit): Indicates “Decreases” only because the truck moves in the −x direction, then identifies the reversing phase as one where the acceleration is negative yet the truck speeds up — earns 2 of 3 points, B1 and B3. Moving in the −x direction does not by itself make v_x fall: a truck moving in −x while slowing down has a rising v_x.

✗ Common error (no credit): Indicates “Increases” because the truck speeds up, and identifies no phase that contradicts the claim, reading the reversing acceleration as positive because the speed rises — v_x is never distinguished from the speed, and a positive reversing acceleration contradicts the change in v_x from 0 to −6.0 m/s, so no criterion is earned.