FRQ: Displacement from Velocity–Time Graphs — The Remote-Controlled Car Race

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Translation Between Representations (TBR)

▤ Scenario

At a school robotics club time trial, two remote-controlled cars — Car A and Car B — are released from the same starting line at time t = 0 and driven along a straight, level 100 m track. A motion sensor mounted beside the track records the velocity of each car every 0.1 s for the full 20.0 s of the trial.

Car A travels at a constant velocity of +3.0 m/s for the entire 20.0 s.

Car B starts from rest, accelerates uniformly to +6.0 m/s over the first 10.0 s, and then travels at a constant +6.0 m/s from t = 10.0 s until t = 20.0 s.

All motion is one-dimensional and along the track. The direction of travel is defined as positive, the starting line is the origin (x = 0), and the ground is the reference frame throughout. Neither car reverses direction, and the velocity–time graphs of the two cars are not provided.

✎ Free Response Questions

(a) On a single set of axes, sketch and label the velocity–time graphs of Car A and Car B for the interval t = 0 to t = 20.0 s. Label both axes with the quantity and its unit, identify which line belongs to which car, and mark the numerical velocity of each car at t = 0, t = 10.0 s and t = 20.0 s.  (3 points)

(b) On the graph you drew in part (a), shade and label the region whose area represents the displacement of each car over the full 20.0 s. Then calculate that displacement for each car. For each car, name the geometric shape or shapes you used and show the area calculation for each shape separately.  (3 points)

(c) Let v_A be the constant velocity of Car A and let a_B be the constant acceleration of Car B during its speeding-up phase. Beginning from a kinematic relationship on the official AP Physics 1 equation sheet, derive a symbolic expression for the time t_meet at which the two cars have travelled equal displacements from the starting line, assuming this occurs while Car B is still accelerating. Then evaluate t_meet numerically for this race and verify that the assumption holds.  (3 points)

(d) A student claims: “Because Car B ends the race with a higher velocity than Car A, Car B must be ahead of Car A at every instant after t = 10.0 s.” On a single set of axes, sketch the position–time graphs of both cars from t = 0 to t = 20.0 s, consistent with your answers to parts (a) through (c). Then write a paragraph-length argument evaluating the student’s claim. Your argument must state whether the conclusion is correct, state whether the reasoning is valid, and refer explicitly to a specific feature of your position–time graph.  (3 points)

❖ Answer Key & Scoring Guide

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

Part (a) — Model Answer

Take the direction of travel as positive. Velocity is plotted on the vertical axis in m/s and time on the horizontal axis in s.

Car A is a horizontal line at v = +3.0 m/s from t = 0 to t = 20.0 s — constant velocity means zero slope.

Car B is a straight line rising from the origin to v = +6.0 m/s at t = 10.0 s, then a horizontal line at +6.0 m/s to t = 20.0 s. Uniform acceleration is a straight line of constant slope on velocity–time axes, not a curve. The slope of that first segment is the acceleration:

a_B = Δv / Δt = (6.0 − 0) / (10.0 − 0) = 0.60 m/s²

The two lines cross at t = 5.0 s, where both cars momentarily have the same velocity of +3.0 m/s. This is the instant of equal velocity, not equal position — a distinction that matters in parts (c) and (d).

Velocity–time graphs for Car A and Car B, with the displacement regions of part (b) shaded.

Scoring (3 points):

▸ 1 point: Car A drawn as a horizontal line at v = +3.0 m/s spanning the full interval t = 0 to t = 20.0 s and identified as Car A.

▸ 1 point: Car B drawn as a straight line from the origin to +6.0 m/s at t = 10.0 s, followed by a horizontal line at +6.0 m/s to t = 20.0 s, and identified as Car B.

▸ 1 point: Both axes labelled with quantity and unit, and the values +3.0 m/s, +6.0 m/s, t = 10.0 s and t = 20.0 s marked on the graph.

⚠︎ Common error (partial credit): both lines correct in shape but the axes unlabelled or the cars not identified — earns 2 of 3 points.

⚠︎ Common error (partial credit): Car B’s second phase drawn as continuing to rise past t = 10.0 s — earns the Car A point and the labelling point only, 2 of 3 points.

✗ Common error (no credit): Car B’s speeding-up phase drawn as an upward-curving line because “it is accelerating.” This imports the position–time shape onto velocity–time axes; uniform acceleration is a straight line here. The Car B point is not earned.

Part (b) — Model Answer

On a velocity–time graph, displacement is the area between the line and the time axis. Both cars start at x = 0 and neither reverses, so displacement equals distance travelled here.

Car A — one rectangle of height +3.0 m/s and width 20.0 s:

Δx_A = (3.0 m/s)(20.0 s) = 60.0 m

Car B — a triangle from t = 0 to t = 10.0 s, plus a rectangle from t = 10.0 s to t = 20.0 s:

Δx_B(triangle) = ½ (10.0 s)(6.0 m/s) = 30.0 m

Δx_B(rectangle) = (6.0 m/s)(10.0 s) = 60.0 m

Δx_B = 30.0 m + 60.0 m = 90.0 m

Car A travels 60.0 m in the positive direction and Car B travels 90.0 m in the positive direction. Car B finishes 30.0 m ahead. The shaded regions are marked on the graph above.

Scoring (3 points):

▸ 1 point: States that displacement is the area under the velocity–time graph, and shades or otherwise marks the correct region for each car on the part (a) graph.

▸ 1 point: Car A — rectangle named and its area evaluated as 60.0 m, with units.

▸ 1 point: Car B — triangle and rectangle both named, evaluated as 30.0 m and 60.0 m, and summed to 90.0 m, with units.

⚠︎ Common error (partial credit): Car B’s first phase treated as a rectangle of height 6.0 m/s, giving 60 m instead of 30 m and a total of 120 m — earns 2 of 3 points if the area principle is stated and Car A is correct.

⚠︎ Common error (partial credit): correct numerical areas but no shapes named or no units given — earns 2 of 3 points.

✗ Common error (no credit): using the slope of the velocity–time line, or reading the final velocity value off the vertical axis, as the displacement. This confuses area with slope or with value and earns no credit for that car.

Part (c) — Model Answer

Both cars start from the same point, so equal displacement means equal position. Start from the equation-sheet relation for motion with constant acceleration:

x = x_0 + v_0 t + ½ a t²

For Car A, a = 0 and v_0 = v_A, with x_0 = 0:

x_A = v_A t

For Car B, x_0 = 0 and v_0 = 0, with a = a_B, valid while Car B is still accelerating:

x_B = ½ a_B t²

Set the two positions equal and solve. The root t = 0 is the start of the race and is discarded, so divide through by t:

v_A t = ½ a_B t²

t_meet = 2 v_A / a_B

Evaluating for this race, with a_B = 0.60 m/s² from part (a):

t_meet = 2(3.0 m/s) / (0.60 m/s²) = 10.0 s

The assumption holds: t_meet = 10.0 s is the last instant of Car B’s accelerating phase, so the expression x_B = ½ a_B t² is still valid at that time. Checking both positions confirms the result — x_A = (3.0)(10.0) = 30.0 m and x_B = ½(0.60)(10.0)² = 30.0 m.

Scoring (3 points):

▸ 1 point: Begins the derivation from a stated fundamental relation on the equation sheet — x = x_0 + v_0 t + ½ a t² (or the constant-velocity form for Car A) — rather than assembling variables without a starting principle. This point is earned for the correct starting relation even if the algebra that follows is flawed.

▸ 1 point: Sets x_A = x_B, discards the trivial root t = 0, and reaches t_meet = 2 v_A / a_B.

▸ 1 point: Evaluates a_B = 0.60 m/s² and t_meet = 10.0 s with units, and verifies that t_meet does not exceed 10.0 s so the accelerating-phase assumption is valid.

⚠︎ Common error (partial credit): correct symbolic expression but a_B taken as 6.0 m/s² (using the final velocity as the acceleration), giving t_meet = 1.0 s — earns 2 of 3 points.

⚠︎ Common error (partial credit): the correct expression t_meet = 2 v_A / a_B written down with no starting relation cited — the first point is not earned; a maximum of 2 of 3 points.

✗ Common error (no credit): solving v_A = a_B t for the time at which the velocities are equal, giving t = 5.0 s. This finds where the two velocity–time lines cross, not where the cars meet, and confuses the value of the graph with the area beneath it. The second and third points are not earned.

Part (d) — Model Answer

The position–time graph of Car A is a straight line from the origin through (20.0 s, 60.0 m), since constant velocity gives constant slope. The position–time graph of Car B is concave up from the origin to (10.0 s, 30.0 m) — uniform acceleration produces a curve of increasing slope — and then a straight line of slope 6.0 m/s from (10.0 s, 30.0 m) to (20.0 s, 90.0 m). The two curves meet at t = 10.0 s.

Position–time graphs for Car A and Car B. The curves cross at t = 10.0 s, x = 30 m.

Evaluation of the claim. The student’s conclusion is correct, but the reasoning offered for it is not valid. The conclusion is correct because at t = 10.0 s the two position–time curves intersect at x = 30 m — the cars are level, not ahead — and for every instant after that, Car B’s curve has slope 6.0 m/s while Car A’s has slope 3.0 m/s, so the vertical gap between the curves grows without ever closing again.

The reasoning is invalid because a higher velocity is a statement about the slope of the position–time graph, not about its height. Position is the accumulated area under the velocity–time graph, so it depends on the entire history of the motion, not on the velocity at one instant.

The graph makes this visible in the interval from t = 5.0 s to t = 10.0 s: throughout that interval Car B’s velocity–time line lies above Car A’s, so Car B’s position–time curve is steeper — yet Car A’s curve is still the higher of the two. At t = 8.0 s, for instance, x_A = (3.0)(8.0) = 24 m while x_B = ½(0.60)(8.0)² = 19.2 m. Car B is closing the gap during that interval but has not yet erased the 15 m lead Car A built up over the first 5.0 s, when Car A was the faster car.

Only at t = 10.0 s, when the accumulated areas finally become equal, do the curves meet. Had the race been stopped at t = 9.0 s, Car B would have had the greater velocity and still been behind — which is precisely what the student’s reasoning cannot account for.

Scoring (3 points):

▸ 1 point: Position–time graph shows Car A as a straight line from the origin reaching 60 m at t = 20.0 s, and Car B as a concave-up curve from the origin reaching 30 m at t = 10.0 s followed by a straight line of greater slope reaching 90 m at t = 20.0 s, with the two curves crossing at t = 10.0 s.

▸ 1 point: States explicitly that the conclusion is correct for all t > 10.0 s but the stated reasoning is invalid, and identifies why — greater velocity is greater slope, not greater position.

▸ 1 point: Supports the argument with a specific graphical or area-based feature — for example the interval 5.0 s < t < 10.0 s where Car B is faster yet Car A is still ahead (x_A = 24 m against x_B = 19.2 m at t = 8.0 s), or the equality of accumulated areas at t = 10.0 s.

⚠︎ Common error (partial credit): correctly rejects the reasoning but also rejects the conclusion, asserting that Car B is not ahead after t = 10.0 s — earns 2 of 3 points.

⚠︎ Common error (partial credit): argument is entirely verbal, with no reference to a graphical feature, an accumulated area, or a numerical position comparison — earns 2 of 3 points.

⚠︎ Common error (partial credit): Car B’s position–time curve drawn concave down, or the crossing placed at t = 5.0 s (the equal-velocity time) instead of t = 10.0 s — the graph point is not earned; a maximum of 2 of 3 points.

✗ Common error (no credit): agreeing with the student on the grounds that “Car B is faster, so Car B is ahead.” This restates the misconception rather than evaluating it; the argument must distinguish the slope of the position–time graph from its height.

✗ Common error (no credit): Car B’s position–time graph drawn as a straight line for 0 ≤ t ≤ 10.0 s. Uniform acceleration is concave up on position–time axes, and a straight line here contradicts the velocity–time graph of part (a).

Total: 12 points (3 + 3 + 3 + 3)