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

Assessments aligned to 2026 AP Physics 1 standards

Translation Between Representations (TBR)  |  MID-LEVEL  |  12 points  |  30 min

▤ 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 each car may be treated as a point object.

Figure 1 shows the velocity–time axes on which the sensor record for Car A has already been drawn.

Figure 1 — Velocity–time axes for the 20.0 s trial, with Car A’s graph drawn.

✎ Free Response Questions

Part A

On Figure 1, draw the velocity–time graph of Car B for the interval t = 0 to t = 20.0 s, on the same axes as the line already shown for Car A.

Part B

Figure 2 is a motion diagram for the same trial. The upper row marks the position of Car A at t = 0, t = 5.0 s, t = 10.0 s, t = 15.0 s and t = 20.0 s. The lower row marks the position of Car B at t = 0 only.

On Figure 2, mark the position of Car B at t = 5.0 s, t = 10.0 s, t = 15.0 s and t = 20.0 s. Label each mark you add with its time.

Figure 2 — Motion diagram on the track’s position scale. Car B is marked at t = 0 only.

Part 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.

Derive a symbolic expression for the time t_meet at which the two cars have travelled equal displacements from the starting line, taking Car B’s acceleration to be a_B for the whole of the interval from t = 0 to t_meet. Express your final answer in terms of v_A, a_B, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Then, using your expression, predict the factor by which t_meet would change if Car B’s acceleration were twice as large, with v_A and the duration of Car B’s speeding-up phase unchanged. Justify your response.

Part D

Indicate whether the positions you marked for Car B in part B are or are not consistent with the velocity–time graph you drew in part A, over the interval from t = 5.0 s to t = 10.0 s, by writing one of the following.

•   ______ Consistent

•   ______ Not consistent

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 — 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; both scales are supplied on Figure 1.

Car A’s line is already drawn — a horizontal line at v = +3.0 m/s across the whole interval, because constant velocity means zero slope.

Car B starts from rest, so its line begins at the origin. Uniform acceleration is a straight line of constant slope on velocity–time axes, not a curve, so the first segment is straight and rises to v = +6.0 m/s at t = 10.0 s. The slope of that segment is Car B’s acceleration:

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

From t = 10.0 s to t = 20.0 s Car B’s velocity is constant at +6.0 m/s, so the second segment is horizontal.

The two lines cross at t = 5.0 s, where both cars momentarily have the same velocity, +3.0 m/s. That crossing is an instant of equal velocity, not of equal position.

Car B’s velocity–time graph added to the supplied axes, with the crossing marked.

Scoring (3 points):

▸ A1 — 1 point: Car B’s line for 0 ≤ t ≤ 10.0 s drawn as a straight segment beginning at the origin, rather than as a curve.

▸ A2 — 1 point: That segment reaching v = +6.0 m/s at t = 10.0 s.

▸ A3 — 1 point: A horizontal segment at v = +6.0 m/s from t = 10.0 s to t = 20.0 s.

⚠︎ Common error (partial credit): Car B’s speeding-up segment 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. A2 and A3 may still be earned, A1 is not — 2 of 3 points.

⚠︎ Common error (partial credit): The second segment drawn continuing to rise past t = 10.0 s, toward the top of the axis, instead of levelling off — A1 and A2 are earned, A3 is not; 2 of 3 points.

Part B — Model Answer

On a velocity–time graph, displacement over an interval is the area between the line and the time axis. Both cars start at x = 0 and neither reverses, so each car’s position at any instant is the area accumulated up to that instant.

Car B’s accumulated area, and so its position, at each of the four marked instants.

Car A — the region under its line is a rectangle of height 3.0 m/s, so at t = 5.0 s

x_A = (3.0 m/s)(5.0 s) = 15.0 m

and the same rectangle gives 30.0 m, 45.0 m and 60.0 m at t = 10.0 s, 15.0 s and 20.0 s. Those marks are already on Figure 2.

Car B — while it is speeding up the region is a triangle, so at t = 5.0 s, where its line has reached 3.0 m/s,

x_B = ½ (5.0 s)(3.0 m/s) = 7.5 m

and at t = 10.0 s the triangle is complete:

x_B = ½ (10.0 s)(6.0 m/s) = 30.0 m

After t = 10.0 s a rectangle of height 6.0 m/s is added to that completed triangle:

x_B = 30.0 m + (6.0 m/s)(5.0 s) = 60.0 m  at t = 15.0 s

x_B = 30.0 m + (6.0 m/s)(10.0 s) = 90.0 m  at t = 20.0 s

So Car B is 7.5 m behind Car A at t = 5.0 s, level with it at t = 10.0 s, and 30.0 m ahead at the end. The spacing between consecutive marks is the area under the velocity–time line for that interval.

The completed motion diagram, with Car B’s four added marks.

Scoring (3 points):

▸ B1 — 1 point: Car B marked behind Car A at t = 5.0 s, at approximately 7.5 m on the supplied scale.

▸ B2 — 1 point: Car B marked at approximately 30 m at t = 10.0 s, level with Car A’s mark to within the tolerance below.

▸ B3 — 1 point: Car B marked at approximately 60 m at t = 15.0 s and at approximately 90 m at t = 20.0 s.

Scoring Note: Marks are read against the supplied 10 m scale; a mark within a quarter of a scale division — 2.5 m — of the value implied by the response’s own part A graph earns its point, and the values stated above are those implied by a correct part A graph. B3 requires both of its marks.

Scoring Note: The time label on each added mark is not separately scored, but a mark that cannot be assigned to an instant cannot earn its criterion; where a response labels only some of its marks, score the marks it labels.

⚠︎ Common error (partial credit): With Car B’s line correctly drawn in part A, the positions at t = 15.0 s and t = 20.0 s obtained from the constant-velocity rectangle alone, omitting the 30.0 m already accumulated while Car B was speeding up — marks at 30 m and 60 m. B1 and B2 are earned, B3 is not; 2 of 3 points.

⚠︎ Common error (partial credit): With Car B’s line correctly drawn in part A, the last two marks placed by continuing the widening spacing, as though Car B kept speeding up after t = 10.0 s — B1 and B2 are earned, B3 is not; 2 of 3 points.

✗ Common error (no credit): With Car B’s line correctly drawn in part A, its speeding-up phase treated as a rectangle of height 6.0 m/s, placing Car B at 30 m at t = 5.0 s and 60 m at t = 10.0 s. This applies the final velocity of the phase to the whole of it; the region is a triangle, not a rectangle, and no mark lands within tolerance.

✗ Common error (no credit): With Car B’s line correctly drawn in part A, positions taken from that graph’s value or its slope rather than the area beneath it — a mark at 6 m at t = 10.0 s read off the velocity axis, or marks at 0.6 m for t = 5.0 s and t = 10.0 s and at 0 m thereafter, read off the slope, which is 0.60 m/s² while Car B speeds up and zero once its line is horizontal. On velocity–time axes the slope is acceleration and the area beneath is displacement, so no mark lands within tolerance.

Part C — Model Answer

Both cars start from the same point, so equal displacement means equal position. Begin from the kinematic relation for motion with constant acceleration, which is in the reference information:

x = x_0 + v_x0 t + ½ a_x t²

For Car A, x_0 = 0, v_x0 = v_A and a_x = 0:

x_A = v_A t

For Car B, x_0 = 0, v_x0 = 0 and a_x = a_B:

x_B = ½ a_B t²

Setting the two positions equal:

v_A t = ½ a_B t²

The root t = 0 is the start of the race, when the cars are level at the starting line. Discarding it and dividing through by t:

t_meet = 2 v_A / a_B

The derived expression depends on a_B inversely, so doubling a_B while holding v_A fixed multiplies t_meet by ½. With the speeding-up phase still lasting 10.0 s, a_B would apply for the whole interval from t = 0 to the meeting at t = 5.0 s, so the expression is still the right one to use. Doubling Car B’s acceleration halves the time it needs to draw level, because the meeting time is inversely proportional to a_B.

Scoring (4 points):

▸ C1 — 1 point: Begins the derivation by writing a fundamental physics principle or an equation from the reference information — for example x = x_0 + v_x0 t + ½ a_x t², the constant-velocity form for Car A, or the statement that displacement is the area under a velocity–time graph. This point is earned for a correct starting relation even where the algebra that follows is flawed.

▸ C2 — 1 point: Writes expressions for the two cars’ displacements from the same origin over the same interval — or for their average velocities over that interval — and sets them equal.

▸ C3 — 1 point: Reaches t_meet = 2 v_A / a_B, rejecting the trivial t = 0 solution where the solution path produces one.

▸ C4 — 1 point: Predicts that t_meet is multiplied by ½.

Scoring Note: A correct, isolated, final expression for t_meet earns points C2 and C3.

Scoring Note: C4 requires the prediction to be justified from the functional dependence of the derived expression on a_B; a bare factor with no reasoning does not earn it. C1, C2 and C3 are scored on the derivation itself and carry no consistency clause, while C4 may be earned for a factor that follows correctly from the expression the response derived, even where that expression is incorrect.

⚠︎ Common error (partial credit): The correct expression written down with no fundamental principle or reference-information equation cited, and the factor correct and justified from the inverse dependence — C1 is not earned; 3 of 4 points.

⚠︎ Common error (partial credit): The derivation correct but the factor given as 2, on the reasoning that a larger acceleration means a longer race — C1, C2 and C3 are earned, C4 is not; 3 of 4 points.

⚠︎ Common error (partial credit): Solving v_A = a_B t instead, which locates the time at which the two velocities are equal rather than the time at which the displacements are equal. This confuses the value of the velocity–time graph with the area beneath it. C1 may still be earned for the starting relation, and C4 for a factor that follows from the expression reached — C2 and C3 are not, so a maximum of 2 of 4 points.

Part D — Model Answer

The two representations are consistent.

Over the interval from t = 5.0 s to t = 10.0 s Car B’s velocity–time line lies above Car A’s, so Car B is the faster car throughout. On the motion diagram, though, Car B’s mark at t = 5.0 s sits 7.5 m behind Car A’s and only draws level at t = 10.0 s. There is no contradiction, because the two representations report different things: the height of a velocity–time line is how fast a car is going at an instant, while its position is the area accumulated beneath that line since t = 0.

Car A accumulated area faster over the first 5.0 s — its rectangle to t = 5.0 s has area 15.0 m against the 7.5 m of Car B’s triangle — so Car B enters the interval 7.5 m behind. Being the faster car from t = 5.0 s onward is what lets Car B close that gap, and the two accumulated areas become equal only at t = 10.0 s. A higher velocity sets how fast a gap is closing, not which car is ahead.

Drawn as position against time, all of this is one picture: Car B’s curve is steeper than Car A’s line from t = 5.0 s onward while still lying below it, the two meet at t = 10.0 s at x = 30 m, and the gap then widens for the rest of the race.

Position against time for both cars. The curves meet at t = 10.0 s, x = 30 m.

The same reasoning settles a claim students often make about this race: that because Car B ends with the higher velocity, Car B must be ahead of Car A at every instant after t = 10.0 s. The conclusion is correct, but the reasoning offered for it is not. Had the race been stopped at t = 8.0 s, Car B would have been the faster car, at x_B = 19.2 m, and still behind Car A, at x_A = 24.0 m.

Scoring (2 points):

▸ D1 — 1 point: Indicates whether the two representations are or are not consistent, where the indication given follows from the graph the response drew in part A and the marks it made in part B.

▸ D2 — 1 point: Supports that indication by identifying that the height of a velocity–time line is the speed at an instant while position is the area accumulated beneath it, so the faster car need not be the car ahead — or by citing a specific feature of the response’s own representations that shows how the car that is faster over the interval can still be the car behind, such as the area under Car A’s line over 0 ≤ t ≤ 5.0 s exceeding the area under Car B’s, so that the two accumulated areas become equal only at t = 10.0 s.

Scoring Note: For a correct part A and a correct part B the two representations are consistent, and a response saying so with either form of support above earns both points.

Scoring Note: A response that recasts the argument on position–time axes should show Car B’s curve concave up and the two curves meeting at t = 10.0 s, not at t = 5.0 s — the equal-velocity instant is not the equal-position instant. A position–time sketch is not required and is not separately scored.

⚠︎ Common error (partial credit): States correctly that the two representations are consistent but supports it only by restating that Car B is faster and catches up, without saying how a faster car can still be the car behind — D1 is earned, D2 is not; 1 of 2 points.

✗ Common error (no credit): Where the response’s own part A graph and part B marks are themselves consistent, concludes that the two representations contradict each other, because Car B is the faster car over the interval yet is still behind. This treats a greater velocity as a greater position; velocity is the height of the velocity–time line and position is the area beneath it, so the two representations agree exactly as drawn. Neither point is earned.