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FRQ: Relative Motion in One Dimension — Two Vehicles on a Straight Highway

Assessments aligned to 2026 AP Physics 1 standards

Qualitative/Quantitative Translation (QQT)  |  MID-LEVEL  |  8 points  |  20 min

▤ Scenario

Car A and Car B travel along a straight, level highway. Both cars move in the same direction, which is defined as the positive direction. A traffic sensor mounted on the road records the velocity of each car relative to the ground. A passenger in Car A watches Car B, and a passenger in Car B watches Car A.

In the first stage of the trip, Car A travels at a constant +22 m/s and Car B travels at a constant +30 m/s, both measured relative to the ground.

Car A then speeds up uniformly from +22 m/s to +28 m/s over an interval of 3.0 s, after which it holds a constant +28 m/s. Car B travels at a constant +30 m/s for the whole trip.

Figure 1 shows the two cars during the first stage of the trip, with the ground-frame velocities recorded by the traffic sensor.

Figure 1 — Ground-frame velocities during the first stage, as recorded by the road sensor.

All motion is one-dimensional. Take the ground as an inertial reference frame, and take every velocity referred to below to be measured at a moment when both cars are travelling at constant velocity.

✎ Free Response Questions

Part A

During the first stage of the trip, the passenger in Car A measures the velocity of Car B, and the passenger in Car B measures the velocity of Car A. Let v_B/A be the velocity of Car B as measured by the passenger in Car A, and let v_A/B be the velocity of Car A as measured by the passenger in Car B.

Indicate whether the magnitude of v_A/B is greater than, less than, or equal to the magnitude of v_B/A by writing one of the following.

•   |v_A/B| > |v_B/A|

•   |v_A/B| < |v_B/A|

•   |v_A/B| = |v_B/A|

Justify your answer using qualitative reasoning beyond referencing equations.

Part B

Starting with the relationship between the velocity of an object measured in one reference frame and the velocity of that same object measured in a second reference frame moving relative to the first, derive an expression for v_A/B in terms of v_B/A.

Express your final answer in terms of v_B/A, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

Part C

Justify how your derived equation in part B is or is not consistent with your reasoning in part A.

❖ 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

|v_A/B| = |v_B/A|

Each passenger is at rest in their own car, so neither one is watching a car moving down a road — each is watching the other car’s position relative to themselves change. There is only one gap between Car A and Car B, and that single gap widens at one definite rate. The passenger in Car A watches that one gap open up ahead of them; the passenger in Car B watches the same one gap open up behind them. Because both passengers are reading the same physical separation changing at the same rate, the two measurements cannot differ in size.

The two passengers do disagree about direction — from Car A the other car pulls ahead, from Car B the other car falls behind — but that is a disagreement about sign, not about magnitude. Nothing in this argument refers to which car is faster, or to how fast either car is going over the ground. Both passengers measure the same single separation changing at the same rate, so the two magnitudes must agree.

Figure 2 — The same separation seen from each car, each observer at rest in their own frame.

Scoring (3 points):

▸ A1 — 1 point: For indicating |v_A/B| = |v_B/A|.

▸ A2 — 1 point: For a qualitative statement that each passenger is at rest in their own reference frame, and so is measuring the motion of the other car relative to themselves rather than relative to the road.

▸ A3 — 1 point: For a qualitative argument that both passengers are reading the same single separation between the two cars, and that this one separation changes at one rate, so the two measurements must agree in magnitude.

Scoring Note: A2 and A3 are content criteria, not route criteria. A2 requires the response to establish that the two measurements are made from interchangeable viewpoints — by stating that each passenger is at rest in their own frame, by an observer-exchange symmetry argument, or by any equivalent. A3 requires a single shared separation, or a single closing or separating rate. The symmetry argument establishes both and earns A2 and A3 together; a rate-of-approach argument establishes A3 only.

⚠︎ Common error (partial credit): Indicates |v_A/B| = |v_B/A| correctly but justifies it entirely by writing the subtraction v_A/B = v_A/G − v_B/G = −(v_B/G − v_A/G) — earns 1 of 3 points, for A1 only. The prompt asks for reasoning beyond referencing equations, and this algebra is what part B derives.

⚠︎ Common error (partial credit): Indicates the correct comparison and gives the single-separation argument, but offers neither the rest-frame statement nor any equivalent route from the Scoring Note above — earns 2 of 3 points, for A1 and A3.

✗ Common error (no credit): Indicates |v_A/B| < |v_B/A| on the grounds that Car B is the faster car, so Car B’s motion must be the bigger of the two. This treats a relative velocity as a property belonging to the faster object rather than as one separation measured from two frames, and the response offers nothing that establishes either A2’s or A3’s content by any route.

Part B — Model Answer

Let G denote the ground frame. Combining the motion of an object with the motion of the observer’s frame in one dimension is a subtraction of velocity components, so for the passenger in Car A watching Car B,

v_B/A = v_B/G − v_A/G

The passenger in Car B watching Car A is the same construction with the observer and the observed object exchanged,

v_A/B = v_A/G − v_B/G

The right-hand side of the second expression is the right-hand side of the first with its two terms interchanged, and interchanging the terms of a difference reverses its sign,

v_A/B = −(v_B/G − v_A/G) = −v_B/A

The two relative velocities are the same pair of ground-frame velocities subtracted in opposite order, so one is the negative of the other.

Scoring (3 points):

▸ B1 — 1 point: For beginning from the frame-combination relationship — writing v_B/A = v_B/G − v_A/G, or the equivalent combination form v_B/G = v_B/A + v_A/G. This point is awarded for the correct starting relationship alone, before any algebra is performed.

▸ B2 — 1 point: For writing the corresponding relationship with the observer and the observed object exchanged — v_A/B = v_A/G − v_B/G, or the equivalent combination form v_A/G = v_A/B + v_B/G.

▸ B3 — 1 point: For correctly obtaining v_A/B = −v_B/A.

Scoring Note: A correct, isolated, final expression for v_A/B earns points B2 and B3.

⚠︎ Common error (partial credit): Writes both frame relationships correctly but reports v_A/B = v_B/A, dropping the sign — earns 2 of 3 points, for B1 and B2.

⚠︎ Common error (partial credit): Begins from v_B/G = v_B/A + v_A/G but then writes v_A/B = v_A/G + v_B/G without ever setting up the exchanged relationship, which would give a relative speed of 52 m/s in the first stage — earns 1 of 3 points, for B1 only.

✗ Common error (no credit): Reports the value −8 m/s and nothing else. The part asks for an expression for v_A/B in terms of v_B/A, and a number evaluated from the first stage is not that expression, so no criterion is met.

Part C — Model Answer

The derived equation is consistent with the reasoning in part A.

The equation v_A/B = −v_B/A multiplies v_B/A by −1 and does nothing else to it. Multiplying by −1 changes a sign and leaves a magnitude untouched, so the equation says |v_A/B| = |v_B/A| — which is the comparison selected in part A. The equation also supplies something the part A argument could only gesture at: it fixes which sign each passenger measures, confirming that the two observers disagree about direction while agreeing about size.

The deeper agreement is that neither one depends on the particular speeds. The part A argument referred only to a single separation changing at a single rate, and never to the value of either car’s ground-frame velocity. The derived equation likewise contains no ground-frame velocity at all — v_A/G and v_B/G cancelled out of it. So the comparison must survive any change in either car’s speed. Figure 3 shows both cars’ ground-frame velocities across the whole trip.

Figure 3 — Ground-frame velocities across the trip; the gap is v_B/A where both are constant.

It does survive. Once Car A has finished speeding up and holds a constant +28 m/s,

v_B/A = v_B/G − v_A/G = (+30 m/s) − (+28 m/s) = +2 m/s
v_A/B = −v_B/A = −2 m/s

The gap between the cars now opens four times more slowly than it did in the first stage, but the two measurements are still equal in magnitude and opposite in sign. The speed-up changed the value of the relative velocity; it did not touch the relationship between the two measurements, because both members of the pair changed together. The derived equation names no particular velocities, so the comparison in part A survives any change in either car’s speed.

Scoring (2 points):

▸ C1 — 1 point: For identifying what the response’s own part B expression implies about the two magnitudes, and stating the verdict — consistent, or not consistent — that follows from it together with the response’s own part A selection. For the correct expression v_A/B = −v_B/A, that it relates the two quantities by a sign change only, so their magnitudes are equal.

▸ C2 — 1 point: For drawing the correct conclusion about whether the response’s own derived expression depends on either car’s ground-frame velocity, and so whether it shares the independence the part A reasoning relied on. For the correct expression v_A/B = −v_B/A, that it contains no ground-frame velocity and therefore does share it. This point is awarded for work consistent with the response’s own part B expression.

Scoring Note: A response that demonstrates the independence by predicting a second case instead of stating it in words — for example evaluating v_B/A = +2 m/s once Car A holds a constant +28 m/s and then obtaining v_A/B = −2 m/s from the derived relationship — earns C2.

⚠︎ Common error (partial credit): States that the two are consistent and identifies that the equation changes only the sign, but never examines whether the derived expression depends on either car’s speed — earns 1 of 2 points, for C1.

✗ Common error (no credit): Concludes that the derived equation is not consistent with the part A reasoning because the relative velocity changes when Car A speeds up. This confuses a change in the value of the relative velocity with a failure of the relationship between the two measurements, and both members of the pair change together.

✗ Common error (no credit): Concludes that the derived equation is not consistent with the part A reasoning because Car A accelerates during the trip, so the two frames are not both inertial. The scenario specifies that every velocity referred to is measured at a moment when both cars travel at constant velocity, so both frames are inertial at every instant the question asks about. What v_A/B = −v_B/A requires is that both velocities be measured at the same instant and referred to the same third frame, and the scenario fixes both.

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