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

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Qualitative/Quantitative Translation (QQT)  |  MID-LEVEL  |  8 points

▤ 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.

All motion is one-dimensional. Take the ground as an inertial reference frame, and evaluate every velocity asked for below at a moment when both cars are travelling at constant velocity.

✎ Free Response Questions

(a) During the first stage of the trip, calculate the velocity of Car B as measured by the passenger in Car A. Show your work.

(b) During the first stage of the trip, calculate the velocity of Car A as measured by the passenger in Car B. State the relationship between this result and your answer to part (a).

(c) After Car A has finished speeding up and is again travelling at a constant velocity, calculate the velocity of Car B as measured by the passenger in Car A. Justify why your answer to part (a) no longer describes what that passenger measures.

(d) Two students discuss these results.

Student 1: "The velocity of Car B relative to Car A must always be exactly the opposite of the velocity of Car A relative to Car B, because both cars are travelling on the same road in the same direction."

Student 2: "That is only true in the first stage. Once Car A speeds up and the two cars are nearly matching speeds, the two relative velocities are no longer opposites."

In a coherent paragraph-length response, evaluate both claims. Identify which student's conclusion is correct, support that conclusion with the values you calculated in parts (a) through (c), explain the relationship that produces it, and state the condition that relationship depends on.

❖ Answer Key & Scoring Guide

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

Part (a) — Model Answer

Define the direction of travel as positive. Let G = ground frame, A = Car A's frame, B = Car B.

Combining the motion of an object with the motion of the observer's frame in one dimension is a subtraction of velocity components:

v_B/A = v_B/G − v_A/G

v_B/A = (+30 m/s) − (+22 m/s) = +8 m/s

Car B moves at +8 m/s relative to the passenger in Car A — forward, in the direction of travel. Car B pulls away from Car A at 8 m/s.

Figure 1 — The two ground-frame velocity components and their difference. The relative velocity v_B/A is the amount by which Car B's arrow overshoots Car A's.

Scoring (2 points):

▸ 1 point: Starts from the relative-velocity relationship 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 substitution is performed.

▸ 1 point: Correct value +8 m/s, with units and with the direction stated (positive, forward, or in the direction of travel).

⚠︎ Common error (partial credit): Writes the correct relationship but then adds the two velocities, giving 52 m/s — earns 1 of 2 points, for the starting relationship only.

✗ Common error (no credit): Reports a bare 8 m/s with no relationship shown and no sign or direction — neither criterion is met.

Part (b) — Model Answer

The same subtraction, with the observer and the observed object exchanged:

v_A/B = v_A/G − v_B/G

v_A/B = (+22 m/s) − (+30 m/s) = −8 m/s

Car A moves at −8 m/s relative to the passenger in Car B — backward, opposite to the direction of travel. The two results are related by v_A/B = −v_B/A: equal in magnitude, opposite in sign.

Figure 2 — The same physical situation seen from each car. Each observer is at rest in their own frame, and each measures the other car moving at 8 m/s in the opposite sense.

Scoring (1 point):

▸ 1 point: Correct value −8 m/s with units and sign (or 8 m/s stated as backward), together with the statement v_A/B = −v_B/A. Both the value and the stated relationship are required for this point.

✗ Common error (no credit): Reports +8 m/s — the subtraction has been carried out in the wrong order, treating the relative velocity as an unsigned rate of separation rather than a signed component.

Part (c) — Model Answer

After the speed-up, Car A travels at a constant v_A/G = +28 m/s while Car B is still at a constant v_B/G = +30 m/s. Both cars are again moving at constant velocity, so the same subtraction applies:

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

Car B now moves at only +2 m/s relative to the passenger in Car A — still forward, but pulling away four times more slowly than before.

The part (a) result no longer describes what that passenger measures, because it was built from Car A's earlier ground-frame velocity of +22 m/s. Car B's ground-frame velocity has not changed at all, so the whole change in the measured relative velocity comes from the change in the observer's own motion.

Figure 3 — Ground-frame velocities of both cars. The vertical separation between the two lines is v_B/A: 8 m/s before the speed-up and 2 m/s after it.

Scoring (2 points):

▸ 1 point: Correct value +2 m/s, with units and direction, obtained from the velocities that apply after the speed-up.

▸ 1 point: Justification identifying that Car A's ground-frame velocity has changed while Car B's has not, so the part (a) value belongs to the earlier stage only.

⚠︎ Common error (partial credit): Correct value +2 m/s with no justification, or with a justification that only restates the arithmetic — earns 1 of 2 points.

✗ Common error (no credit): Answers +8 m/s by carrying part (a) forward — the observer's own velocity has changed, so the earlier relative velocity no longer describes what the passenger in Car A measures.

Part (d) — Model Answer

Student 1's conclusion is correct and Student 2's is not. In the first stage the pair is v_B/A = +8 m/s and v_A/B = −8 m/s. After the speed-up the same subtraction gives v_B/A = +2 m/s and v_A/B = (+28 m/s) − (+30 m/s) = −2 m/s. In both stages the two measurements are equal in magnitude and opposite in sign, so the speed-up changed the value of the relative velocity but not the relationship between the two measurements.

The relationship follows from how the two quantities are defined. Both are the same pair of ground-frame velocities, subtracted in opposite order:

v_B/A = v_B/G − v_A/G         and         v_A/B = v_A/G − v_B/G

Exchanging the two terms reverses the sign of the result, so v_A/B = −v_B/A follows by algebra alone. Nothing in that derivation refers to the road, to the two cars sharing a direction, or to which car is faster — which is why Student 1's stated reason does not hold up even though the conclusion does. The same result would follow for two objects moving in opposite directions, or for two objects that share no road at all.

The condition the relationship depends on is that both velocities are measured at the same instant and referred to the same third frame — here, the ground. Ignore that condition and the relationship appears to fail: pairing v_B/A = +8 m/s from the first stage with v_A/B = −2 m/s from the second is not a sign reversal, because the two numbers describe different moments. That is the grain of truth in Student 2's claim — the relative velocity genuinely does change when Car A's velocity changes — but a change in the value is not a failure of the relationship, because both members of the pair change together.

Scoring (3 points):

▸ 1 point: Identifies that Student 1's conclusion is correct and Student 2's is not, and supports it with the calculated pairs — +8 m/s with −8 m/s in the first stage, +2 m/s with −2 m/s after the speed-up.

▸ 1 point: Explains the relationship that produces it — v_B/A = v_B/G − v_A/G and v_A/B = v_A/G − v_B/G are the same two velocities subtracted in opposite order, so one is the negative of the other by algebra — and notes that Student 1's stated reason, the shared road and shared direction, plays no part in that derivation.

▸ 1 point: States the condition the relationship depends on — both velocities measured at the same instant and referred to the same frame — and shows what goes wrong without it, for example that +8 m/s from the first stage and −2 m/s from the second are not a sign-reversed pair.

⚠︎ Common error (partial credit): Names the correct student and quotes the values, but never states the relationship or the condition — earns 1 of 3 points.

⚠︎ Common error (partial credit): Gives the full algebraic argument but leaves Student 1's shared-road reasoning unchallenged, so an incorrect reason is allowed to stand as the cause — earns 2 of 3 points.

✗ Common error (no credit): Sides with Student 2 on the grounds that 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.