FRQ: Two Vehicles on a Straight Highway 

Assessments aligned to 2026 AP Physics 1 standards

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

Scenario

Car A and Car B travel along a straight, level highway. Both cars move in the same direction, which is defined as positive. Car A moves at a constant +22 m/s relative to the ground. Car B moves at a constant +30 m/s relative to the ground. A traffic sensor mounted on the road (the ground frame) records both velocities. A passenger in Car A also observes Car B.

All motion is one-dimensional. Use the ground as the reference frame unless otherwise specified.

(a) Calculate the velocity of Car B as measured by the passenger in Car A. Show your work using the relative velocity framework.

(b) Calculate the velocity of Car A as measured by a passenger in Car B. Show your work. State the relationship between your answer here and your answer in part (a), and explain why that relationship holds physically.

(c) Car A now accelerates from +22 m/s to +28 m/s over 3.0 s, while Car B continues at constant +30 m/s. During Car A's acceleration phase, predict the velocity of Car B as measured from Car A at the end of the 3.0 s interval. Justify your answer.

(d) A student argues: 'The velocity of Car B relative to Car A must always equal the velocity of Car A relative to Car B, just with the opposite sign, because both cars are moving on the same road.' A second student says: 'This is only true if both cars are in inertial frames.' Evaluate both students' claims. Identify which student is correct and construct a complete argument in support of that position.

Answer Key & Scoring Guide

Part (a) — Model Answer

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

Using the relative velocity relationship:

v_B/A = v_B/G − v_A/G

v_B/A = (+30) − (+22) = +8 m/s

Car B moves at +8 m/s (forward) relative to the passenger in Car A. Car B is slowly pulling ahead.

Scoring (3 points):

✔️1 point: Correct setup — v_B/A = v_B/G − v_A/G (or equivalent derivation from the addition rule)

✔️1 point: Correct substitution with signs

✔️1 point: Correct answer +8 m/s with direction stated (units required)

✔️Common error: Adding instead of subtracting (30 + 22 = 52 m/s) — earns 1 point for correct formula setup only.

Part (b) — Model Answer

Using the relative velocity relationship:

v_A/B = v_A/G − v_B/G

v_A/B = (+22) − (+30) = −8 m/s

Car A moves at −8 m/s (backward) relative to a passenger in Car B.

Relationship: v_A/B = −v_B/A. The two velocities are equal in magnitude and opposite in sign. This holds physically because relative motion is symmetric: if Car A sees Car B advancing at +8 m/s, then Car B must see Car A receding at −8 m/s. The physical situation is a mirror image from each observer's perspective — the gap between the two cars closes (or opens) at the same rate regardless of who is watching.

Scoring (3 points):

✔️1 point: Correct calculation: −8 m/s

✔️1 point: States that v_A/B = −v_B/A (equal magnitude, opposite sign)

✔️1 point: Physical justification — symmetry of relative motion, both observers see the same rate of separation/approach

✔️No credit for the relationship without a physical explanation.

Part (c) — Model Answer

After 3.0 s, Car A's velocity relative to the ground:

v_A/G (final) = +28 m/s

Car B still moves at +30 m/s relative to the ground throughout (constant velocity).

Velocity of Car B relative to Car A at the end of the interval:

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

Justification: Even though Car A was accelerating during the 3.0 s, the question asks for the velocity at the end of the interval — at that instant, Car A is moving at +28 m/s and we apply the standard relative velocity formula using the instantaneous velocities. Car B is still +2 m/s ahead of Car A in terms of their relative velocity.

Scoring (2 points):

✔️1 point: Correct final velocity of Car A used (+28 m/s)

✔️1 point: Correct answer +2 m/s with justification that instantaneous velocities are used at the end of the interval

✔️Common error: Using Car A's initial velocity (+22 m/s) giving +8 m/s — earns 0 points; student did not account for the acceleration.

Part (d) — Model Answer

The first student is correct that v_A/B = −v_B/A always — but for the wrong reason. The second student's additional condition (inertial frames) is not required for this relationship to hold.

The relationship v_A/B = −v_B/A is a mathematical consequence of the definition of relative velocity: v_A/B = v_A/G − v_B/G, and v_B/A = v_B/G − v_A/G. These two expressions are always negatives of each other by algebra, regardless of whether the frames are inertial or not. The first student's conclusion is correct, though the reasoning ('both on the same road') is not precise — it should reference the definition of relative velocity, not the physical medium.

The second student is incorrect in claiming the inertial frame condition is required for this specific relationship. The sign-reversal symmetry v_A/B = −v_B/A is a kinematic identity, not a dynamic law. It does not depend on Newton's Laws and therefore does not require inertial frames.

Scoring (2 points):

✔️1 point: Correctly identifies that the first student's conclusion is right (v_A/B = −v_B/A always holds) and that the second student's condition is unnecessary for this specific relationship

✔️1 point: Correct argument — the relationship follows from the algebraic definition of relative velocity and is a kinematic identity, independent of whether frames are inertial

✔️Partial credit (1 point): Correctly identifies which student is right without a full algebraic or physical justification.

✔️Note: A student who says the second student is correct because 'inertial frames are always required for relative velocity' confuses the kinematic identity with the acceleration-invariance principle — this is a critical conceptual error, earning 0 points.