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FRQ: Instantaneous Velocity from a Shrinking Interval — The Sonic-Ranger Record

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

In a school laboratory, a motorised toy car runs along a straight, horizontal 10 m test track. A sonic ranger mounted at the start line records the car’s position x, in metres, measured from the start line, at time t, in seconds, after release.

The car is released from rest at the start line, so x_0 = 0 and v_x0 = 0. Over the whole of the recorded run the data are well described by

x = Ct²

where C is a positive constant with units of m/s². For this particular car the recording gives C = 0.5 m/s². The ranger records from release until the car reaches the 10 m mark. Figure 1 shows the recorded position–time curve.

Figure 1 — The recorded position–time curve, release to the 10 m mark, with t = 2.0 s marked.

All motion is one-dimensional along the track. Take the positive x-direction as the direction of the car’s motion, use the ground as the reference frame, and treat the car as a point object. Write the car’s average velocity over an interval as v_avg and its instantaneous velocity at a single instant as v_x.

✎ Free Response Questions

Part A

A student analysing the recording picks the interval that begins at t = 2.0 s and lasts 1.0 s. Write the car’s instantaneous velocity at t = 2.0 s as v_2.

Indicate whether the car’s average velocity v_avg over that interval is greater than, less than, or equal to v_2 by writing one of the following.

•   v_avg > v_2

•   v_avg < v_2

•   v_avg = v_2

Justify your answer using qualitative reasoning beyond referencing equations.

Part B

Now consider a general interval within the recorded run: one that begins at an arbitrary time t and lasts for a time Δt.

Starting with the definition of average velocity, derive an expression for the car’s average velocity over that interval. Express your final answer in terms of C, t, Δt, 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_avg > v_2

The instantaneous velocity at t = 2.0 s is the slope of the tangent to the position–time record there. The average velocity over the interval is the slope of the chord, the straight line joining the record at the two ends of the interval. Both lines start from the same point, at t = 2.0 s.

Figure 2 — The chord across the part A interval rises more steeply than the tangent at 2.0 s.

The record curves upward: the car is speeding up, so at every later instant of the interval it moves faster than v_2. Because the record steepens throughout the interval, it bends up away from the tangent, so the chord’s far end, on the record at t = 3.0 s, lies above the tangent line and the chord rises more steeply. In terms of the motion, the average over the interval is made up of velocities that are all, after its first instant, greater than v_2. When a car’s velocity increases throughout an interval, its average velocity over the interval is greater than its velocity at the start of it.

Scoring (3 points):

▸ A1 — 1 point: For indicating v_avg > v_2.

▸ A2 — 1 point: For a justification that indicates that the car’s velocity increases during the interval — that it is speeding up (it moves in the positive direction throughout, so speeding up is an increase in its velocity), or that the record’s growing steepness shows the car moving faster as time increases.

▸ A3 — 1 point: For a justification that indicates that the average velocity over the interval reflects the car’s faster motion after t = 2.0 s: that it is set by the velocities during the interval, which after t = 2.0 s all exceed v_2; that it is the slope of the chord across the interval, which rises more steeply than the tangent at its start, whose slope is v_2; or that, with a constant acceleration, it equals the mean of the velocities at the interval’s two ends, or the velocity at its middle, either of which exceeds v_2.

Scoring Note: A2 and A3 are content criteria, not route criteria, and they may be earned from different routes. An equation or a calculation alone — for example v_avg = 2.50 m/s against v_2 = 2.0 m/s — does not meet A2 or A3; the response must also state in words the physical idea it is used to show. A1, A2 and A3 are scored independently of one another.

⚠︎ Common error (partial credit): Indicates v_avg > v_2 because the car is speeding up, but never relates the average over the interval to the motion within it — not to the velocities during the interval or at its two ends, the chord across it or the velocity at its middle — earns 2 of 3 points, A1 and A2.

⚠︎ Common error (partial credit): Indicates v_avg > v_2 from the values v_avg = 2.50 m/s and v_2 = 2.0 m/s alone, with no reasoning in words — earns 1 of 3 points, A1. The prompt asks for reasoning beyond referencing equations.

⚠︎ Common error (partial credit): Notes that the car is speeding up but indicates v_avg = v_2, because a 1.0 s interval is short enough for the average velocity to be the instantaneous one — earns 1 of 3 points, A2. The average approaches the instantaneous velocity only as the interval shrinks toward the instant; over this full second the car’s velocity rises by 1.0 m/s.

✗ Common error (no credit): Reads the record as a picture of the track — a hill that gets steeper — and indicates v_avg < v_2 because the car slows down as it climbs — the steepening is read as the shape of the track, not as the car moving faster, and the record is a graph of position against time, so no criterion is earned.

Part B — Model Answer

Average velocity is the displacement divided by the time interval over which it occurs, and the displacement is the change in position between the two instants that bound the interval:

v_avg = Δx / Δt         Δx = x(t + Δt) − x(t)

Substitute the recorded position function x = Ct² at each instant, then expand the square, keeping the cross term:

Δx = C(t + Δt)² − Ct²
Δx = C(t² + 2tΔt + Δt²) − Ct²
Δx = 2CtΔt + CΔt²

Divide every term of that displacement by Δt:

v_avg = (2CtΔt + CΔt²) / Δt = 2Ct + CΔt = C(2t + Δt)

The average velocity over an interval of length Δt beginning at time t is C(2t + Δt), in the positive x-direction. Δt cancels from the first term but survives in the second because the car’s velocity changes during the interval. Average velocity is the change in position divided by the time interval, so it needs the position at both ends of the interval, not the velocity at one instant.

Scoring (3 points):

▸ B1 — 1 point: For a multistep derivation that includes the definition of average velocity, v_avg = Δx/Δt, or an equivalent form such as Δx = v_avg Δt. This point is earned for the starting principle alone, even if the steps that follow are incorrect; a final expression for v_avg on its own is not a multistep derivation.

▸ B2 — 1 point: For the displacement over the interval using only C, t and Δt, Δx = C(t + Δt)² − Ct² or any equivalent form, however it is reached, such as 2CtΔt + CΔt² or CΔt(2t + Δt); Δx = x(t + Δt) − x(t) with x = Ct² never applied does not earn this point. A correct final expression for v_avg in terms of C, t and Δt also indicates this relation, unless the response’s own displacement contradicts it.

▸ B3 — 1 point: For a correct expression for v_avg in terms of C, t and Δt, such as 2Ct + CΔt or C(2t + Δt), consistent with the indicated displacement. Where B2 was not earned, an expression using only C, t and Δt that follows correctly from dividing by Δt the response’s own displacement for the interval from t to t + Δt, obtained from x = Ct² directly or through a velocity and acceleration the response takes from it, also earns this point; an incorrect expression with any other error does not.

Scoring Note: A correct, isolated, final expression for v_avg in terms of C, t and Δt earns points B2 and B3. A final expression preceded only by words or by unapplied equations from the reference information is isolated and is not a multistep derivation; words or equations that state an expression for the displacement over the interval, correct or not, make a final expression not isolated. B1 and B2 carry no consistency credit, and B1, B2 and B3 do not depend on the response in part A.

⚠︎ Common error (partial credit): Writes v_avg = Δx/Δt and Δx = C(t + Δt)² − Ct², but expands (t + Δt)² as t² + Δt², dropping the cross term 2tΔt, and gives v_avg = CΔt — earns 2 of 3 points, B1 and B2. That average would not depend on when the interval begins.

⚠︎ Common error (partial credit): Writes v_avg = Δx/Δt and reaches Δx = 2CtΔt + CΔt² correctly, but divides only the first term by Δt, giving v_avg = 2Ct + CΔt² — earns 2 of 3 points, B1 and B2. The second term then has the units of a length, not of a velocity.

⚠︎ Common error (partial credit): Averages the velocities at the two ends of the interval, ½[2Ct + 2C(t + Δt)] = C(2t + Δt), which is valid because the acceleration is constant, but never writes the definition of average velocity — earns 2 of 3 points, B2 and B3. B2 is met through its final-expression clause; the prompt names the definition of average velocity as the starting point.

⚠︎ Common error (partial credit): Writes v_avg = Δx/Δt but takes C as the acceleration, so that v_x = Ct at the start of the interval and Δx = CtΔt + ½CΔt², giving v_avg = Ct + ½CΔt — earns 2 of 3 points, B1 and B3. x = Ct² matches x = ½a_x t² with a_x = 2C, so this is half the true average velocity.

⚠︎ Common error (partial credit): Applies v_avg = Δx/Δt to the interval from release to t + Δt instead, giving C(t + Δt)²/(t + Δt) = C(t + Δt) — earns 1 of 3 points, B1. The interval the question defines begins at t, not at release.

✗ Common error (no credit): Matches x = Ct² with x = x_0 + v_x0 t + ½a_x t² to get a_x = 2C, then gives v_x = v_x0 + a_x t = 2Ct as the average velocity over the interval — 2Ct is the velocity at the single instant t, no displacement is divided by a time interval, and Δt never appears, so no criterion is earned.

Part C — Model Answer

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

Written out, v_avg = 2Ct + CΔt. The first term contains no Δt, and the second is directly proportional to Δt. As the interval is made shorter, CΔt shrinks toward zero while 2Ct is untouched, so the average velocity approaches 2Ct: the value an average over a very short interval approaches is the instantaneous velocity at time t, v_x = 2Ct. For any interval of nonzero length CΔt is positive, so the average over the interval is greater than the instantaneous velocity at its start. At t = 2.0 s, 2Ct = 2.0 m/s is v_2, and the 1.0 s interval gives v_avg = 0.5(4.0 + 1.0) = 2.50 m/s, greater than v_2, as part A concluded from the car speeding up.

For intervals beginning at t = 2.0 s, v_avg plotted against Δt is a straight line of slope C that heads for v_2 = 2.0 m/s on the v_avg axis, a value approached as the interval shrinks but never reached.

Figure 3 — v_avg for intervals beginning at t = 2.0 s, against the interval length Δt.

The term CΔt is the whole gap between the average velocity and the velocity at the start of the interval, and it shrinks in proportion to the interval. It comes from the speeding up part A reasoned from: v_x = 2Ct rises at a steady a_x = 2C = 1.0 m/s², twice the fitted constant and not C itself, and CΔt = ½a_xΔt is half the velocity the car gains during the interval — 0.50 m/s for the part A interval.

Scoring (2 points):

▸ C1 — 1 point: For attempting to address the functional dependence of v_avg on Δt in the equation the response derived in part B.

Scoring Note: It is not necessary to use the functional dependence correctly to earn this point. The response only needs functional-dependence language — such as proportional, depends on, increases, decreases, contains Δt or does not contain Δt — to relate v_avg and Δt.

▸ C2 — 1 point: For correctly using the dependence of v_avg on Δt in the equation the response derived in part B to compare v_avg with the instantaneous velocity at the start of the interval, and stating the verdict — consistent or not consistent — that follows from it together with the response’s own part A answer or reasoning. Attributing the difference between v_avg and that instantaneous velocity to the Δt term is enough; the trend with Δt need not be stated. The instantaneous velocity may be taken as the value the response’s expression approaches as the interval shrinks, from the constant-acceleration relationships, or from the slope of the tangent, and the verdict must follow from the value taken, so for v_avg = CΔt, C(t + Δt) or Ct + ½CΔt either verdict can earn this point. For the correct expression, v_avg exceeds 2Ct, the instantaneous velocity at time t, by CΔt, which is positive for any interval, so v_avg > v_2: consistent with a part A answer of v_avg > v_2, not with v_avg < v_2 or v_avg = v_2.

Scoring Note: The comparison may be shown by evaluating the expression for the part A interval against v_2 = 2.0 m/s and attributing the difference to the Δt term, or by reading v_avg = 2C(t + Δt/2) as the velocity at the later instant t + Δt/2. The acceleration a_x = 2C is not required for either point.

⚠︎ Common error (partial credit): States how v_avg depends on Δt, but gives no verdict on whether this agrees with part A — earns 1 of 2 points, C1.

⚠︎ Common error (partial credit): States that v_avg increases with Δt and calls this consistent with part A, but never ties that dependence to v_2 by any of C2’s routes — earns 1 of 2 points, C1.

⚠︎ Common error (partial credit): Reads the whole of the correct expression, C(2t + Δt), as shrinking to zero with the interval, and concludes that the car’s velocity at an instant is zero, so the equation is not consistent with part A — earns 1 of 2 points, C1. The 2Ct term contains no Δt and does not shrink.

✗ Common error (no credit): Says the two parts agree because both give a larger average velocity, or quotes 2.50 m/s against 2.0 m/s, with nothing said about how v_avg depends on Δt — earns 0 of 2 points.

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