What Is the Velocity at a Turning Point? Position-Time Graph FRQ
Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.
FRQ: Velocity from a Position–Time Graph — The Proving-Ground Run
Assessments aligned to 2026 AP Physics 1 standards
Mathematical Routines (MR) | MID-LEVEL | 10 points | 25 min
▤ Scenario
At a vehicle proving ground, an engineer runs a prototype electric shuttle along a straight, level test lane. The controller holds the same constant acceleration for the whole 8.0-second run.
The onboard data logger records the shuttle’s position x, in metres, as a function of time t, in seconds:
x = −2t² + 16t
valid over 0 ≤ t ≤ 8.0 s. Figure 1 shows the logged position–time graph.
Figure 1 — The logged position of the shuttle against time for the whole run.
All motion is one-dimensional along the x-axis. Take the shuttle’s initial direction of travel as positive x, the origin at the start line, and the test lane (the ground) as the reference frame. Treat the shuttle as a point object.
✎ Free Response Questions
Part A
i. On Figure 1, draw a line tangent to the logged curve at each of t = 2.0 s, t = 4.0 s and t = 6.0 s. Then, for each of the three instants, state the direction of the shuttle’s motion, if it has one, and name the feature of the tangent that tells you this.
ii. The engineer wants the shuttle’s velocity at the single instant t = 4.0 s. Calculate the shuttle’s average velocity over the interval from t = 3.0 s to t = 5.0 s, and again over the narrower interval from t = 3.9 s to t = 4.1 s. Using both results, state what the instantaneous velocity at t = 4.0 s must be, and explain the reasoning that leads to that value.
iii. Derive an expression for the shuttle’s velocity v_x as a function of time t. Express your final answer in terms of t, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
Part B
A student examines Figure 1 and claims: “The shuttle is slowing down for the entire 8.0-second run.”
Indicate whether the speed of the shuttle at t = 6.0 s is greater than, less than, or equal to its speed at t = 2.0 s by writing one of the following.
• speed at t = 6.0 s > speed at t = 2.0 s
• speed at t = 6.0 s < speed at t = 2.0 s
• speed at t = 6.0 s = speed at t = 2.0 s
Justify your response.
Identify the part of the run, if any, for which the student’s claim is incorrect, and explain why it is incorrect there.
❖ 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 (i) — Model Answer
At t = 2.0 s the curve rises to the right, so the tangent slopes upward. A positive slope on a position–time graph is a positive velocity, so the shuttle is moving in the +x direction.
At t = 4.0 s the curve reaches its maximum and the tangent is horizontal. A zero slope is a zero velocity, so the shuttle is momentarily at rest — at a single instant, not over an interval.
At t = 6.0 s the curve falls to the right, so the tangent slopes downward. A negative slope is a negative velocity, so the shuttle is moving in the −x direction, back toward the start line. The slope of the tangent is the velocity, and its sign is the direction of motion. The height of the curve is not: at t = 6.0 s the position is still +24 m, positive, while the velocity is negative.
Figure 2 — The three tangents on Figure 1. Their slopes give the velocity at each instant.
Scoring (2 points):
▸ A1 — 1 point: For a straight tangent line at each of the three instants: sloping upward at t = 2.0 s, horizontal at the curve’s maximum at t = 4.0 s, and sloping downward at t = 6.0 s.
▸ A2 — 1 point: For the direction at each instant, each attributed to the slope at that instant: +x at t = 2.0 s, no direction at t = 4.0 s (at rest at that instant), and −x at t = 6.0 s.
Scoring Note: Accept “slope”, “gradient”, “sign of the slope”, “positive, zero and negative slope”, or whether the tangent rises or falls to the right, as naming the feature; “forward” or “away from the start line” for +x, and “backward” or “back toward the start line” for −x; and “at rest”, “v_x = 0” or “turning around” at t = 4.0 s. A1 and A2 are scored independently of each other.
⚠︎ Common error (partial credit): Draws the three tangents correctly but describes the instants as “slowing down, stopped, speeding up” — earns 1 of 2 points, A1. Those words describe the speed, which falls to zero and rises again; the prompt asks for the direction of motion along the x-axis, which the sign of the slope gives.
⚠︎ Common error (partial credit): Draws the three tangents correctly but gives +x as the direction at t = 6.0 s because the position there, +24 m, is positive — earns 1 of 2 points, A1. The sign of the position says where the shuttle is, not which way it is moving.
⚠︎ Common error (partial credit): Gives all three directions correctly, each attributed to the slope, but draws curved arrows along the curve instead of straight tangent lines — earns 1 of 2 points, A2. A tangent is a straight line that touches the curve at its instant.
✗ Common error (no credit): Reads the curve as a picture of the lane — a hill the shuttle drives up and down — drawing curved arrows along it and describing the instants as uphill, at the summit and downhill — no straight tangent is drawn and no direction along the x-axis is named, so neither A1 nor A2 is earned.
Part A (ii) — Model Answer
Average velocity is the displacement divided by the time interval. Evaluate the logged position equation at each end of each interval:
x(3.0) = −2(3.0)² + 16(3.0) = −18 + 48 = +30 m
x(5.0) = −2(5.0)² + 16(5.0) = −50 + 80 = +30 m
v_avg = [(+30 m) − (+30 m)] / (5.0 s − 3.0 s) = 0 m/s
x(3.9) = −2(3.9)² + 16(3.9) = −30.42 + 62.4 = +31.98 m
x(4.1) = −2(4.1)² + 16(4.1) = −33.62 + 65.6 = +31.98 m
v_avg = [(+31.98 m) − (+31.98 m)] / (4.1 s − 3.9 s) = 0 m/s
Both intervals are centred on t = 4.0 s, and each gives exactly 0 m/s because the positions at its two ends are equal. That holds for every interval centred on t = 4.0 s: for a half-width h, x(4.0 − h) and x(4.0 + h) are both 32 − 2h². As an interval closes on an instant, its average velocity approaches the instantaneous velocity at that instant; here every centred average is already 0 m/s, so the instantaneous velocity at t = 4.0 s is 0 m/s. Zero displacement over an interval is not the same claim as zero velocity at an instant: over the wider interval the shuttle moved 2 m out and 2 m back.
Scoring (2 points):
▸ A3 — 1 point: For both average velocities, 0 m/s, each obtained as the change in position over its interval divided by the interval’s duration.
▸ A4 — 1 point: For stating the instantaneous velocity at t = 4.0 s with a reason drawn from the averages — that the average velocity approaches the instantaneous velocity as the interval narrows toward t = 4.0 s, or that, with a constant acceleration (stated or used), the average velocity over an interval centred on t = 4.0 s equals the velocity at t = 4.0 s (equivalently, that equal end positions put the turning point at the interval’s centre) — consistent with the response’s own averages, each found as a change in position divided by the interval’s duration. The expected value is 0 m/s.
Scoring Note: A value taken only from the horizontal tangent of part A (i) or from the shape of the curve, or justified only by the two averages being equal with nothing linking an average to the velocity at an instant, does not meet A4. A3 and A4 do not depend on the response in part A (i).
⚠︎ Common error (partial credit): Obtains both averages correctly but concludes that the shuttle “did not move” during either interval, and gives that as the reason the velocity at t = 4.0 s is 0 m/s — earns 1 of 2 points, A3. A zero displacement over an interval is not a zero velocity: over the wider interval the shuttle moved 2 m out and 2 m back.
⚠︎ Common error (partial credit): Obtains both averages correctly and gives 0 m/s for the instantaneous velocity, citing only the horizontal tangent of part A (i) — earns 1 of 2 points, A3. The prompt asks what value the two averages lead to.
✗ Common error (no credit): Reads the height of the curve as the velocity — 30 m/s at t = 3.0 s and t = 5.0 s, rising to 32 m/s at t = 4.0 s, “the fastest point of the run” — and averages those heights; the height of a position–time graph is a position, not a velocity, and no change in position is divided by a time interval, so neither A3 nor A4 is earned.
Part A (iii) — Model Answer
The acceleration is constant, so the position follows the relationship from the reference information
x = x_0 + v_x0 t + ½a_x t²
Matching its constant term, its t term and its t² term against the logged equation x = −2t² + 16t:
x_0 = 0 v_x0 = +16 m/s ½a_x = −2 m/s² ⇒ a_x = −4.0 m/s²
The velocity relationship from the same reference information then gives
v_x = v_x0 + a_x t = 16 − 4.0t (v_x in m/s, t in s)
The coefficient of t² is half the acceleration, so the acceleration is −4.0 m/s², not −2 m/s². As a check, v_x(4.0) = 16 − 4.0(4.0) = 0 m/s, the zero velocity that the horizontal tangent in part A (i) and the averages in part A (ii) both gave.
Scoring (3 points):
▸ A5 — 1 point: For a multistep derivation that includes either a constant-acceleration relationship from the reference information applied to this motion — for example x = x_0 + v_x0 t + ½a_x t² compared term by term with the logged equation, or v_x = v_x0 + a_x t with the shuttle’s values substituted — or the definition of average velocity applied to an interval centred on t, or to an interval that is then narrowed toward t. The point is earned once the relationship is applied to this motion, even if the steps that follow are incorrect; a final expression for v_x on its own is not a multistep derivation.
▸ A6 — 1 point: For the shuttle’s acceleration. The expected value is a_x = −4.0 m/s², twice the coefficient of t² in the logged equation, and any valid route to it earns the point. A correct final expression reached by a valid route that never needs the acceleration as a separate value, such as an average velocity over an interval centred on t, also meets A6.
▸ A7 — 1 point: For a correct expression for v_x in terms of t: v_x = 16 − 4.0t (v_x in m/s, t in s), or any algebraic equivalent; units need not be written. Where A6 was not earned, an expression v_x = 16 + a_x t built from the response’s own acceleration, found by applying a relationship from the reference information, or the definition of acceleration, to the logged motion rather than read back off the response’s own velocity expression, also earns this point; no other incorrect expression does.
Scoring Note: A correct, isolated, final expression for v_x earns points A6 and A7. A final expression preceded only by words or by unapplied equations from the reference information is isolated and is not a multistep derivation. The average velocity over the whole interval from t = 0 to t (v_x = x/t) is neither centred on t nor narrowed toward it, and does not meet A5. A5, A6 and A7 do not depend on parts A (i) and A (ii).
⚠︎ Common error (partial credit): Matches x = x_0 + v_x0 t + ½a_x t² to the logged equation but reads a_x = −2 m/s² straight off the coefficient of t², dropping the ½, and writes v_x = 16 − 2t — earns 2 of 3 points, A5 and A7. That expression gives v_x(4.0) = +8 m/s, which contradicts both earlier parts.
⚠︎ Common error (partial credit): Finds v_x0 = +16 m/s and a_x = −4.0 m/s² by matching x = x_0 + v_x0 t + ½a_x t² to the logged equation, but assembles the velocity as v_x = 16t − 4.0t², which is not v_x0 + a_x t — earns 2 of 3 points, A5 and A6.
✗ Common error (no credit): Writes v_x = x/t, dividing the position by the time since the start, to give v_x = 16 − 2t — x/t is the average velocity over the whole interval from t = 0 to t, which is neither centred on t nor narrowed toward it, so it is not the velocity at the instant t; no constant-acceleration relationship is applied either, so no criterion is earned.
Part B — Model Answer
speed at t = 6.0 s = speed at t = 2.0 s
The tangents at t = 2.0 s and t = 6.0 s in Figure 2 are equally steep, one rising and one falling. The two instants lie the same 2.0 s either side of the turning point at t = 4.0 s, and with the acceleration constant, as the Scenario states, the speed lost in the 2.0 s before the turn is exactly the speed regained in the 2.0 s after it. The velocities are equal in magnitude and opposite in sign, and speed is the magnitude of velocity, so the two speeds are equal. The expression from part A (iii) confirms it: v_x(2.0) = +8.0 m/s and v_x(6.0) = −8.0 m/s, both speeds 8.0 m/s.
Figure 3 — Velocity and speed against time. v_x falls all run; the speed falls, then rises.
The claim is correct for 0 ≤ t < 4.0 s and incorrect for 4.0 s < t ≤ 8.0 s. Over that second stretch v_x is negative and becoming more negative, so the velocity keeps decreasing while the speed climbs from 0 m/s at the turning point to 16 m/s at the end of the run. “Slowing down” describes the speed, |v_x|, not the velocity v_x. With a constant acceleration, the shuttle slows while v_x and a_x have opposite signs and speeds up once they have the same sign.
Scoring (3 points):
▸ B1 — 1 point: For indicating that the speed at t = 6.0 s is equal to the speed at t = 2.0 s.
▸ B2 — 1 point: For a justification that indicates that the velocities at the two instants have equal magnitudes, based on one of the following: the values +8.0 m/s and −8.0 m/s identified as having the same magnitude; the tangents at t = 2.0 s and t = 6.0 s equally steep, one rising and one falling; the two instants equally far before and after the turning point while the acceleration is constant (stated or used, for example “the velocity changes by the same amount each second”); or equal positions at the two instants linked to equal speeds by a constant-acceleration relationship, such as v_x² = v_x0² + 2a_x(x − x_0).
▸ B3 — 1 point: For identifying the part of the run after t = 4.0 s (4.0 s < t ≤ 8.0 s) as where the claim is incorrect, with the reason that the speed increases there — for example that v_x and a_x have the same sign after the turning point, or that the speed climbs from 0 m/s to 16 m/s. An interval written as 4.0 s ≤ t ≤ 8.0 s, “from t = 4.0 s to the end of the run” or “after the turning point” is accepted as the interval; the reason is still required, and the instant t = 4.0 s itself is not scored.
Scoring Note: B1, B2 and B3 are scored independently of one another and of part A, and carry no consistency credit from part A (iii): the comparison can be made from Figure 1 directly. Equal heights of the curve (x = +24 m at both instants) on their own do not meet B2, since the height of a position–time graph is a position, not a speed; nor does listing +8.0 m/s and −8.0 m/s without identifying their magnitudes as equal. A statement that the shuttle turns around or moves in the −x direction after t = 4.0 s, with nothing said about its speed, does not meet B3.
⚠︎ Common error (partial credit): Indicates equal speeds, justified by +8.0 m/s and −8.0 m/s having the same magnitude, but says the claim fails after t = 4.0 s only because the shuttle is then moving backward — earns 2 of 3 points, B1 and B2. Moving in the −x direction is not speeding up: the claim fails there because the speed rises.
⚠︎ Common error (partial credit): Indicates equal speeds because the curve is at the same height, 24 m, at both instants, and says the claim fails before t = 4.0 s, where the curve rises — earns 1 of 3 points, B1. The height of a position–time graph is a position; the speed is the steepness of the tangent, which is the same at the two instants.
✗ Common error (no credit): Indicates that the speed at t = 6.0 s is less than the speed at t = 2.0 s because −8.0 m/s is less than +8.0 m/s, and agrees with the student because the acceleration is negative for the whole run — the sign of a_x alone does not decide whether the speed rises or falls, only its sign relative to v_x does, so no criterion is earned.
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