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FRQ: Average Acceleration in Uniform Circular Motion — Tetherball at the Park

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Mathematical Routines (MR)  |  CHALLENGE  |  10 points

▤ Scenario

A child strikes a tetherball at the park, sending it into a horizontal circle at the end of a rope tied to the top of a vertical pole. After a few irregular swings the ball settles into uniform circular motion: it travels a circle of radius r = 1.20 m at a constant speed v = 3.0 m/s, completing every revolution in the same period.

Take the instant at which the ball passes the easternmost point of its circle as the start of the interval of interest. At that instant the +x axis points along the radius from the pole outward toward the ball, and the +y axis points in the direction the ball is moving. The quarter-revolution that follows therefore carries the ball from the +x axis around to the +y axis.

All motion lies in the horizontal x–y plane; use the ground as the reference frame. Treat the rope as massless and inextensible, and assume that over the interval of interest the rope neither winds onto nor unwinds from the pole, so the radius remains fixed at 1.20 m. Air resistance is negligible.

✎ Free Response Questions

(a) On an x–y coordinate system, sketch the ball's velocity vector at the start of the quarter-revolution and its velocity vector at the end of the quarter-revolution. On the same axes, construct the change in velocity Δv by reversing the initial velocity and adding it tip-to-tail to the final velocity. Label each vector with its magnitude, and indicate the direction of the ball's average acceleration over the quarter-revolution, both relative to your axes and relative to the circle.

(b) Starting from the definition of average acceleration, derive a symbolic expression for the magnitude of the ball's average acceleration over a quarter-revolution, in terms of only v and r. Then derive an expression for the ratio of that magnitude to the magnitude of the ball's centripetal acceleration, and state what the form of that ratio tells you about how the two compare for any object in uniform circular motion.

(c) Calculate the period of the ball's motion and the duration of one quarter-revolution. Then calculate the magnitude of the ball's average acceleration over that quarter-revolution and the magnitude of its centripetal acceleration.

(d) A second child watching says, “The ball is moving at constant speed, so it is not accelerating.” Evaluate this claim. Identify the specific physics error in it, justify your correction by reasoning from the definition of acceleration as a vector quantity, and state the condition under which the second child’s rule would in fact be correct, giving one example of motion that satisfies that condition.

❖ Answer Key & Scoring Guide

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

Part (a) — Model Answer

At the start of the quarter-revolution the ball sits on the +x axis, and by the definition of the axes its velocity there points along +y. A quarter-revolution later it sits on the +y axis, where its velocity is tangent to the circle and points along −x. Both velocities have magnitude 3.0 m/s.

v_i = (0, +3.0) m/s
v_f = (−3.0, 0) m/s

Reversing the initial velocity gives −v_i = (0, −3.0) m/s. Placing that vector tip-to-tail on v_f closes the triangle, and the resultant drawn from the tail of v_f to the tip of −v_i is the change in velocity:

Δv = v_f − v_i = (−3.0, −3.0) m/s
|Δv| = √((−3.0)² + (−3.0)²) = √18 = 4.24 m/s

Because a_avg = Δv/Δt and Δt is a positive scalar, the average acceleration points in exactly the same direction as Δv — into the third quadrant, at 225° measured counter-clockwise from +x. Relative to the circle, that is the inward radius drawn at the midpoint of the arc: the midpoint of the quarter-arc lies at 45°, and the line from there to the pole points along 225°. The average acceleration therefore points at the pole from the middle of the arc travelled — not from where the ball starts, and not from where it ends.

Figure 1. Left: the quarter-revolution seen from above, with v_i at the start, v_f at the end, and a_avg drawn inward at the midpoint of the arc. Right: the tip-to-tail construction Δv = v_f + (−v_i), of magnitude 4.24 m/s at 225° from the +x axis.

Scoring (2 points):

▸ 1 point: v_i and v_f drawn in the correct directions — along +y at the start and along −x at the end — and each labelled 3.0 m/s.

▸ 1 point: Δv drawn as the tip-to-tail resultant of v_f and −v_i and labelled 4.24 m/s, with a_avg identified as parallel to Δv and directed at the pole from the midpoint of the arc.

⚠︎ Common error (partial credit): Δv constructed correctly but a_avg stated to point at the pole from the ball's starting position — earns 1 of 2 points; the inward direction is exact only at the midpoint of the arc.

✗ Common error (no credit): drawing the resultant from the tip of v_f to the tip of v_i, which computes v_i − v_f and points the average acceleration outward, away from the pole.

Part (b) — Model Answer

Average acceleration is defined as the change in velocity divided by the interval over which that change occurs:

a_avg = Δv / Δt

The speed is constant, so v_i and v_f both have magnitude v, and a quarter-revolution turns the velocity through 90°, making the two vectors perpendicular. Their difference is therefore the hypotenuse of a right triangle whose legs are both v. The period is the circumference divided by the speed, and a quarter-revolution lasts one quarter of it:

|Δv| = √(v² + v²) = √2 v
Δt = T/4 = (2πr/v)/4 = πr/(2v)

Dividing one by the other gives the magnitude of the average acceleration; dividing that in turn by the centripetal acceleration a_c = v²/r gives the ratio:

|a_avg| = |Δv| / Δt = (√2 v) × (2v)/(πr) = 2√2 v²/(πr)
|a_avg| / a_c = [2√2 v²/(πr)] / [v²/r] = 2√2/π ≈ 0.900

Both v and r cancel. The ratio is a pure number, so for any object in uniform circular motion — any speed, any radius — the average acceleration over a quarter-revolution is 90.0% of the instantaneous centripetal acceleration. The shortfall is not an artifact of this ball's particular values. The instantaneous acceleration keeps a constant magnitude v²/r but sweeps through 90° of direction across the arc, and inward directions that differ in orientation partly cancel when they are combined. Only as the interval shrinks toward zero does the direction stop turning appreciably, and the average then approaches the instantaneous value.

Scoring (3 points):

▸ 1 point: begins from the definition a_avg = Δv/Δt, with Δv understood as a vector subtraction. This point is awarded for the correct starting principle alone, before any algebra is performed.

▸ 1 point: obtains both |Δv| = √2 v, from the perpendicularity of v_i and v_f at equal speed, and Δt = πr/(2v), from T = 2πr/v or equivalently from v = rω with ω = 2π/T.

▸ 1 point: combines them into |a_avg| = 2√2 v²/(πr), forms the ratio 2√2/π, and states that because v and r cancel the result holds for every uniform circular motion, not only this one.

⚠︎ Common error (partial credit): derives |a_avg| correctly but reports the ratio only as the decimal 0.900, without noting that v and r cancel — earns 2 of 3 points; the generality is what the derivation is for.

⚠︎ Common error (partial credit): uses the arc length ¼(2πr) in place of the change in velocity, confusing distance travelled with change in velocity — earns 1 of 3 points if the starting definition is stated correctly.

✗ Common error (no credit): setting |Δv| = v_f − v_i = 0 because both speeds equal v, which subtracts magnitudes instead of vectors and collapses the whole derivation.

Part (c) — Model Answer

The period is the circumference divided by the speed, and a quarter-revolution is a quarter of it:

T = 2πr/v = 2π(1.20 m)/(3.0 m/s) = 2.51 s
Δt = T/4 = 0.628 s

Substituting into the expression derived in part (b), and evaluating the centripetal acceleration directly:

|a_avg| = 2√2 v²/(πr) = 2√2 (3.0 m/s)²/(π × 1.20 m) = 6.75 m/s²
a_c = v²/r = (3.0 m/s)²/(1.20 m) = 7.50 m/s²

The same average follows from |Δv|/Δt = (4.24 m/s)/(0.628 s) = 6.75 m/s². The average over the quarter-revolution is smaller than the instantaneous value, and 6.75/7.50 = 0.900 — the ratio derived in part (b).

Scoring (2 points):

▸ 1 point: correct period T = 2.51 s and correct quarter-revolution time Δt = 0.628 s.

▸ 1 point: correct |a_avg| = 6.75 m/s² and a_c = 7.50 m/s², both with units.

⚠︎ Common error (partial credit): a_c = 7.50 m/s² found correctly but the same value reported for |a_avg|, on the assumption that averaging an acceleration of constant magnitude must return that magnitude — earns 1 of 2 points.

Part (d) — Model Answer

The second child's error is treating speed, a scalar, as though it were velocity, a vector. Constant speed fixes only the magnitude of the velocity. The ball's direction of motion changes continuously as it swings around the pole, so its velocity is changing and the ball is accelerating.

The definition settles it. Acceleration is a_avg = Δv/Δt, and Δv = v_f − v_i is a vector subtraction, so a velocity can change while its magnitude stays fixed, provided its direction changes. That is exactly what parts (a) and (c) show: v_i and v_f both have magnitude 3.0 m/s, yet Δv has magnitude 4.24 m/s and the ball's average acceleration over the quarter-revolution is 6.75 m/s², directed at the pole.

The second child's rule is not simply wrong; it is a correct rule applied outside its condition. “Constant speed means no acceleration” holds only when the direction of motion is constant as well — that is, for motion along a straight line. A ball rolling across level ground in a straight line at a steady 3.0 m/s does have zero acceleration. Bend that same path into a circle of radius 1.20 m and the speed is unchanged while the acceleration becomes 7.50 m/s².

Scoring (3 points):

▸ 1 point: names the specific error — speed (a scalar) treated as equivalent to velocity (a vector) — and concludes that the ball is accelerating.

▸ 1 point: justifies the correction from the vector definition a = Δv/Δt, stating that a change in direction alone makes Δv non-zero.

▸ 1 point: states the condition under which the claim would hold — motion in a straight line, where the direction is constant too — and gives an example of such motion.

⚠︎ Common error (partial credit): concludes correctly that the ball is accelerating “because it is moving in a circle”, without distinguishing speed from velocity — earns 1 of 3 points.

⚠︎ Common error (partial credit): identifies the error and applies the vector definition, but never names the straight-line condition, leaving “constant speed means no acceleration” standing as a general rule — earns 2 of 3 points.

✗ Common error (no credit): arguing that the ball is not accelerating because Δv over a complete revolution is zero — that shows only that the average acceleration over a full revolution is zero, not that the instantaneous acceleration is ever zero.