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

Assessments aligned to 2026 AP Physics 1 standards

Mathematical Routines (MR)  |  CHALLENGE  |  10 points  |  25 min

▤ 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. Use axes fixed to the ground with their origin at the centre of the circle, on the pole: the +x axis points from the pole toward the easternmost point, and the +y axis points in the direction the ball is moving as it passes that point. The quarter-revolution that follows therefore carries the ball from the +x axis around to the +y axis. Figure 1 shows the circle seen from above and, beside it, a velocity grid on which v_i, the ball’s velocity at the start of the quarter-revolution, has already been drawn.

Figure 1 — Left: the circle seen from above. Right: velocity grid, 1 square = 1.0 m/s, with v_i.

Throughout, v_i and v_f name the ball’s velocity at the start and at the end of the quarter-revolution, Δv the change in its velocity over the quarter-revolution, and a_avg its average acceleration over the quarter-revolution. Vertical bars mark a magnitude, so |Δv| is the magnitude of Δv, and v alone is the ball’s speed; symbols that a later part defines as magnitudes, such as a_c, carry no bars.

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 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

Part A

i.    On the velocity grid in Figure 1, draw an arrow representing v_f and an arrow representing Δv.

•   Draw the length of each arrow to represent the magnitude of each vector, consistent with the scale used in Figure 1.

•   Label the arrows v_f and Δv.

Then describe the direction of a_avg relative to the circle.

ii.   Starting from the definition of average acceleration, derive an expression for |a_avg|.

Express your final answer in terms of v, r, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

iii.  Calculate the numerical value of |a_avg|. Then calculate the numerical value of a_c, the magnitude of the ball’s centripetal acceleration.

Part B

Let a_rev be the magnitude of the ball’s average acceleration over the complete revolution that begins at the easternmost point, and let a be the magnitude of the ball’s acceleration at any instant during that revolution.

Indicate whether a_rev is greater than, less than, or equal to a by writing one of the following.

•   a_rev > a

•   a_rev < a

•   a_rev = a

Justify your response.

❖ 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 the end of the quarter-revolution the ball is on the +y axis. Its velocity there is tangent to the circle in the direction of travel, so v_f points along −x. The speed is constant, so v_f has the same 3.0 m/s magnitude as v_i and is drawn three squares long.

v_i = (0, +3.0 m/s)         v_f = (−3.0 m/s, 0)
Δv = v_f − v_i = (−3.0 m/s, −3.0 m/s)
|Δv| = √((−3.0 m/s)² + (−3.0 m/s)²) = 4.24 m/s

On the grid, adding −v_i to v_f tip to tail gives Δv as the arrow from the origin to the point (−3.0 m/s, −3.0 m/s). Drawing it instead from the tip of v_i to the tip of v_f, since v_i + Δv = v_f, gives the same arrow in a different place. Either way Δv points 45° below the −x axis and is about 4.2 squares long, longer than either velocity.

Because a_avg = Δv/Δt and Δt is a positive scalar, a_avg points the same way as Δv: 45° below the −x axis, or 225° counterclockwise from +x. The midpoint of the quarter-arc lies at 45° from +x, and the radius drawn from that midpoint to the pole points along 225°. So the average acceleration points at the pole from the middle of the arc travelled, not from where the ball starts or where it ends.

Figure 2 — Left: v_i, v_f and a_avg on the circle. Right: Δv = v_f + (−v_i), 4.24 m/s, toward 225°.

Scoring (3 points):

▸ A1 — 1 point: For an arrow v_f pointing in the −x direction with the same length as v_i, representing 3.0 m/s on the scale. A1 carries no consistency clause: v_i is supplied in Figure 1.

▸ A2 — 1 point: For an arrow Δv drawn either correctly or with components equal to those of the response’s own v_f minus v_i. The correct Δv has components −3.0 m/s and −3.0 m/s, so it points 45° below the −x axis and is longer than v_i.

Scoring Note: Δv may be drawn anywhere on the grid — from the origin, or from the tip of v_i to the tip of v_f. A response that draws no Δv arrow does not earn A2.

▸ A3 — 1 point: For relating the direction of a_avg to the circle, either correctly or consistently with a_avg having the direction of the response’s own Δv. The correct relation is that a_avg points toward the pole along the radius through the midpoint of the quarter-arc.

Scoring Note: A3 requires a direction for a nonzero a_avg, stated relative to the circle; a statement that a_avg is zero or has no direction does not earn A3, whatever Δv the response draws. Any description that fixes that direction relative to the circle earns A3 — for example, perpendicular to the straight line joining the ball’s start and end positions, pointing toward the pole, or along the inward bisector of the angle between the start and end radii. A direction stated only relative to the axes, such as 225° counterclockwise from +x, or a statement that a_avg points “toward the centre” that does not identify the radius along which it does so, does not earn A3.

⚠︎ Common error (partial credit): Draws v_f and Δv correctly but states that a_avg points at the pole from the ball’s starting position, along −x — earns 2 of 3 points, A1 and A2. The inward direction at the start is −x, but a_avg has the direction of Δv, which is the inward direction only along the radius through the midpoint of the arc.

⚠︎ Common error (partial credit): Draws v_f correctly but draws Δv from the tip of v_f to the tip of v_i, which gives v_i − v_f, pointing 45° above the +x axis, and consistently states that a_avg points away from the pole along the radius through the midpoint of the arc — earns 2 of 3 points, A1 and A3. The subtraction is reversed, so the average acceleration is drawn outward when it points inward.

✗ Common error (no credit): Draws v_f identical to v_i, pointing along +y, because the speed has not changed; draws no Δv arrow; and states that the ball has no average acceleration, so there is no direction to describe — earns none of A1, A2 or A3. A constant speed fixes only the length of the velocity arrow; its direction turns with the ball, so v_f points along −x.

Part A (ii) — Model Answer

Average acceleration is the change in velocity divided by the time 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°, so the two are perpendicular. Δv = v_f − v_i is therefore the hypotenuse of a right triangle whose legs both have length 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 the magnitude of the change in velocity by the time interval,

|a_avg| = |Δv| / Δt = √2·v × 2v/(πr) = 2√2·v²/(πr)

The two velocities have the same magnitude, so the whole of Δv, and with it the whole of the average acceleration, comes from the 90° change in the velocity’s direction.

Scoring (2 points):

▸ A4 — 1 point: For beginning from the definition of average acceleration in any equivalent form, for example a_avg = Δv/Δt or a_avg = (v_f − v_i)/Δt. This point is awarded for the correct starting principle alone, before any algebra is performed. A4 carries no consistency clause.

Scoring Note: A constant-acceleration kinematic equation such as v_x = v_x0 + a_x·t earns A4 only where its acceleration is the ball’s average acceleration over the quarter-revolution; used with a constant acceleration of magnitude v²/r it does not, because the ball’s acceleration changes direction.

▸ A5 — 1 point: For a correct expression for |a_avg| in terms of v and r. The expected expression is |a_avg| = 2√2·v²/(πr).

Scoring Note: A correct, isolated, final expression for |a_avg| earns point A5. Any expression equal to 2√2·v²/(πr) is correct. Δt may be found from T = 2πr/v, from the arc length πr/2 travelled at speed v, or from v = rω with ω = 2π/T. A5 carries no consistency clause with part A (i): the correct expression depends only on the given v and r, not on any value found in part A (i).

⚠︎ Common error (partial credit): Writes a_avg = Δv/Δt but takes |Δv| as v − v = 0 because the two speeds are equal, subtracting magnitudes instead of vectors, and obtains |a_avg| = 0 — earns 1 of 2 points, A4. Equal speeds do not make the velocities equal: v_i and v_f are perpendicular, so |Δv| = √2·v.

⚠︎ Common error (partial credit): Writes a_avg = Δv/Δt but substitutes the arc length πr/2 for Δv, obtaining |a_avg| = v — earns 1 of 2 points, A4. The arc length is the distance the ball travels, not the change in its velocity, and it does not even carry the units of velocity.

✗ Common error (no credit): Writes |a_avg| = v²/r from the reference information, treating the average acceleration over the quarter-revolution as the centripetal acceleration, and never writes the definition of average acceleration — earns neither A4 nor A5. v²/r is the magnitude of the ball’s acceleration at each instant; the average over the quarter-revolution depends on the velocities at its two ends, whose directions differ by 90°, and on its duration.

Part A (iii) — Model Answer

Substituting v = 3.0 m/s and r = 1.20 m into the expression derived in part A (ii),

|a_avg| = 2√2·(3.0 m/s)² / (π × 1.20 m) = 6.75 m/s²

The same value follows from part A (i). The period is T = 2πr/v = 2π(1.20 m)/(3.0 m/s) = 2.51 s, so the quarter-revolution lasts Δt = T/4 = 0.628 s, and |a_avg| = |Δv|/Δt = (4.24 m/s)/(0.628 s) = 6.75 m/s². The magnitude of the centripetal acceleration is

a_c = v²/r = (3.0 m/s)² / (1.20 m) = 7.50 m/s²

The average over the quarter-revolution is the smaller of the two, 0.900 of the centripetal value. The ball’s acceleration keeps the magnitude v²/r but turns through 90° of direction across the arc, so the changes in velocity over successive short stretches of the arc point in different directions and partly cancel when they are added. Only as the interval shrinks toward zero does the direction stop turning appreciably, and the average then approaches the instantaneous value.

Nothing about this ball’s values makes the ratio 0.900. Dividing 2√2·v²/(πr) by v²/r cancels both v and r and leaves 2√2/π ≈ 0.900, so over a quarter-revolution the ratio is the same for every object in uniform circular motion, whatever its speed and radius.

Scoring (2 points):

▸ A6 — 1 point: For a correct value of |a_avg|, with units. The expected value is 6.75 m/s². A value obtained by correctly substituting v = 3.0 m/s and r = 1.20 m into the response’s own part A (ii) expression, or by dividing the magnitude of the response’s own part A (i) Δv by a correct Δt = 0.628 s, also earns A6.

Scoring Note: A6 may be earned independently of the response in part A (ii). Consistency credit by either route requires a nonzero value; consistency with part A (ii) also requires an expression that depends on v or r and has the units of acceleration.

▸ A7 — 1 point: For a correct value of a_c, with units. The expected value is 7.50 m/s². A7 carries no consistency clause: a_c depends only on the given v and r.

⚠︎ Common error (partial credit): Evaluates a_c = 7.50 m/s² correctly but reports the same 7.50 m/s² for |a_avg|, on the assumption that averaging an acceleration of constant magnitude must return that magnitude, while the response’s own part A (ii) expression is 2√2·v²/(πr) — earns 1 of 2 points, A7. The acceleration keeps its magnitude but turns through 90° across the arc, so the average over the quarter-revolution, |Δv|/Δt, is smaller than v²/r.

⚠︎ Common error (partial credit): Uses the full period, 2.51 s, as the time interval, obtaining |a_avg| = 1.69 m/s², and evaluates a_c = 7.50 m/s² correctly, while the response’s own part A (ii) expression is correct — earns 1 of 2 points, A7. The change in velocity found in part A (i) occurs over a quarter-revolution, so it is divided by that interval’s duration, T/4 = 0.628 s, not by T.

Part B — Model Answer

a_rev < a

An average acceleration is the change in velocity divided by the duration of the interval. At the end of the complete revolution the ball is back at the easternmost point, moving in the +y direction at 3.0 m/s — the same speed and the same direction as at the start. Its velocity at the end equals its velocity at the start, so the change in velocity over the revolution is zero and a_rev = 0.

The acceleration at each instant is not zero, and this is where the familiar claim “the ball is moving at constant speed, so it is not accelerating” fails. Constant speed fixes only the magnitude of the ball’s velocity. The ball’s direction of motion changes continuously as it swings around the pole, so its velocity is changing at every instant and the ball is accelerating throughout the revolution. With its speed constant it has no tangential acceleration, so its acceleration is entirely centripetal: a = v²/r = 7.50 m/s², directed toward the pole. Part A shows the change directly — over the quarter-revolution |Δv| is 4.24 m/s and |a_avg| is 6.75 m/s². Over the complete revolution the changes in velocity cancel, because the velocity turns all the way round and returns to the value it started with. So a_rev = 0 is less than a = 7.50 m/s².

Figure 3 — Velocity (tangent) and acceleration (toward the pole) at four instants of a revolution.

That claim is a correct rule applied outside its condition. Constant speed means zero acceleration only when the direction of motion is constant too — that is, for motion in a straight line. The centre of 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. A zero average acceleration over an interval shows only that the velocity at its end matches the velocity at its start, not that the ball’s acceleration was zero in between.

Scoring (3 points):

▸ B1 — 1 point: For indicating a_rev < a.

▸ B2 — 1 point: For establishing, by any correct route, that a_rev = 0 — for example, that the ball’s velocity at the end of the revolution equals its velocity at the start, so its change in velocity over the revolution is zero.

Scoring Note: Constant speed alone does not establish B2, because it fixes only the magnitude of the velocity. A return to the starting position establishes B2 only together with a statement that the velocity there is the same as at the start; a zero displacement over the revolution does not by itself establish B2.

▸ B3 — 1 point: For establishing that a is not zero, on a supporting ground: that the ball’s direction of motion, and so its velocity, changes continuously; that the ball’s acceleration has magnitude v²/r, which is not zero; that constant speed means zero acceleration only for motion in a straight line, and the ball’s path is a circle; or any equivalent ground.

Scoring Note: Asserting that a is not zero, or that the ball is accelerating, states the conclusion B3 asks for, not a ground for it. A statement that the ball moves in a circle does not by itself earn B3; it must be joined to one of the grounds above.

Scoring Note: B1, B2 and B3 carry no consistency clause with part A: a_rev is zero and a is v²/r whatever part A reported, so the correct comparison does not depend on any value found in part A.

⚠︎ Common error (partial credit): Indicates a_rev = a, reasoning that the ball’s acceleration always has magnitude v²/r, so its average over a complete revolution must have that magnitude too — earns 1 of 3 points, B3. An average acceleration is the change in velocity divided by the duration of the interval; over a complete revolution the velocity returns to its starting value, so a_rev = 0. The constant magnitude v²/r belongs to the acceleration at each instant, not to its average.

⚠︎ Common error (partial credit): Indicates a_rev = a, showing that the ball’s velocity at the end of the revolution equals its velocity at the start, so a_rev = 0, but then concluding that the ball is therefore not accelerating, so a = 0 as well — earns 1 of 3 points, B2. A zero average shows only that the velocities at the two ends of the revolution match; the ball accelerates toward the pole at every instant in between.

⚠︎ Common error (partial credit): Indicates a_rev < a, reasoning that because the average over the quarter-revolution in part A (iii) is 0.900 of the instantaneous magnitude, the average over a complete revolution is too, so a_rev is about 0.900 × 7.50 m/s² = 6.75 m/s², while a = v²/r = 7.50 m/s² — earns 2 of 3 points, B1 and B3. The factor 0.900 belongs to a quarter-revolution only; over the complete revolution the velocity returns to its starting value, so a_rev is zero.

⚠︎ Common error (partial credit): Indicates a_rev < a, arguing that a_rev is zero because the ball’s displacement over a complete revolution is zero, and that a is not zero because the ball’s direction of motion changes — earns 2 of 3 points, B1 and B3. Zero displacement makes the average velocity zero; a_rev is zero because the velocity, not the position, returns to its starting value.

✗ Common error (no credit): Indicates a_rev = a, reasoning that a ball moving at constant speed is not accelerating, so both a and a_rev are zero — earns none of B1, B2 or B3. This treats speed as though it were velocity: the speed is constant, but the direction of motion, and so the velocity, changes at every instant.

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