Questions on kinematics with variable acceleration

Question 1

Explain why the suvat equations cannot be used when the acceleration depends on time, and state the calculus relationships between displacement, velocity and acceleration.

Accepted Answer

Every suvat equation is derived on the assumption that a is constant, so once a varies those formulae are simply false rather than approximate. In their place, v = dx/dt and a = dv/dt, with integration running the chain backwards and the constants of integration fixed by the given starting conditions. B1 for the reason, B1 B1 for the two relationships.

Question 2

A particle moves in a straight line with velocity v = 4t³ − 6t m s⁻¹ at time t seconds. Find its acceleration at t = 2.

Accepted Answer

a = dv/dt = 12t² − 6, so at t = 2, a = 48 − 6 = 42 m s⁻². M1 for differentiating, A1 for the expression, A1 for 42.

Question 3

A particle moves in a straight line with velocity v = 6t² − 2t m s⁻¹ at time t seconds. Find the displacement of the particle between t = 1 and t = 3.

Accepted Answer

Displacement over an interval is the definite integral of v. ∫(6t² − 2t) dt = 2t³ − t², and between the limits this is (54 − 9) − (2 − 1) = 45 − 1 = 44 m. M1 for integrating v, A1 for 2t³ − t², A1 for 44. A definite integral needs no constant: it cancels between the two limits. Substituting t = 3 alone gives 45 m, which is the displacement from t = 0, not from t = 1.

Question 4

A particle P moves in a plane so that at time t seconds its position vector is r = t³ i + (4t − t²) j metres. Find the time at which P is moving parallel to i.

Accepted Answer

Differentiate each component: v = 3t² i + (4 − 2t) j. Moving parallel to i means the j component of the velocity is zero, so 4 − 2t = 0 and t = 2 s. At that instant v = 12i, so the i component is not zero and P is indeed moving, in the direction of i. M1 for differentiating r, M1 for setting the j component of v to zero, A1 for t = 2. The condition is on the velocity, not on the position: setting 4t − t² = 0 finds when P is on the line of i, which is a different question.

Question 5

A particle starts at the origin and moves in a straight line with velocity v = 8t − 2t² m s⁻¹ at time t seconds. Find its maximum velocity, and its displacement from the origin at t = 3.

Accepted Answer

Maximum velocity occurs where a = 0. Since a = dv/dt = 8 − 4t, that is at t = 2, where v = 16 − 8 = 8 m s⁻¹. It is a maximum because a is positive before t = 2 and negative after. For the displacement, integrate: x = 4t² − (2/3)t³ + c, and the start at the origin gives c = 0, so x(3) = 36 − 18 = 18 m. M1 for setting a = 0, A1 for t = 2, A1 for v = 8, M1 for integrating with a constant, A1 for 18. Maximum velocity comes from a = 0, not from v = 0. Confusing them here gives t = 4 and a velocity of zero.

Question 6

A particle moves in a straight line with acceleration a = 6t − 12 m s⁻² at time t seconds, and has velocity 9 m s⁻¹ when t = 0. Find v in terms of t, and the times at which the particle is at rest.

Accepted Answer

Integrating, v = 3t² − 12t + c, and v(0) = 9 gives c = 9. Factorising, v = 3(t² − 4t + 3) = 3(t − 1)(t − 3), so the particle is at rest at t = 1 and t = 3. M1 for integrating with a constant, A1 for v, M1 for solving v = 0, A1 for both times. Use the initial condition to evaluate c before anything else is asked of v; an answer left with an unevaluated + c is not a particular solution.

Question 7

A particle moves in a straight line with velocity v = 3t² − 12t + 9 m s⁻¹ at time t seconds, starting from the origin. Find its displacement from the origin at t = 3, and the total distance it has travelled by that time.

Accepted Answer

Integrating gives x = t³ − 6t² + 9t, with no constant since x(0) = 0. Then x(3) = 27 − 54 + 27 = 0, so the displacement is 0 and the particle is back where it started. Distance is a different matter. Since v = 3(t − 1)(t − 3), the particle changes direction at t = 1, where x(1) = 1 − 6 + 9 = 4. It travels 4 m forwards from t = 0 to t = 1 and 4 m backwards from t = 1 to t = 3, so the total distance is 8 m. M1 A1 for the displacement function, A1 for 0, M1 for finding the turning time, M1 for splitting the journey, A1 for 8 m. Integrating v straight through from 0 to 3 lets the two legs cancel and gives 0 m for the distance.

Question 8

A particle moves in a straight line with acceleration a = 12 − 6t m s⁻² at time t seconds. When t = 0 the particle is at the origin O and has velocity 3 m s⁻¹. Find the displacement of the particle from O at the instant when its acceleration is zero.

Accepted Answer

Integrate once: v = 12t − 3t² + c, and v(0) = 3 gives c = 3, so v = 3 + 12t − 3t². Integrate again: x = 3t + 6t² − t³ + k, and x(0) = 0 gives k = 0. The acceleration is zero when 12 − 6t = 0, so t = 2, and then x = 6 + 24 − 8 = 22 m. M1 for integrating a with a constant, A1 for v, M1 for integrating v with a second constant, A1 for x, A1 for t = 2 and 22 m.

Question 9

A particle P moves in a plane with i and j pointing east and north. At time t seconds its position vector is r = (t² + t) i + ((1/3)t³ − 3t) j metres. Find the value of t at which P is moving north-east.

Accepted Answer

v = dr/dt = (2t + 1) i + (t² − 3) j. Moving north-east means the two components of the velocity are equal and both positive, so t² − 3 = 2t + 1, that is t² − 2t − 4 = 0, giving t = 1 ± √5. The negative root is rejected, so t = 1 + √5 = 3.24 s (3 s.f.). At that time 2t + 1 = 7.47 > 0, so both components are positive and the direction is north-east rather than south-west. M1 for differentiating, A1 for v, M1 for equating the components, A1 for t = 3.24. Equating the components of r instead of v gives the time at which P is north-east of the origin, which is not what was asked.

Question 10

A particle moves in a straight line with velocity v = kt(6 − t) m s⁻¹ for 0 ≤ t ≤ 6, where k is a positive constant. The particle travels 72 m in these six seconds. Find the value of k.

Accepted Answer

For 0 ≤ t ≤ 6, t(6 − t) ≥ 0, so v does not change sign and the distance travelled equals the displacement. ∫₀⁶ k(6t − t²) dt = k[3t² − (1/3)t³]₀⁶ = k(108 − 72) = 36k. So 36k = 72 and k = 2. M1 for integrating v between 0 and 6, A1 for 36k, M1 for equating to 72, A1 for k = 2.

Question 11

A particle moves in a straight line with velocity v = t² − 7t + 10 m s⁻¹ at time t seconds. Find the acceleration of the particle at each instant when it is at rest.

Accepted Answer

Rest means v = 0: t² − 7t + 10 = (t − 2)(t − 5) = 0, so t = 2 and t = 5. The acceleration is a = dv/dt = 2t − 7, which is −3 m s⁻² at t = 2 and 3 m s⁻² at t = 5. M1 for solving v = 0, A1 for both times, M1 for differentiating and substituting, A1 for both accelerations. The particle is at rest at those instants but its acceleration is not zero: the velocity is passing through zero, decreasing at t = 2 and increasing at t = 5, which is why the particle reverses direction at each.

Question 12

A particle P moves in a plane so that at time t seconds its velocity is v = (3t² − 4)i + 6t j m s⁻¹. At t = 0, P has position vector (2i − j) m. Find the position vector of P at t = 2, and the speed of P at that instant.

Accepted Answer

Integrate each component separately: r = (t³ − 4t)i + 3t² j + C, and putting t = 0 gives C = 2i − j. So r = (t³ − 4t + 2)i + (3t² − 1)j. At t = 2, r = (8 − 8 + 2)i + (12 − 1)j = 2i + 11j metres. For the speed, v(2) = (12 − 4)i + 12j = 8i + 12j, so the speed is √(64 + 144) = √208 = 4√13 = 14.4 m s⁻¹ to 3 significant figures. M1 for integrating, A1 for the components, M1 for using the initial position, A1 for 2i + 11j, M1 for the magnitude, A1 for 14.4. The constant of integration is a vector, so it needs both components.

Question 13

A particle moves along a straight line so that its displacement from the origin O at time t seconds is x = t³ − 9t² + 24t metres, t ≥ 0. Find the times at which the particle is instantaneously at rest, and show that the particle never returns to O after t = 0.

Accepted Answer

v = dx/dt = 3t² − 18t + 24 = 3(t − 2)(t − 4), so the particle is at rest at t = 2 and t = 4. For the second part, x = t(t² − 9t + 24), so x = 0 requires either t = 0 or t² − 9t + 24 = 0. That quadratic has discriminant 81 − 96 = −15, which is negative, so it has no real roots and t = 0 is the only time at which x = 0. M1 for differentiating, A1 for the factorised velocity, A1 for both times, M1 for factorising out t and testing the quadratic, A1 for the conclusion.

Question 14

A particle P moves in a plane so that at time t seconds its position vector is r = (t² − 4t) i + 2t j metres, t ≥ 0. Show that the least speed of P is 2 m s⁻¹, and find the position vector of P at the instant when its speed is least.

Accepted Answer

v = (2t − 4) i + 2 j, so the speed satisfies |v|² = (2t − 4)² + 4. The squared term is never negative and is zero only at t = 2, so |v|² ≥ 4 with equality at t = 2: the least speed is √4 = 2 m s⁻¹. At t = 2, r = (4 − 8) i + 4 j = −4i + 4j m. M1 for differentiating, A1 for v, M1 for forming |v|² as a function of t, A1 for the argument that its minimum is 4 at t = 2, A1 for the position. Minimising a squared expression by inspection is a complete argument; differentiating |v|² is an alternative, but must still show the value is a minimum.

Question 15

A car starts from rest at time t = 0 and moves in a straight line. For 0 ≤ t ≤ 10 its velocity is v = 0.3t² m s⁻¹. For t ≥ 10 its acceleration is zero. Find the total distance travelled between t = 0 and t = 20, and explain why a constant-acceleration formula may be used for the second stage but not for the first.

Accepted Answer

First stage: the velocity is positive throughout, so the distance is ∫₀¹⁰ 0.3t² dt = [0.1t³]₀¹⁰ = 100 m. At t = 10 the velocity is 0.3 × 100 = 30 m s⁻¹, and with zero acceleration the car keeps that velocity, so the second stage covers 30 × 10 = 300 m. Total distance 400 m. M1 for integrating v over the first stage, A1 for 100 m, B1 for v = 30 at t = 10, M1 for the constant-velocity distance, A1 for 400 m. B1 for the reason: the constant-acceleration formulae are derived assuming a is constant, and in the first stage a = 0.6t varies with t, whereas in the second stage a is the constant 0. Using s = ½(u + v)t on the first stage gives 150 m, which is wrong because the velocity does not grow linearly.