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Kinematics with variable acceleration questions
When acceleration is not constant, the suvat equations no longer apply and calculus takes over. Differentiate down the ladder from displacement to acceleration, integrate back up, and let the boundary conditions pin the constants.
7 original questions · 32 marks · the kinematics with variable acceleration notes · Mechanics
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Explain why the suvat equations cannot be used when the acceleration depends on time, and state the calculus relationships between displacement, velocity and acceleration.
Worked 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.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.
Worked 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. One differentiation is all that is needed, and the only real care required is not integrating by reflex because the question mentions velocity.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.
Worked 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. The two conditions answer completely different questions, and confusing them here gives t = 4 and a velocity of zero.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.
Worked 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. The initial condition has to be used before anything else can be asked of v, and an answer left as +c scores the M mark and nothing beyond it.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.
Worked 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. Splitting the journey at the turning time is compulsory rather than optional.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.
Worked 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. Speed is the magnitude of velocity, so the answer is a scalar with no i or j in it.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.
Worked 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. Solving x = 0 numerically and reporting 'no solutions found' proves nothing. The discriminant is what turns a search into a proof.
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