Questions on further kinematics: acceleration as a function of x, t or v

Question 1

State which form of the acceleration you would use if it depends on the displacement, and why the other form fails.

Accepted Answer

Use v dv/dx. Writing dv/dt instead leaves an equation containing v, t and x, which cannot be separated because t and x are both unknown functions of each other. B1 for choosing v dv/dx, B1 for why dv/dt fails.

Question 2

A particle has acceleration (6t − 2) m/s² and speed 1 m/s at t = 0. Find its speed at t = 3.

Accepted Answer

The acceleration depends on t, so dv/dt = 6t − 2 and v = 3t² − 2t + c. With v = 1 at t = 0, c = 1, so v = 3t² − 2t + 1. At t = 3: 27 − 6 + 1 = 22 m/s. M1 for integrating, A1 for the expression for v, A1 for 22 m/s.

Question 3

A particle moves along the x-axis with acceleration 4x m/s² directed away from the origin, and has speed 2 m/s at x = 0. Find its speed at x = 1.

Accepted Answer

The acceleration depends on x, so use v dv/dx = 4x. Integrating: ½v² = 2x² + c, and v = 2 at x = 0 gives c = 2, so v² = 4x² + 4. At x = 1: v² = 8 and v = 2.83 m/s. M1 for v dv/dx = 4x, A1 for v² = 4x² + 4, A1 for 2.83 m/s. The speed grows without limit as the particle moves out, since the acceleration grows with the distance.

Question 4

A particle moving at 16 m/s decelerates at 2√v m/s², where v is its speed. Find the time it takes to come to rest.

Accepted Answer

dv/dt = −2√v. Separating: ∫v^−½ dv = ∫−2 dt, so 2√v = −2t + c, and v = 16 at t = 0 gives c = 8. So √v = 4 − t, and v = 0 when t = 4 s. M1 for separating the variables, A1 for √v = 4 − t, A1 for 4 s. Unlike a resistance proportional to v, which gives an exponential that never reaches zero, this one stops the particle in a finite time, because the deceleration does not fade away as fast as the speed does.

Question 5

A particle moving at 20 m/s decelerates at 0.02v² m/s². Find its speed after it has travelled 50 m.

Accepted Answer

The acceleration depends on v and distance is wanted, so use v dv/dx = −0.02v². Cancelling one v, which is valid while the particle is moving, gives dv/v = −0.02 dx, so ln v = −0.02x + c. With v = 20 at x = 0, v = 20e^−0.02x. At x = 50: v = 20e⁻¹ = 7.36 m/s. M1 for using v dv/dx, M1 for separating and integrating, A1 for v = 20e^−0.02x, A1 for 7.36 m/s.

Question 6

A particle moving at 15 m/s decelerates at 0.4v m/s². Find how long it takes for the speed to halve.

Accepted Answer

Time is wanted and the acceleration depends on v, so dv/dt = −0.4v, giving v = 15e^−0.4t. Setting v = 7.5: e^−0.4t = 0.5, so 0.4t = ln2 and t = 1.73 s. M1 for dv/dt = −0.4v, A1 for the expression for v, M1 for setting v = 7.5, A1 for 1.73 s. The same time would halve it again, so the speed never actually reaches zero.

Question 7

A body has acceleration (2 − 0.05v²) m/s². Find its terminal speed.

Accepted Answer

At the terminal speed the acceleration is zero, so 2 = 0.05v² and v² = 40, giving v = 6.32 m/s. M1 for setting the acceleration to zero, A1 for v² = 40, A1 for 6.32 m/s. No integration is needed.

Question 8

A particle moving at 12 m/s decelerates at 0.3v m/s². Find the total distance it travels before coming to rest.

Accepted Answer

Distance is wanted and the acceleration depends on v, so use v dv/dx = −0.3v. M1 for choosing the pairing that separates against x. Cancelling one v, valid while the particle is moving, leaves dv/dx = −0.3, a constant, so the speed falls in a straight line against distance: v = 12 − 0.3x. M1 for the cancellation with its condition, A1 for the linear expression. Setting v = 0 gives x = 40 m. A1. Against time the same motion is the exponential decay v = 12e^−0.3t, which never quite reaches rest: a finite stopping distance and an infinite stopping time describing one journey, with no contradiction between them.

Question 9

A particle moves along the x-axis with velocity (2x + 1) m/s, where x is its displacement from O in metres, and is at O when t = 0. Find its displacement when t = 1, and its acceleration at that instant.

Accepted Answer

dx/dt = 2x + 1. Separating: ∫dx/(2x + 1) = ∫dt, so ½ ln(2x + 1) = t + c, and x = 0 at t = 0 gives c = 0. Hence 2x + 1 = e^2t and x = ½(e^2t − 1). At t = 1: x = ½(e² − 1) = 3.19 m. The acceleration is v dv/dx = (2x + 1) × 2 = 2v, and v = e² = 7.39 there, so a = 14.8 m/s². M1 for separating, A1 for x = ½(e^2t − 1), A1 for 3.19 m, B1 for the acceleration by v dv/dx. Differentiating v = e^2t with respect to t gives the same 2e² = 14.8, as it must.

Question 10

A raindrop falls from rest with acceleration (9.8 − 0.5v) m/s², where v is its speed. Find its speed after 2 seconds and its terminal speed.

Accepted Answer

dv/dt = 9.8 − 0.5v. Separating: ∫dv/(9.8 − 0.5v) = ∫dt, so −2 ln(9.8 − 0.5v) = t + c. With v = 0 at t = 0, c = −2 ln 9.8, so ln[(9.8 − 0.5v)/9.8] = −t/2 and v = 19.6(1 − e^−0.5t). At t = 2: v = 19.6(1 − e⁻¹) = 12.4 m/s. The terminal speed is where the acceleration is zero, 0.5v = 9.8, so 19.6 m/s, which is also the limit of the expression as t grows. M1 for separating with the constant term kept, A1 for v = 19.6(1 − e^−0.5t), A1 for 12.4 m/s, B1 for the terminal speed. The drop has reached 63% of its terminal speed after 2 s, and the fraction is fixed by the 0.5 alone.

Question 11

A particle starts from rest at a point O and moves in a straight line with acceleration (12t − 3t²) m/s² at time t seconds. Find the time at which it is next at rest, and its distance from O at that instant.

Accepted Answer

dv/dt = 12t − 3t², so v = 6t² − t³ + c, and c = 0 from rest at t = 0. Then v = t²(6 − t), which is zero again at t = 6 s and positive in between, so the particle moves one way throughout. Integrating again, x = 2t³ − t⁴/4, with x = 0 at t = 0. At t = 6: x = 432 − 324 = 108 m. M1 for integrating to v with c = 0, A1 for t = 6 s, M1 for integrating again, A1 for 108 m.

Question 12

A particle P moves along a straight line through a fixed point O. When P is x metres from O its acceleration is 6/x² m/s² directed towards O. At x = 2 the particle is moving away from O with speed 2 m/s. Find the speed of P when x = 4, and the greatest distance of P from O.

Accepted Answer

The acceleration depends on the displacement, so write it as v dv/dx. Using dv/dt here leaves three variables in one equation and goes nowhere, so choose the form of the acceleration that matches the variables given.
Taking the direction of motion as positive, the acceleration is negative: v dv/dx = −6/x².
Separating and integrating: ∫v dv = ∫−6x⁻² dx, so v²/2 = 6/x + c.
At x = 2, v = 2, giving 2 = 3 + c and c = −1. Hence v² = 12/x − 2. Find the constant before going any further, and keep it exact.
At x = 4: v² = 3 − 2 = 1, so the speed is 1 m/s.
The particle is furthest from O when it is instantaneously at rest, so set v = 0: 12/x = 2 and x = 6 m. Beyond that point v² would be negative, which is the confirmation that 6 m is the limit.
M1 for v dv/dx = −6/x², M1 A1 for integrating, A1 for c = −1, A1 for the speed at x = 4, A1 for the greatest distance.

Question 13

A student models a particle slowing under resistance with dv/dt = 0.5v, and later cancels the v in v dv/dx = −0.5v without comment. Explain the error in the first equation, the condition that makes the cancellation legitimate, and what solution the cancelled equation no longer contains.

Accepted Answer

A resistance opposes the motion, so a particle slowing down needs a negative right-hand side: dv/dt = −0.5v. With the plus sign the speed grows exponentially, and the equation describes a runaway rather than a stopping particle. B1 for the sign error and its meaning. Dividing both sides by v is legitimate only while v ≠ 0, that is, while the particle is actually moving, and the condition is worth stating in the working. B1 for the condition. What the division discards is the solution v = 0 itself: a particle already at rest satisfies the original equation and stays at rest, and the cancelled equation cannot say so. B1 for the lost rest solution.

Question 14

A body falls from rest against air resistance, so that its acceleration when its speed is v m/s is (9.8 − 0.02v²) m/s² downwards. By writing the acceleration as v dv/dx, find v² as a function of the distance fallen x, and hence find the terminal speed and the speed after the body has fallen 50 m.

Accepted Answer

Distance is wanted, so v dv/dx = 9.8 − 0.02v². Separating: ∫v/(9.8 − 0.02v²) dv = ∫dx. The numerator is a multiple of the derivative of the denominator, so the left side is −25 ln(9.8 − 0.02v²) = x + c. From rest at x = 0, c = −25 ln 9.8, so ln[(9.8 − 0.02v²)/9.8] = −x/25 and v² = 490(1 − e^−0.04x).
As x grows the exponential dies away and v² → 490, so the terminal speed is √490 = 22.1 m/s, the same as setting the acceleration to zero. At x = 50: v² = 490(1 − e⁻²) = 423.7, so v = 20.6 m/s.
M1 for v dv/dx = 9.8 − 0.02v², M1 for separating and recognising the log integral, A1 for −25 ln(9.8 − 0.02v²) = x + c, A1 for v² = 490(1 − e^−0.04x), A1 for the terminal speed, A1 for 20.6 m/s. Writing u = v² turns the equation into ½ du/dx = 9.8 − 0.02u, a linear equation in u, which is the same integral seen from the other side.

Question 15

A particle moving at u m/s is subject only to a resistance that produces a deceleration kv², where v is its speed and k is a positive constant. Show that the distance over which its speed halves is the same whatever the value of u, but that the time taken to halve is not. Evaluate both for k = 0.02 and u = 20.

Accepted Answer

Against distance: v dv/dx = −kv², and cancelling one v (valid while v ≠ 0) gives dv/v = −k dx, so ln v = −kx + ln u and v = ue^−kx. Setting v = u/2: e^−kx = ½, so x = (ln 2)/k, which contains no u: every starting speed halves over the same distance.
Against time: dv/dt = −kv², so ∫v⁻² dv = −∫k dt and −1/v = −kt − 1/u, that is 1/v = 1/u + kt. Setting v = u/2: 2/u = 1/u + kt, so t = 1/(ku), which does depend on u: a faster particle halves its speed sooner.
With k = 0.02 and u = 20: x = (ln 2)/0.02 = 34.7 m and t = 1/(0.02 × 20) = 2.5 s.
M1 for v dv/dx with the cancellation, A1 for v = ue^−kx, A1 for x = (ln 2)/k with the comment that u is absent, M1 for separating dv/dt = −kv², A1 for 1/v = 1/u + kt and t = 1/(ku), A1 for 34.7 m and 2.5 s.