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Further kinematics: acceleration as a function of x, t or v questions
Two ways of writing the acceleration, and the whole skill is picking the one that separates. Get the pairing right and the integration is routine.
6 original questions · 22 marks · the further kinematics: acceleration as a function of x, t or v notes · Further Mechanics 2
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State which form of the acceleration you would use if it depends on the displacement, and why the other form fails.
Worked 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. Matching the form to the variable the acceleration depends on is what makes the equation separable. B1 for choosing v dv/dx, B1 for why dv/dt fails.A particle has acceleration (6t − 2) m/s² and speed 1 m/s at t = 0. Find its speed at t = 3.
Worked 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.A particle moving at 20 m/s decelerates at 0.02v² m/s². Find its speed after it has travelled 50 m.
Worked 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.A particle moving at 15 m/s decelerates at 0.4v m/s². Find how long it takes for the speed to halve.
Worked 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.A body has acceleration (2 − 0.05v²) m/s². Find its terminal speed.
Worked 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. The terminal speed is simply where the driving effect and the resistance balance, and setting up an integral here wastes most of the time available.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.
Worked 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, and choosing the right form is the first method mark.
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; carrying it symbolically to the end is where the arithmetic usually breaks.
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.
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