Mechanics: Kinematics Depth
In A-level Mechanics, kinematics at depth rarely means a single suvat substitution: questions combine two or more phases of motion, mix constant-acceleration algebra with calculus for variable acceleration, or work in vector form with i and j components. The marks reward a clearly stated positive direction, choosing the correct suvat formula or calculus tool for what is given, and care at the points where a question changes character - when a particle is momentarily at rest, when its direction reverses (so displacement and distance travelled differ), or where two stages of motion must be linked through a shared value of time, velocity or position.
Before you start
Make sure you're comfortable with these topics first:
Method
- Decide whether the acceleration is constant or given as a function of t. Constant acceleration means suvat or a velocity-time graph; acceleration as a function of t means calculus.
- For suvat, list the three known quantities from s, u, v, a, t and pick the single formula containing exactly those three plus the unknown, to avoid an unnecessary two-step solve.
- For a velocity-time graph, read the gradient of a section as the acceleration on that section, and the area under a section as the distance travelled in that section; total distance is not the same as displacement if the particle reverses direction (some area lies below the axis).
- For variable acceleration, differentiate displacement to get velocity and velocity to get acceleration; integrate acceleration to get velocity and velocity to get displacement, using a given initial condition to find the constant of integration each time.
- To find when a particle is at rest, set v = 0 and solve. To find total distance travelled (as opposed to displacement) when direction reverses, split the motion into intervals either side of each v = 0 point, and sum the modulus of the displacement change in each interval.
- For two-stage or two-particle problems (a particle meeting or catching up with another, or motion changing at a boundary time), write an expression for each stage or each particle separately, then equate the shared quantity (the same time, or the same position) to link them.
- For vector kinematics, apply the same suvat or calculus relations component-wise in i and j, then find speed as the magnitude of the velocity vector and direction using the arctan of its components.
- State the positive direction explicitly before substituting signed values, keep g = 9.8 m/s^2 unless told otherwise, and give exact (surd or fraction) answers unless the question asks for a decimal.
Worked example
A particle P moves in a straight line so that its velocity at time t seconds (0 <= t <= 6) is given by v = t^2 - 5t + 4 m/s, where the positive direction is the direction in which P is initially moving. (a) Find the values of t at which P is instantaneously at rest. (b) Find the total distance travelled by P in the first 4 seconds.
- Factorise to find when v = 0: t^2 - 5t + 4 = (t - 1)(t - 4) = 0, so P is at rest at t = 1 and t = 4.
- Check the sign of v either side of these roots: v(0) = 4 > 0 (moving in the positive direction), but v(2) = 4 - 10 + 4 = -2 < 0 (moving in the negative direction). So P reverses direction at t = 1, meaning displacement and distance differ over [0, 4].
- Integrate v to get displacement: s = int v dt = t^3/3 - (5/2)t^2 + 4t + C. Since P starts at the origin, s = 0 when t = 0, so C = 0.
- Find the displacement at each turning point: s(1) = 1/3 - 5/2 + 4 = 11/6 m; s(4) = 64/3 - 40 + 16 = -8/3 m.
- Find the distance covered in each stage: from t = 0 to t = 1, distance = |s(1) - s(0)| = 11/6 m; from t = 1 to t = 4, distance = |s(4) - s(1)| = |-8/3 - 11/6| = 9/2 m.
- Total distance = 11/6 + 9/2 = 11/6 + 27/6 = 38/6 = 19/3 m. So the total distance travelled in the first 4 seconds is 19/3 m (= 6 1/3 m, or 6.33 m to 3 s.f.).
Practice questions
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Q1State the two modelling conditions that must both hold for the constant-acceleration (suvat) formulae to be used.Show answer
Answer: The acceleration must be constant (uniform), and the moving object must be modelled as a particle (so its size and any rotation can be ignored).
Q2A particle moves with constant acceleration. It has velocity 3 m/s at time t = 0 and velocity 15 m/s at time t = 4 s. Find its displacement in this time.Show answer
Answer: 36 m (s = 0.5(u+v)t = 0.5 x 18 x 4).
Q3A particle moves in a straight line with acceleration a = (6t - 4) m/s^2. At t = 0 its velocity is 2 m/s. Find its velocity when t = 3 s.Show answer
Answer: 17 m/s (v = 3t^2 - 4t + C, C = 2 from v(0) = 2, so v(3) = 27 - 12 + 2).
Q4A cyclist accelerates uniformly from rest to 8 m/s in 5 seconds, then travels at this constant speed for a further 10 seconds. By sketching a velocity-time graph, find the total distance travelled.Show answer
Answer: 100 m (triangle area 0.5 x 5 x 8 = 20 m, plus rectangle area 10 x 8 = 80 m).
Q5Explain why, when a ball thrown vertically upward returns to its point of projection, the equation s = ut + 0.5at^2 with s = 0 has two solutions for t, but only one is a sensible answer to 'find the time taken to return to the point of projection'.Show answer
Answer: The equation factorises as t(u + 0.5at) = 0, which is satisfied both at t = 0 (the instant of projection itself, before any motion has happened) and again when the ball returns to the same height after going up and coming back down. The trivial root t = 0 must be rejected, leaving the non-zero root as the physically meaningful time of flight.
Q6A particle has initial velocity (2i - 3j) m/s and constant acceleration (i + 2j) m/s^2. Find its velocity after 3 seconds, and its speed at this time, giving the speed as an exact surd.Show answer
Answer: Velocity = 5i + 3j m/s (v = u + at = (2i-3j) + 3(i+2j)). Speed = sqrt(5^2 + 3^2) = sqrt(34) m/s (exact), approximately 5.83 m/s.
Q7A particle moves in a straight line so that its displacement from a fixed point O at time t seconds is s = t^3 - 12t^2 + 36t metres (t >= 0). Find the particle's acceleration when it is instantaneously at rest for the second time.Show answer
Answer: 12 m/s^2. (v = 3t^2 - 24t + 36 = 3(t-2)(t-6), at rest at t = 2 and t = 6; a = 6t - 24, and a(6) = 36 - 24 = 12.)
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
A train accelerates uniformly from rest at 0.5 m/s^2 until it reaches its maximum speed after 20 seconds. It then travels at this maximum speed for a further 40 seconds, before decelerating uniformly to rest in the next 10 seconds. (a) Find the maximum speed of the train. (2) (b) Find the total distance travelled by the train while it is accelerating and while it is decelerating (i.e. excluding the constant-speed phase). (4) (c) Find the total distance travelled by the train for the whole journey. (2)
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A particle P moves in a plane such that its position vector relative to a fixed origin O at time t seconds (t >= 0) is r = (t^2 - 4t)i + 3tj metres. (a) Find an expression for the velocity of P at time t seconds. (2) (b) Find the value of t at which P is moving parallel to the vector (i + j). (2) (c) Find the speed of P at this value of t, giving your answer as an exact surd. (3)
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Points A and B lie 400 m apart on a straight horizontal road. At time t = 0, a cyclist P sets off from A towards B with initial speed 2 m/s and constant acceleration 0.4 m/s^2. At the same instant, a cyclist Q sets off from B towards A with constant speed 6 m/s (zero acceleration). (a) Write down an expression, in terms of t, for the distance of P from A, and for the distance of Q from A. (2) (b) Show that P and Q meet when 0.2t^2 + 8t - 400 = 0. (2) (c) Find the time at which P and Q meet, giving your answer to 3 significant figures. (3)
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