Mechanics: Quantities, Units and Kinematics
Kinematics is the study of motion in a straight line or plane without considering the forces that cause it, describing quantities such as displacement, velocity, acceleration and time. A Level Mechanics uses SI units (metres, seconds, kilograms), the constant acceleration (SUVAT) formulae, velocity-time graphs, and calculus for variable acceleration.
Before you start
Make sure you're comfortable with these topics first:
Method
- Identify which SUVAT formula to use based on which of s, u, v, a, t are given and which is required (e.g. v = u + at, s = ut + (1/2)at^2, v^2 = u^2 + 2as, s = (1/2)(u+v)t).
- For motion under gravity, use a = g = 9.8 m/s^2 (taking downward as positive, or as stated in the question), and treat the object as a particle.
- For a velocity-time graph, read the gradient of a section as the acceleration, and read the area under the graph (or under a section) as the distance travelled.
- For variable acceleration given as a function of time, differentiate displacement to get velocity, and differentiate velocity to get acceleration; integrate acceleration to get velocity, and integrate velocity to get displacement, using given initial conditions to find the constant of integration.
- To find when a particle is at rest, set v = 0 and solve; to find the total distance travelled (as opposed to displacement), consider each interval separately and take the modulus of displacement in any interval where the direction reverses.
- For vector kinematics in two dimensions, apply v = u + at and r = r0 + ut + (1/2)at^2 component-wise (i and j components separately).
Worked example
A ball is thrown vertically upwards from ground level with initial speed 14 m/s, and is modelled as a particle moving freely under gravity with g = 9.8 m/s^2. (a) Find the greatest height reached by the ball. (b) Find the total time the ball is in the air before it returns to ground level.
- Taking upwards as positive, at the greatest height v = 0, u = 14, a = -9.8. Use v^2 = u^2 + 2as: 0 = 14^2 + 2(-9.8)s.
- 0 = 196 - 19.6s, so s = 196/19.6 = 10 m.
- For the total time in the air, use s = 0 (the ball returns to its starting height) in s = ut + (1/2)at^2: 0 = 14t - 4.9t^2.
- Factorise: 0 = t(14 - 4.9t), giving t = 0 (the start) or t = 14/4.9 = 2.857... s.
- Round the non-zero solution to 3 significant figures: t = 2.86 s.
Practice questions
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Q1State the SI unit of acceleration.Show answer
Answer: metres per second squared (m/s^2).
Q2A cyclist accelerates uniformly from 2 m/s to 10 m/s in 4 seconds. Find the acceleration.Show answer
Answer: 2 m/s^2 ((10-2)/4).
Q3A car travels at a constant 18 m/s for 25 seconds. Find the distance travelled.Show answer
Answer: 450 m (18 x 25).
Q4A particle starts from rest and accelerates uniformly at 3 m/s^2 for 6 seconds. Find the distance travelled.Show answer
Answer: 54 m (s = ut + 0.5at^2 = 0 + 0.5(3)(36)).
Q5A stone is dropped from rest from a height of 20 m, modelled as a particle moving freely under gravity with g = 9.8 m/s^2. Find the speed with which it hits the ground, to 3 significant figures.Show answer
Answer: 19.8 m/s (v^2 = 0 + 2(9.8)(20) = 392, v = sqrt(392)).
Q6A particle moves in a straight line with displacement x = t^3 - 6t^2 + 9t metres from O at time t seconds. Find an expression for the velocity, and find the values of t at which the particle is instantaneously at rest.Show answer
Answer: v = 3t^2 - 12t + 9; at rest when t = 1 and t = 3 (factorise 3(t-1)(t-3) = 0).
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
A particle P moves in a straight line with constant acceleration. It passes through a point A with speed 5 m/s and, 8 seconds later, passes through a point B with speed 21 m/s. (a) Find the acceleration of the particle. (2) (b) Find the distance AB. (2)
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A particle moves in a straight line such that its acceleration at time t seconds is a = 4 - 2t m/s^2. When t = 0, the particle is at a fixed origin O and moving with velocity 3 m/s in the positive direction. (a) Find an expression for v, the velocity of the particle, in terms of t. (3) (b) Find the maximum velocity of the particle and the value of t at which it occurs. (2)
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At time t = 0, car A passes a fixed point O travelling in a straight line with speed 6 m/s and constant acceleration 2 m/s^2. At the same instant, car B is 20 m ahead of O on the same straight line, travelling in the same direction as A with constant speed 14 m/s (zero acceleration). Both cars are modelled as particles. (a) Write down an expression, in terms of t, for the distance of A from O at time t. (1) (b) Write down an expression, in terms of t, for the distance of B from O at time t. (1) (c) Show that, at the time when A catches up with B, t^2 - 8t - 20 = 0. (2) (d) Hence find the time at which A catches up with B. (2)
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Free printable worksheet
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