Kinematics of motion in a straight line

AS Level Mathematics · Mechanics. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Kinematics describes how a particle moves along a line. Displacement, velocity and acceleration are vectors: choose one direction as positive and give the opposite direction a minus sign. Distance and speed are scalars and are never negative.

On a displacement–time graph the gradient is the velocity. On a velocity–time graph the gradient is the acceleration and the area under the graph is the displacement.

If the acceleration is constant, use the constant acceleration formulae. If s or v is given as a function of t, the acceleration is usually not constant, so you must differentiate or integrate instead.

Rules to remember

Common mistake

Using the constant acceleration formulae when v or s is given as a function of t. These formulae work only when a is constant; otherwise differentiate or integrate.

Worked example

A cyclist accelerates uniformly from 4 m s−1 to 10 m s−1 while travelling 35 m. Find the acceleration.

  1. Known: u = 4, v = 10, s = 35. Time is not given, so use v2 = u2 + 2as.
  2. 102 = 42 + 2 × a × 35, so 100 = 16 + 70a.
  3. 70a = 84, so a = 84 ÷ 70 = 1.2.

Answer: 1.2 m s−2

Practice questions

Try each one, then open the answer.

1. A stone is released from rest at the top of a cliff and hits the sea 3 s later. Find the height of the cliff. (g = 10 m s−2)

  1. A
    30 m
  2. B
    90 m
  3. C
    45 m
  4. D
    15 m
Show answer

Answer: C. s = ut + (1/2)gt2 = 0 + (1/2)(10)(9) = 45 m. 30 m s−1 is the speed on impact, not the height.

2. A car passes point A at 10 m s−1 accelerating at 2 m s−2. At the same instant a motorcycle starts from rest at A with acceleration 4 m s−2 in the same direction. Find the time when the motorcycle draws level with the car.

  1. A
    5 s
  2. B
    20 s
  3. C
    2.5 s
  4. D
    10 s
Show answer

Answer: D. Car: s = 10t + t2. Motorcycle: s = 2t2. Equal displacements: 2t2 = 10t + t2, so t2 − 10t = 0 and t = 10 s (t = 0 is the start). Both are then 200 m from A.

3. A ball is thrown vertically upwards at 25 m s−1. Find the greatest height reached. (g = 10 m s−2)

  1. A
    62.5 m
  2. B
    12.5 m
  3. C
    31.25 m
  4. D
    125 m
Show answer

Answer: C. At the top v = 0. v2 = u2 − 2gs: 0 = 625 − 20s, so s = 31.25 m. Forgetting the 2 in 2gs gives 62.5 m.

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