Kinematics – Differentiation, Velocity and Maximum Height | Pearson Edexcel International GCSE Maths

Kinematics uses differentiation to connect displacement, velocity and acceleration. In this example, a stone is projected vertically upwards and its displacement is given as a function of time. Differentiation allows us to find its velocity and determine when it reaches its maximum height.

The Displacement Function

A stone is projected vertically upwards from the ground. After t seconds, its height above the ground, s metres, is given by:

s(t) = 15t − 4.9t²

0 ≤ t ≤ 4

Question (a) – Find ds/dt

Differentiate the displacement function with respect to time.

Working

s(t) = 15t − 4.9t²

ds/dt = 15 − 4.9 × 2 × t
= 15 − 9.8t
Answer: ds/dt = 15 − 9.8t

Question (b) – Velocity at t = 0.5

Find the velocity of the stone when t = 0.5 seconds.

Working

v = ds/dt

v(t) = 15 − 9.8t
v(0.5) = 15 − 9.8(0.5)
= 10.1

Answer: 10.1 m/s
Answer: 10.1 m/s

Displacement and Velocity

Velocity is the rate of change of displacement with respect to time. Therefore, if displacement is represented by s(t), velocity is obtained by differentiating:

v(t) = ds/dt

For this example:

s(t) = 15t − 4.9t²
v(t) = 15 − 9.8t

Question (c) – Maximum Height

Find the maximum height of the stone above the ground, giving the answer correct to 1 decimal place.

Working

s(t) = 15t − 4.9t²
= −4.9t² + 15t

s′(t) = 15 − 9.8t

s′(t) = 0

⇒ 15 − 9.8t = 0
9.8t = 15
t = 15 / 9.8
= 75 / 49

s(75 / 49) = 15(75 / 49) − 4.9(75 / 49)²

= 11.5 m
Answer: 11.5 m

Why Set the Derivative Equal to Zero?

As the stone travels upwards, its velocity is positive. Gravity causes the stone to slow down until, at the highest point, its instantaneous velocity becomes zero.

v(t) = s′(t) = 0

Solving this equation gives the time at which the maximum height occurs:

t = 75/49 seconds

This value of t is then substituted back into the original displacement function to find the maximum height.

Key Results

Displacement:
s(t) = 15t − 4.9t²

Velocity:
v(t) = 15 − 9.8t

Velocity at t = 0.5:
10.1 m/s

Maximum height:
11.5 m

Source: Question adapted from Harry Smith, Revise Pearson Edexcel International GCSE (9–1) Mathematics A – Higher Tier Revision Guide, Pearson.

Solutions and workings: Tiago Hands.

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