Rates of Change: Velocity and Acceleration | IEB Grade 12 Maths

The derivative measures how fast one quantity changes compared to another. In Paper 1 this shows up as rates of flow, growth, velocity and acceleration, usually in an unfamiliar context the IEB writes specially for the paper.

Instantaneous rate of change

If a quantity $Q$ depends on time $t$ , then $Q^{'} (t)$ is the instantaneous rate of change of $Q$ at time $t$ . The units are "units of $Q$ per unit of $t$ ".

Example. Water flows into a tank so that the volume after $t$ minutes is $V (t) = 0, 5 t^{2} + 3 t$ litres.

$V^{'} (t) = t + 3$

At $t = 4$ : $V^{'} (4) = 7$ litres per minute.

Compare this with the average rate of change between two times, $\frac{V ( b ) - V ( a )}{b - a}$ , which is a secant gradient. Read the question carefully to see which one is asked.

Velocity and acceleration

If $s (t)$ is the displacement (position) of an object at time $t$ :

Velocity is $v (t) = s^{'} (t)$
Acceleration is $a (t) = v^{'} (t) = s^{''} (t)$

The object is at rest when $v (t) = 0$ . It reaches its maximum height when its upward velocity is 0.

Example. A ball is thrown upwards so that $h (t) = 1, 5 + 19, 6 t - 4, 9 t^{2}$ metres.

$v (t) = 19, 6 - 9, 8 t = 0 \Rightarrow t = 2 s$

$h (2) = 1, 5 + 39, 2 - 19, 6 = 21, 1 m$

Acceleration is $a (t) = - 9, 8$ m/s², which is gravity acting downwards.

Key terms

Instantaneous rate of change: The derivative at a particular moment.
Average rate of change: The change in a quantity divided by the change in time over an interval.
Velocity: The rate of change of displacement; it has a sign showing direction.
Acceleration: The rate of change of velocity.

Common mistakes

Giving the value of the quantity, such as height, when the rate is asked for.
Giving the average rate when the instantaneous rate is asked for.
Leaving out or mixing up units such as m/s and m/s².
Forgetting that a negative velocity means moving in the opposite direction.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

routine

Water flows into a tank. The volume after $t$ minutes is $V (t) = 0, 5 t^{2} + 3 t$ litres.

Determine the rate at which water flows into the tank after 4 minutes.

Worked solution

Hidden - try the question first

Solution

$V^{'} (t) = t + 3$
$V^{'} (4) = 4 + 3$
$= 7$ litres per minute

Question 2

5 marks total

routine

A particle moves along a straight line. Its displacement after $t$ seconds is $s (t) = t^{3} - 6 t^{2} + 9 t$ metres.

(a) Determine its velocity after 2 seconds. (3)

(b) At what times is the particle at rest? (2)

Worked solution

Hidden - try the question first

Solution

(a) $v (t) = s^{'} (t) = 3 t^{2} - 12 t + 9$
$v (2) = 12 - 24 + 9$
$= - 3$ m/s
(b) $3 t^{2} - 12 t + 9 = 0 \Rightarrow 3 (t - 1) (t - 3) = 0$
$t = 1$ s or $t = 3$ s

Question 3

6 marks total

complex

A ball is thrown upwards from the top of a wall. Its height above the ground after $t$ seconds is $h (t) = 1, 5 + 19, 6 t - 4, 9 t^{2}$ metres.

(a) Determine the maximum height of the ball. (3)

(b) Determine the velocity of the ball when it hits the ground. Give your answer to one decimal place. (3)

Worked solution

Hidden - try the question first

Solution

(a) $v (t) = s^{'} (t) = 3 t^{2} - 12 t + 9$
$v (2) = 12 - 24 + 9$
$= 21, 1$ m
(b) $4, 9 t^{2} - 19, 6 t - 1, 5 = 0 \Rightarrow t = \frac{19 , 6 + 19 , 6 ^{2} + 4 ( 4 , 9 ) ( 1 , 5 )}{9 , 8}$
$t \approx 4, 08$ s (the negative root is rejected)
$h^{'} (4, 08) \approx - 20, 3$ m/s, so it hits the ground at about 20,3 m/s downwards

Question 4

8 marks total

problem-solving

After a patient takes a tablet, the concentration of the medicine in the blood after $t$ hours is $C (t) = 6 t^{2} - t^{3}$ mg/ℓ, for $0 \leq t \leq 6$ .

(a) At what rate is the concentration changing after 1 hour? (2)

(b) When is the concentration at its highest, and what is it? (3)

Worked solution

Hidden - try the question first

Solution

(a) $v (t) = s^{'} (t) = 3 t^{2} - 12 t + 9$ , so $C^{'} (1) = 9$ mg/ℓ per hour
(b) $12 t - 3 t^{2} = 0 \Rightarrow 3 t (4 - t) = 0$
$t = 4$ hours (at $t = 0$ the concentration is 0)
$C (4) = 96 - 64 = 32$ mg/ℓ
(c) The rate $C^{'} (t)$ is greatest where its derivative is 0: $C^{''} (t) = 12 - 6 t = 0$
$t = 2$ hours
$C^{''}$ changes from positive to negative there, so $C^{'} (2) = 12$ mg/ℓ per hour is the maximum rate

FAQs

Is physics knowledge needed for velocity questions in IEB Maths?

No. Everything you need is in the question. Velocity is the derivative of displacement and acceleration is the derivative of velocity.

How do I know if a question wants average or instantaneous rate?

'At' a moment means instantaneous: use the derivative. 'Between' two times or 'over' an interval means average: use the gradient formula.

What does it mean when the rate is fastest?

The rate itself is a function, so you maximise it by setting its derivative, the second derivative of the original, equal to 0.

Rates of change

Instantaneous rate of change

Velocity and acceleration

Key terms

Common mistakes

Practice questions

FAQs

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