Rates and motion applications | Year 12

01 / LEARN

The idea, step by step

Position, velocity and acceleration

For straight-line motion with position $s (t)$ , velocity is $s^{'} (t)$ and acceleration is $s^{''} (t)$ . Velocity can be negative; speed is its absolute value. A zero velocity is a rest instant, not automatically a change of direction.

A direction reversal requires the velocity to change sign. A particle speeds up when velocity and acceleration have the same sign and slows down when their signs differ.

Rates involving a changing measurement

If a radius or length depends on time, differentiate the geometric relationship with respect to time. For $A = π r^{2}$ , the rate is $d A / d t = 2 π r d r / d t$ .

Substitute the current measurement and its rate after differentiating. Use square units per time for area change and cubic units per time for volume change.

A common trap

Treating negative acceleration as always meaning a particle slows down.

Instead: Compare acceleration with velocity. Negative velocity and negative acceleration produce increasing speed.

02 / WORKED EXAMPLES

See the thinking

Example 1

Rest times and reversals

A particle has $s (t) = t^{3} - 6 t^{2} + 9 t$ m for $0 \leq t \leq 4$ s. Find rest times and direction changes.

$v = 3 (t - 1) (t - 3)$ , so it is at rest at 1 and 3 seconds.
The velocity signs are positive, negative, positive across those times.

It reverses direction at both

t = 1

s and

t = 3

Example 2

A changing circular area

A circle has radius increasing at 0.4 m/s. Find its area rate when $r = 5$ m.

Differentiate $A = π r^{2}$ : $d A / d t = 2 π r d r / d t$ .
Substitute $r = 5$ and $d r / d t = 0.4$ .

4 π m^{2} /s

Check your understanding

If $v = - 3$ m/s and $a = - 2$ m/s squared, is speed increasing or decreasing?

Show answer

Increasing.

The two signs agree: velocity is becoming more negative, so its magnitude increases.

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $s (t) = 2 t^{2} - 5 t + 1$ metres, with t in seconds, find velocity and acceleration.
2 marks
For $s (t) = t^{3} - 3 t$ metres on $t \geq 0$ seconds, find the rest time and whether direction changes.
3 marks
For $s (t) = t^{3} - 6 t^{2} + 9 t$ metres on $0 \leq t \leq 4$ seconds, find displacement and total distance.
4 marks
For velocity $v (t) = t^{2} - 4 t + 3$ metres per second, with t in seconds, find the speed and its instantaneous rate of change at $t = 2$ seconds.
3 marks
A cube has side length increasing at 0.2 cm/s. Find the volume rate when the side is 3 cm.
3 marks
A cost model is $C (q) = 0.02 q^{2} + 5 q + 100$ dollars. Find and interpret marginal cost at 50 units.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For $s (t) = 2 t^{2} - 5 t + 1$ metres, with t in seconds, find velocity and acceleration.
2 marks
Differentiate position once for velocity.
Differentiate again for acceleration.
$v = 4 t - 5$ m/s; $a = 4$ m/s squared.
For $s (t) = t^{3} - 3 t$ metres on $t \geq 0$ seconds, find the rest time and whether direction changes.
3 marks
The velocity is $3 (t - 1) (t + 1)$ .
On the physical domain it changes sign at 1.
$t = 1$ s; direction changes from negative to positive.
For $s (t) = t^{3} - 6 t^{2} + 9 t$ metres on $0 \leq t \leq 4$ seconds, find displacement and total distance.
4 marks
The reversal times are 1 and 3.
Positions at $0, 1, 3, 4$ are $0, 4, 0, 4$ .
Add absolute position changes for distance.
Displacement 4 m; distance 12 m.
For velocity $v (t) = t^{2} - 4 t + 3$ metres per second, with t in seconds, find the speed and its instantaneous rate of change at $t = 2$ seconds.
3 marks
The velocity at 2 is $- 1$ .
Acceleration is $2 t - 4$ , which is zero at 2.
Speed 1 m/s; acceleration is zero at that instant, so the instantaneous rate of change of speed is zero.
A cube has side length increasing at 0.2 cm/s. Find the volume rate when the side is 3 cm.
3 marks
Use $V = x^{3}$ and differentiate with respect to time.
Then $d V / d t = 3 x^{2} d x / d t = 3 (9) (0.2)$ .
$5.4 {cm}^{3} /s$
A cost model is $C (q) = 0.02 q^{2} + 5 q + 100$ dollars. Find and interpret marginal cost at 50 units.
3 marks
Differentiate with respect to quantity: $C^{'} (q) = 0.04 q + 5$ .
At 50, this is 7; it is a local rate, not the exact finite cost of the next unit.
$7$ per additional unit approximately.

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Curriculum reference & lesson scope

SACE Stage 2 Mathematical Methods

Further differentiation and applications from the current SACE Methods outline. Logarithmic derivatives have their own chapter; integration follows later. Trigonometric calculus uses radians.

Official curriculum source ↗

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