Displacement, velocity and acceleration models | Year 11

01 / LEARN

The idea, step by step

Connected motion quantities

Displacement is a signed position relative to an origin. Differentiate it to find velocity, then differentiate velocity to find acceleration. State the positive direction and the units in a physical interpretation.

Speed is the magnitude of velocity. Negative velocity means motion opposite to the positive direction, not negative speed. The sign of acceleration alone does not tell you the direction of motion.

v (t) = s' (t), a (t) = v' (t), speed = ∣ v (t) ∣ .

Rest, turning and distance

A particle is at rest when v=0, but changes direction only if v changes sign. A zero at an endpoint or without a sign change does not alone establish a turn.

Displacement is final minus initial position. To find total distance, split the interval at direction changes and add absolute position changes. A particle can travel a positive distance and have zero displacement.

s(t)=t²−6t+8 metres on 0≤t≤5 seconds
t(s)	0	3	5
s(m)	8	−1	3
Motion	Negative direction	Turns	Positive direction

A common trap

Using final position minus initial position as total distance after a turn.

Instead: Find sign changes in velocity and add the absolute position changes of each section.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

For $s (t) = t^{2} - 6 t + 8$ metres on $0 \leq t \leq 5$ , find velocity, acceleration, displacement and distance.

$v = 2 t - 6$ and $a = 2$ ; velocity changes − to + at t=3.
Positions at t=0,3,5 are 8,−1,3 metres.
Displacement=3−8=−5m; distance=|−1−8|+|3−(−1)|=13m.

v=2t−6m/s; a=2m/s²; displacement −5m; distance 13m.

Example 2

Worked example 2

$s (t) = 10 - 2 t$ metres. Describe the motion.

$v = - 2$ m/s and $a = 0$ .
The negative velocity indicates motion in the negative direction; its magnitude is 2.

Constant speed 2m/s in the negative direction.

Check your understanding

At an instant v=−7m/s and a=2m/s². What is speed?

Show answer

7m/s.

Speed is the magnitude |v|, independently of the acceleration value.

03 / WORKSHEET

Your turn

6 questions · 23 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $s (t) = 3 t^{2} + 2 t$ metres, find v, a and v(2).
3 marks
For $s (t) = t^{2} - 4 t$ metres, with t in seconds and $0 \leq t \leq 5$ , find the rest time and total distance.
5 marks
For $s (t) = t^{3} - 6 t^{2} + 9 t$ metres, with t in seconds and $0 \leq t \leq 4$ , find rest times, displacement and total distance.
6 marks
Velocity is −7m/s. State the speed and direction.
2 marks
For $s (t) = t^{2}$ , can velocity be zero while acceleration is nonzero?
3 marks
For $s (t) = 10 - 2 t$ , find displacement from 1s to 4s and total distance.
4 marks

Read the worked solutions

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

For $s (t) = 3 t^{2} + 2 t$ metres, find v, a and v(2).
3 marks
Differentiate s, then v; substitute 2 into v.
v=6t+2m/s; a=6m/s²; v(2)=14m/s.
For $s (t) = t^{2} - 4 t$ metres, with t in seconds and $0 \leq t \leq 5$ , find the rest time and total distance.
5 marks
$v = 2 t - 4 = 0$ at 2 with sign change − to +.
Positions 0,−4,5 give distance 4+9=13.
Rest at t=2s; distance 13m.
For $s (t) = t^{3} - 6 t^{2} + 9 t$ metres, with t in seconds and $0 \leq t \leq 4$ , find rest times, displacement and total distance.
6 marks
$v = 3 (t - 1) (t - 3)$ changes sign at 1 and 3.
Positions at 0,1,3,4 are 0,4,0,4; displacement 4 and distance 4+4+4=12.
Rest at 1s and 3s; displacement 4m; distance 12m.
Velocity is −7m/s. State the speed and direction.
2 marks
Use the magnitude for speed and sign for direction.
Speed 7m/s; negative direction.
For $s (t) = t^{2}$ , can velocity be zero while acceleration is nonzero?
3 marks
$v = 2 t$ but $a = 2$ .
Zero instantaneous velocity does not imply zero acceleration.
Yes: at t=0, v=0 and a=2.
For $s (t) = 10 - 2 t$ , find displacement from 1s to 4s and total distance.
4 marks
Positions are 8m and 2m, so displacement=−6m.
Velocity has no sign change, giving distance 6m.
Displacement −6m; distance 6m.

Checking saved progress…

Ready for the topic test?

Review the chapter, then choose a practice test.

Topic review

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topic 6: rates of change, derivatives and applications. Introductory polynomial calculus with explicit modelling and interpretation.

Official curriculum source ↗

Back to Introduction to calculus