Displacement, velocity and acceleration models | Year 11

01 / LEARN

Understand the idea.

Connected motion quantities

Displacement gives signed position relative to an origin along a chosen positive direction. Its time derivative is velocity; differentiating again gives acceleration. With displacement in metres and time in seconds, these rates have units $m / s$ and $m / s^{2}$ .

Velocity records direction as well as rate. Speed is its nonnegative magnitude, $∣ v ∣$ . Negative velocity means movement opposite to the chosen positive direction. Acceleration describes changes in velocity, so positive acceleration can slow a particle that currently has negative velocity.

\begin{aligned} v (t) & = s^{'} (t) \\ a (t) & = v^{'} (t) \\ speed & = ∣ v (t) ∣ \end{aligned}

Rest, turning and distance

To locate possible changes of direction, solve $v (t) = 0$ within the given time domain and check the velocity sign on either side. Rest at one instant does not guarantee a reversal: velocity may reach zero and then keep the same sign.

Displacement over an interval is final position minus initial position. Total distance counts travel in both directions positively. Split at every actual direction change, calculate each section’s position change, then add their absolute values. Returning to the starting point gives zero displacement, not necessarily zero distance.

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

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

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

Worked solution

1
Differentiate to obtain velocity and acceleration.
$\begin{aligned} v (t) & = s^{'} (t) = (2 t - 6) m / s \\ a (t) & = v^{'} (t) = 2 m / s^{2} \end{aligned}$
Successive time derivatives give the two requested rate functions.
2
Locate and verify the direction change.
$\begin{aligned} v (t) = 0 & ⟹ t = 3 s \\ 0 \leq t < 3 : & v (t) < 0 \\ 3 < t \leq 5 : & v (t) > 0 \end{aligned}$
The velocity changes from negative to positive inside the stated interval, so distance must be split at 3 seconds.
3
Evaluate positions at the endpoints and turn.
$\begin{aligned} s (0) & = 8 m \\ s (3) & = 9 - 18 + 8 = - 1 m \\ s (5) & = 25 - 30 + 8 = 3 m \end{aligned}$
These positions determine the travel on each section without needing to integrate speed.
4
Find the signed displacement.
$s (5) - s (0) = 3 - 8 = - 5 m$
The final position lies 5 metres in the negative direction from the initial position.
5
Add the distances on both sections.
$\begin{aligned} D & = ∣ s (3) - s (0) ∣ + ∣ s (5) - s (3) ∣ \\ = ∣ - 1 - 8 ∣ + ∣ 3 - (- 1) ∣ \\ = 9 + 4 = 13 m \end{aligned}$
There is no reversal within either section, so each absolute position change equals its distance travelled.

Answer

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

EXAMPLE 02

Worked example 2

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

Worked solution

1
Identify the starting position.
$s (0) = 10 m$
At time zero the particle is 10 metres on the positive side of the origin.
2
Find the constant velocity.
$v (t) = s^{'} (t) = - 2 m / s$
A linear displacement function has a constant rate, with the negative sign indicating the negative direction.
3
Find acceleration and speed.
$a (t) = 0 m / s^{2}, ∣ v (t) ∣ = 2 m / s$
Constant velocity has zero acceleration, and its magnitude gives constant speed 2 metres per second in the negative direction.

Answer

Constant speed 2m/s in the negative direction.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

7m/s.

Speed is $∣ v ∣ = ∣ - 7 ∣ = 7 m / s$ . The positive acceleration does not change this instantaneous magnitude; here it acts opposite to the negative velocity, so the particle is slowing at that instant.

Keep practising

6 worksheet questions · 23 marks

Open worksheet 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 $v$ .
$v = 6 t + 2$ m/s; $a = 6$ m/s²; $v (2) = 14$ m/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 $t = 2$ s; 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 $= - 6$ m.
Velocity has no sign change, giving distance 6m.
Displacement −6m; distance 6m.

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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 ↗

In this lesson

Understand the idea.

Connected motion quantities

Rest, turning and distance

See the method in action.

Worked example 1

Differentiate to obtain velocity and acceleration.

Locate and verify the direction change.

Evaluate positions at the endpoints and turn.

Find the signed displacement.

Add the distances on both sections.

Worked example 2

Identify the starting position.

Find the constant velocity.

Find acceleration and speed.

Try it yourself.

Check your understanding

Keep practising