Displacement and total distance | Year 12

01 / LEARN

Understand the idea.

Direction is carried by the sign of velocity

Velocity includes direction. Its signed integral is displacement, the difference between final and initial position. Speed is $∣ v (t) ∣$ , so integrating speed gives total distance, counting movement in either direction positively.

Find zeros of velocity and check its sign on either side. A sign change indicates a reversal; touching zero without changing sign is only a stop. Initial position is needed for final position, but not for displacement or distance.

The velocity v(t)=t-2 metres per second is plotted for zero to five seconds. It increases from minus two to three, crossing zero at two seconds. No area values or solved displacement and distance are marked.

Work interval by interval

Split the time interval at every reversal. Add the signed interval integrals for displacement; add their magnitudes for distance. With movement in both directions, the distance exceeds the magnitude of displacement because opposite movements cancel only in displacement.

If acceleration is given, integrate once and use the initial velocity before analysing direction. Keep time units consistent: integrating metres per second with respect to seconds gives metres. A velocity-time sketch helps check the signs and relative contributions.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

For $v (t) = t - 2$ m/s on $0 \leq t \leq 5$ , find displacement and distance.

Worked solution

1
Locate the reversal.
$t - 2 = 0 \Rightarrow t = 2 s$
Velocity is negative before 2 seconds and positive afterwards.
2
Integrate each signed contribution.
$\begin{aligned} I_{1} & = {[\frac{t^{2}}{2} - 2 t]}_{0}^{2} = - 2 m \\ I_{2} & = {[\frac{t^{2}}{2} - 2 t]}_{2}^{5} = \frac{9}{2} m \end{aligned}$
Splitting at the reversal separates movement in the two directions.
3
Add signed changes for displacement.
$Δ s = - 2 + \frac{9}{2} = \frac{5}{2} m$
Displacement allows movement in opposite directions to cancel.
4
Add magnitudes for distance.
$d = 2 + \frac{9}{2} = \frac{13}{2} m$
Every metre travelled contributes positively to total distance.

Answer

Displacement $5 / 2$ m; distance $13 / 2$ m.

EXAMPLE 02

Worked example

A particle has $v (t) = (t - 1) (t - 3)$ m/s on $[0, 4]$ . Find distance.

Velocity v(t)=(t-1)(t-3) metres per second over zero to four seconds. The curve is above the time axis on zero to one and three to four, and below on one to three. Shading distinguishes the signed regions without supplying their areas.

Worked solution

1
Find the velocity zeros and signs.
$\begin{aligned} v (t) & = (t - 1) (t - 3) \\ v (t) & > 0 on [0, 1) and (3, 4] \\ v (t) & < 0 on (1, 3) \end{aligned}$
The two linear factors change sign at 1 and 3, giving two reversals.
2
Find an antiderivative of velocity.
$F (t) = \frac{t^{3}}{3} - 2 t^{2} + 3 t$
Expanding velocity gives $t^{2} - 4 t + 3$ , which can be integrated term by term.
3
Evaluate all three signed pieces.
$\begin{aligned} F (1) - F (0) & = \frac{4}{3} \\ F (3) - F (1) & = - \frac{4}{3} \\ F (4) - F (3) & = \frac{4}{3} \end{aligned}$
The values $F (0) = 0$ , $F (1) = \frac{4}{3}$ , $F (3) = 0$ and $F (4) = \frac{4}{3}$ determine the interval changes.
4
Sum their magnitudes.
$d = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4 m$
The negative middle displacement represents travel in reverse, which still contributes positive distance.

Answer

$4$ m

03 / YOUR TURN

Try it yourself.

Check your understanding

$v (t) = 2 - t$ on $[0, 4]$ . Find displacement and distance.

Show answer & reasoning

$0$ m and $4$ m

Velocity changes sign at $t = 2$ . The signed integrals are $2$ m on $[0, 2]$ and $- 2$ m on $[2, 4]$ , so displacement is $0$ m and distance is $2 + 2 = 4$ m.

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

$v (t) = 3$ m/s on $[0, 5]$ . Find displacement and distance.
3 marks
$v (t) = - 2$ m/s on $[0, 4]$ . Find displacement and distance.
3 marks
$v (t) = t - 1$ m/s on $[0, 3]$ . Find displacement and distance.
3 marks
$v (t) = {(t - 2)}^{2}$ m/s on $[0, 4]$ . Does it change direction at 2? Find distance.
3 marks
$v (t) = 2 t$ m/s and initial position is 5 m. Find position at 3 s.
3 marks
$v (t) = cos t$ m/s for $0 \leq t \leq π$ . Find distance.
3 marks

Read the worked solutions

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

$v (t) = 3$ m/s on $[0, 5]$ . Find displacement and distance.
3 marks
Velocity is positive throughout; integrate 3 over 5 seconds.
Both 15 m.
$v (t) = - 2$ m/s on $[0, 4]$ . Find displacement and distance.
3 marks
Velocity remains negative; integrate speed 2 for distance.
Displacement $- 8$ m; distance 8 m.
$v (t) = t - 1$ m/s on $[0, 3]$ . Find displacement and distance.
3 marks
Split at 1: contributions $- \frac{1}{2}$ and 2.
Add signed values or magnitudes as needed.
$\frac{3}{2}$ m and $\frac{5}{2}$ m.
$v (t) = {(t - 2)}^{2}$ m/s on $[0, 4]$ . Does it change direction at 2? Find distance.
3 marks
Velocity is nonnegative on both sides of 2.
${[\frac{{(t - 2)}^{3}}{3}]}_{0}^{4} = \frac{16}{3}$ .
No; distance $\frac{16}{3}$ m.
$v (t) = 2 t$ m/s and initial position is 5 m. Find position at 3 s.
3 marks
Displacement $= \int_{0}^{3} 2 t d t = 9$ m.
Position $5 + 9 = 14$ m.
14 m
$v (t) = cos t$ m/s for $0 \leq t \leq π$ . Find distance.
3 marks
Split at $\frac{π}{2}$ . Integrals are 1 and $- 1$ ; add their magnitudes.
$2$ m

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods integration and logarithmic antiderivatives; original learning and assessment material, with by-hand methods and explicitly described numerical models.

Official curriculum source ↗

In this lesson

Understand the idea.

Direction is carried by the sign of velocity

Work interval by interval

See the method in action.

Worked example

Locate the reversal.

Integrate each signed contribution.

Add signed changes for displacement.

Add magnitudes for distance.

Worked example

Find the velocity zeros and signs.

Find an antiderivative of velocity.

Evaluate all three signed pieces.

Sum their magnitudes.

Try it yourself.

Check your understanding

Keep practising