Displacement and total distance | Year 12

01 / LEARN

The idea, step by step

Direction is carried by the sign of velocity

Displacement is final position minus initial position. Its signed value is $\int_{a}^{b} v (t) d t$ . Total distance counts all movement positively, so it is $\int_{a}^{b} ∣ v (t) ∣ d t$ . These agree only when velocity is nonnegative throughout; if velocity is always negative, distance is the negative of displacement.

Solve $v (t) = 0$ and test signs to locate direction changes. A zero velocity without a sign change is a stop, not a reversal. If initial position is given, add displacement to find final position.

Work interval by interval

Calculate the signed integral between each pair of direction-change times. Add these signed values for displacement; add their magnitudes for distance. Use a velocity-time sketch as a check on where the positive and negative contributions occur.

Do not integrate acceleration directly for distance. First obtain velocity, using its initial condition; then decide its sign over the time interval. Keep time units consistent so that the integral has the intended length unit.

A common trap

Taking the absolute value of total displacement to get distance.

Instead: Split at every genuine direction change before taking magnitudes.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

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

Velocity changes sign at $t = 2$ .
The signed integrals on $[0, 2]$ and $[2, 5]$ are $- 2$ and $9 / 2$ .
Displacement $= - 2 + 9 / 2 = 5 / 2$ ; distance $= 2 + 9 / 2 = 13 / 2$ .

Displacement

5 / 2

m; distance

13 / 2

Example 2

Worked example

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

Zeros at 1 and 3 give signs+,-,+.
An antiderivative is $t^{3} / 3 - 2 t^{2} + 3 t$ .
Signed contributions are $4 / 3, - 4 / 3, 4 / 3$ ; their magnitudes sum to 4.

4

Check your understanding

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

Show answer

$0$ m and $4$ m

The two signed triangular contributions are 2 and $- 2$ .

03 / WORKSHEET

Your turn

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

Worksheet PDF 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 $- 1 / 2$ and 2.
Add signed values or magnitudes as needed.
$3 / 2$ m and $5 / 2$ 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.
${[{(t - 2)}^{3} / 3]}_{0}^{4} = 16 / 3$ .
No; distance $16 / 3$ 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 $π / 2$ . 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.

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