Velocity, acceleration and speed | Year 12

01 / LEARN

The idea, step by step

Rates are vectors; speed is a scalar

Velocity is $v = r^{'}$ and acceleration is $a = r^{''}$ . Speed is $∣ v ∣ = \sqrt{x^{' 2} + y^{' 2} + z^{' 2}}$ and cannot be negative.

Rest requires every velocity component to be zero at the same time. A particle can pass its greatest height with nonzero horizontal velocity. Check the entire vector.

Acceleration need not change speed

For nonzero speed, $d ∣ v ∣ / d t = (v \cdot a) / ∣ v ∣$ . A positive numerator means speeding up and a negative one slowing down.

In uniform circular motion acceleration is perpendicular to velocity. Speed stays constant even though velocity changes direction. At rest, the displayed quotient cannot be used directly because its denominator vanishes.

A common trap

Speed is the derivative of distance from the origin.

Instead: Speed is the magnitude of the position derivative; radial distance can remain constant during motion.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

For $(t^{2}, 2 t)$ m, find velocity, acceleration and speed at t=1.

Differentiate once to (2 t, 2).
Differentiate again to (2, 0).
The magnitude of (2, 2) is $2 \sqrt{2}$ .

(2, 2)

m/s;

(2, 0)

m/s

2

;

2 \sqrt{2}

m/s.

Example 2

Worked example

For $(3 cos t, 3 sin t)$ , show speed is constant but acceleration is nonzero.

Velocity is $(- 3 sin t, 3 cos t)$ .
Its squared magnitude simplifies to 9.
Acceleration magnitude is 3 and points inward.

Speed 3; acceleration

(- 3 cos t, - 3 sin t)

Check your understanding

Find the magnitude of velocity (3, -4, 12).

Show answer

Take $\sqrt{9 + 16 + 144}$ .

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find speed for $(3 t, 4 t)$ m.
2 marks
Find velocity and acceleration at 2 for $(t^{2}, t^{3})$ .
2 marks
When is $(t^{2} - 2 t, t^{3} - 3 t)$ momentarily at rest?
2 marks
Given velocity (2, 1) and acceleration (-1, 0), find the speed rate.
2 marks
Find speed for $(cos 2 t, sin 2 t, 3 t)$ .
2 marks
Give an example of nonzero constant speed with zero acceleration.
2 marks

Read the worked solutions

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

Find speed for $(3 t, 4 t)$ m.
2 marks
Velocity is (3, 4).
Its magnitude is 5.
5 m/s.
Find velocity and acceleration at 2 for $(t^{2}, t^{3})$ .
2 marks
Velocity is $(2 t, 3 t^{2})$ .
Acceleration is (2, 6 t); substitute 2.
$(4, 12)$ and (2, 12).
When is $(t^{2} - 2 t, t^{3} - 3 t)$ momentarily at rest?
2 marks
The first velocity component $2 t - 2$ vanishes only at 1.
The second $3 t^{2} - 3$ also vanishes there.
t=1.
Given velocity (2, 1) and acceleration (-1, 0), find the speed rate.
2 marks
The dot product is-2.
Divide by speed $\sqrt{5}$ ; the particle slows down.
$- 2 / \sqrt{5}$
Find speed for $(cos 2 t, sin 2 t, 3 t)$ .
2 marks
Velocity is $(- 2 sin 2 t, 2 cos 2 t, 3)$ .
The squared speed is 4+9.
$\sqrt{13}$
Give an example of nonzero constant speed with zero acceleration.
2 marks
Velocity is the constant vector (3, 4).
Its derivative is zero.
$(3 t, 4 t)$ has speed 5 and zero acceleration.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics topic 6.3–6.5: polynomial and trigonometric parameterisations, related coordinate rates, motion, tangent directions and arc length. This is school Vector calculus, not vector-field calculus.

Official curriculum source ↗

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