The SUVAT equations apply only to motion with constant (uniform) acceleration

Motion & Kinematics - OCR A-Level Physics

The SUVAT equations apply only to motion with constant (uniform) accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻².
s = displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m)., u = initial velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹., v = final velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹., a = accelerationThe rate of change of velocity. A vector quantity. Measured in m s⁻²., $t = time$
To select the correct equation, identify which of the five variables are known (you need three) and which you want to find
If an object decelerates, a is negative (opposing the direction of motion)

$$v = u + at$$

s = ut + \frac{1}{2}at^{2}

v^{2} = u^{2} + 2as

s = \frac{(u + v)}{2} \times t

Examiner Tips and Tricks

Always list your known SUVAT variables before choosing an equation. 'Starts from rest' means u = 0. 'Comes to rest' means $v = 0$.
This step prevents wrong equation selection.

Worked Example [5 marks]

A car accelerates from rest at 2.0 $m s^{-2}$ for 10 s, then brakes at -4.0 $m s^{-2}$ until it stops. Find the total distance travelled.

Show Solution

Stage 1

$u = 0, a = 2.0 \text{ m s}^{-2}, t = 10 \text{ s}. Use v = u + at: v = 0 + 2.0 \times 10 = 20 \text{ m s}^{-1}$

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Stage 1 distance

$s_1 = ut + \frac{1}{2}at^{2} = 0 + \frac{1}{2}(2.0)(10^{2}) = 100 \text{ m}$

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Stage 2

$u = 20, v = 0, a = -4.0. Use v^{2} = u^{2} + 2as: 0 = 400 + 2(-4.0)s$

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$s_2 = \frac{400}{8.0} = 50 \text{ m}$

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$Total distance = s_1 + s_2 = 100 + 50 = 150 \text{ m}$

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Answer

150 m