3.4.1.3

The four SUVAT equations describe motion under constant acceleration

Motion Along a Straight Line — AQA A-Level Physics

$$v = u + at$$
  • $v$: final velocity (m s⁻¹)
  • $u$: initial velocity (m s⁻¹)
  • $a$: accelerationThe rate of change of velocity. A vector quantity. Measured in m s⁻². (m s⁻²)
  • $t$: time (s)
$$s = ut + \frac{1}{2}at^2$$
  • $s$: displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). (m)
  • $u$: initial velocity (m s⁻¹)
  • $a$: acceleration (m s⁻²)
  • $t$: time (s)
$$s = \frac{(v + u)}{2} t$$
  • $s$: displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). (m)
  • $v$: final velocity (m s⁻¹)
  • $u$: initial velocity (m s⁻¹)
  • $t$: time (s)
$$v^2 = u^2 + 2as$$
  • $v$: final velocity (m s⁻¹)
  • $u$: initial velocity (m s⁻¹)
  • $a$: acceleration (m s⁻²)
  • $s$: displacement (m)
Worked Example
A train travelling at 50 m s^-1 must decelerate uniformly to 10 m s^-1 in 20 s. How far apart should markers 1 and 2 be placed?
Show Solution
1
List known values
  • $u = 50$ m s$^{-1}$
  • $v = 10$ m s$^{-1}$
  • $t = 20$ s
  • $s = ?$
2
Choose the equation with s, u, v, t

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

3
Substitute and calculate

$$s = \frac{(50 + 10)}{2} \times 20 = 30 \times 20 = 600 \text{ m}$$

Answer
$s = 600$ m
Examiner Tips and Tricks
  • This is the most examined part of mechanics.
  • Step 1: write out s, u, v, a, t and identify what you know and what you need.
  • Step 2: pick the equation that contains those four quantities.
  • Step 3: substitute and solve.
Motion Along a Straight Line Overview