Kinematics – Describing Motion

Learn to describe and solve problems involving position, velocity, and acceleration using the SUVAT equations.

The Language of Motion

Kinematics describes how objects move without asking why they move.

Key quantities:

Position $x$ (m): where the object is
Velocity $v$ (m/s): rate of change of position
Acceleration $a$ (m/s²): rate of change of velocity

SUVAT Equations

For constant acceleration, use the SUVAT equations:

Variable	Symbol	Units
Displacement	$s$	m
Initial velocity	$u$	m/s
Final velocity	$v$	m/s
Acceleration	$a$	m/s²
Time	$t$	s

Equations:

$v = u + at$ $s = ut + \tfrac{1}{2}at^2$ $v^2 = u^2 + 2as$ $s = \tfrac{1}{2}(u+v)t$

Solving Strategy

1
List the known and unknown variables
Write down what you know (s, u, v, a, t) and what you need to find.
2
Choose the right equation
Pick the equation that contains your unknown and uses only known values.
3
Solve and check units
Substitute and calculate. Always check the units make sense.

Car from rest, acceleration 2 m/s². Find speed and distance after 10 s.

Known: $u = 0$ , $a = 2$ , $t = 10$ . Find $v$ and $s$ .

$v = u + at = 0 + 2 \times 10 = 20 \text{ m/s}$

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

Real world: car braking distance

A car travelling at $u = 30\text{ m/s}$ applies brakes giving deceleration $a = -6\text{ m/s}^2$ . Stopping distance: $v^2 = u^2 + 2as \Rightarrow 0 = 900 - 12s \Rightarrow s = 75\text{ m}$ . Road safety engineers use SUVAT to set speed limits and safe following distances on different road types.

Real world: landing aircraft

A plane touches down at $u = 70\text{ m/s}$ and decelerates at $2\text{ m/s}^2$ along the runway. Using $v^2 = u^2 + 2as$ with $v = 0$ : $s = \frac{70^2}{2 \times 2} = 1225\text{ m}$ . Runway length must exceed this distance. Pilots and airport designers use kinematics equations to ensure safe landings.

Real world: ball thrown upward (sports)

A volleyball player spikes the ball upward at $u = 12\text{ m/s}$ from a height of 2 m. Time to reach maximum height: $t = \frac{v - u}{a} = \frac{0 - 12}{-9.8} \approx 1.22\text{ s}$ . Additional height gained: $s = \frac{u^2}{2g} = \frac{144}{19.6} \approx 7.3\text{ m}$ . Peak height = $2 + 7.3 = 9.3\text{ m}$ — well above any block attempt.

Tip List all five SUVAT variables and mark what you know and what you need. Then choose the equation that includes only those variables — this removes the need to guess which formula to use.

Practice Problems

A ball is thrown upward at 15 m/s. When does it reach maximum height (g = 10 m/s²)?

Show answer

At max height v = 0: t = (v−u)/a = (0−15)/(−10) = 1.5 s

A car decelerates from 20 m/s to rest over 40 m. Find deceleration.

Show answer

v² = u² + 2as → 0 = 400 + 2a(40) → a = −5 m/s²

Free fall for 3 s from rest (g = 9.8 m/s²). Find distance fallen.

Show answer

s = ½ × 9.8 × 9 = 44.1 m