Acceleration and Equations of Motion

Uniformly accelerated motion described by the kinematic equations relating displacement, velocity, acceleration, and time.

Uniformly Accelerated Motion

When acceleration $a$ is constant:

Acceleration-time graph showing constant acceleration

Acceleration Definition

a = \dfrac{\Delta v}{\Delta t} = \dfrac{v - u}{t}

a = acceleration (m/s²), v = final velocity, u = initial velocity, t = time

The Four Kinematic Equations (SUVAT)

Equation	Missing
$v = u + at$	$s$
$s = ut + \frac{1}{2}at^2$	$v$
$v^2 = u^2 + 2as$	$t$
$s = \frac{1}{2}(u+v)t$	$a$

Note

SUVAT variables: $s$ = displacement, $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $t$ = time.

A ball is thrown upward at 20 m/s. Find its maximum height (g = 10 m/s²)

At max height, $v = 0$ . Use $v^2 = u^2 + 2as$ :

$0 = 400 + 2(-10)s \implies s = 20 \text{ m}$

Real world: Car accelerating from traffic lights

A car starts from rest at a red light and reaches 60 km/h (≈ 16.7 m/s) in 8 seconds. Its acceleration:

$a = \dfrac{16.7 - 0}{8} \approx 2.1 \text{ m/s}^2$

Distance covered during that acceleration: $s = \frac{1}{2} \times 2.1 \times 64 \approx 67 \text{ m}$

Real world: Train braking into a station

A train travelling at 90 km/h (25 m/s) needs to stop within 500 m as it approaches a station. Finding the required deceleration:

$v^2 = u^2 + 2as \implies 0 = 625 + 2a(500) \implies a = -0.625 \text{ m/s}^2$

The negative sign shows it is decelerating (braking).

Real world: Cyclist speeding up on a downhill

A cyclist starts a descent at 5 m/s and after 10 seconds is travelling at 15 m/s. Their acceleration:

$a = \dfrac{15 - 5}{10} = 1 \text{ m/s}^2$

Tip

Where do you see this in real life? Car manufacturers advertise "0–100 km/h in X seconds" — that is an acceleration figure. Anti-lock braking systems (ABS) control deceleration to keep it safe. Roller coasters are designed around exactly what accelerations the human body can tolerate.

Practice Problems

A car accelerates from rest at 3 m/s² for 8 s. Find final velocity and distance.

Show answer

v = 0 + 3×8 = 24 m/s. s = ½ × 3 × 64 = 96 m

An object decelerates from 30 m/s to 0 in 5 s. Find the deceleration.

Show answer

a = (0 − 30)/5 = −6 m/s²