IGCSE 0625 · Section 1.2 · Extended
Supplement (Extended) content

The change in velocity, per second.

Acceleration is not about how fast you are going. It is about how fast your velocity is changing. Get the small word "change" right, and the equation and the velocity-time graph fall into place together.

The Key Idea

Acceleration is the change in velocity per unit time: a = Δv ÷ t = (v − u) ÷ t, measured in m/s². On a velocity-time graph it is the gradient of the line. A negative value means deceleration (slowing down).

SECTION 01

Acceleration is a gradient.

Drag the start and end points of the line on the graph below. The line's steepness is the acceleration. Notice the trap built into the readout: dividing the final velocity by time (v ÷ t) gives the wrong answer whenever the object did not start from rest.

SECTION 02

Speeding up, slowing down.

The single most common acceleration error

Acceleration is Δv ÷ t, not v ÷ t. If a car speeds up from 4 m/s to 20 m/s in 8 s, the acceleration is (20 − 4) ÷ 8 = 2.0 m/s², not 20 ÷ 8. Only when the object starts from rest (u = 0) do the two happen to agree. Always subtract the starting velocity first.

Worked Example

A train slows from 30 m/s to 18 m/s in 6.0 s. Calculate its acceleration and state what the sign tells you.

Step 1 · Change in velocity Δv = v − u = 18 − 30 = −12 m/s
Step 2 · Divide by time a = Δv ÷ t = −12 ÷ 6.0 = −2.0 m/s²
Step 3 · Interpret the sign The acceleration is −2.0 m/s². The negative sign shows the train is decelerating (slowing down).
Practice this topic →
Six original Cambridge-style questions.
Defining acceleration, the change-in-velocity calculation, deceleration and its sign, reading acceleration off a graph, and the v / t versus Δv / t trap. Attempt each, then reveal the worked solution.