Equations of motion: s = ut + ½at²

Beginner

Displacement under constant acceleration.

Key relations = ut + ½at²

Displacement under constant acceleration is s = ut + ½at², the area under the v–t trapezoid. The first term is what you would cover at constant u, the second is the extra from accelerating.

Worked example

From rest, a = 2 m/s² for 5 s: s = ½·2·25 = 25 m.

Common mistake

Keep sign convention: if a opposes u, a is negative.

In more detail

The displacement under constant acceleration is s = ut + ½at², which combines the initial velocity and the effect of acceleration over time.

How to solve it

List u, a and t, and fix a positive direction for displacement.
Confirm that the acceleration is constant, since the formula assumes it.
Substitute into s = ut + ½at², keeping the sign of a.
Check that the answer is in metres and that its sign matches your chosen direction.

Worked examples

Example 1: A car is moving at 4 m/s and accelerates at 1.5 m/s² for 6 s. How far does it travel?

s = (4)(6) + ½(1.5)(6)² = 24 + ½(1.5)(36) = 24 + 27 = 51 m.

Example 2: A sprinter starts from rest with acceleration 3 m/s² for 4 s. How far has she run?

u = 0, so s = ½(3)(4)² = ½(3)(16) = 24 m.

Checking your answer

The average velocity over the interval is halfway between the start and end speeds, so the distance should equal that average times the time. In the first example the final speed is 4 + 9 = 13 m/s, the average is 8.5 m/s, and 8.5 × 6 = 51 m. This matches.

Real-world uses

Finding how far a car travels while accelerating from a traffic light.
Estimating the distance a sprinter covers before reaching top speed.
A dropped ball or a skier starting from rest on a slope uses this equation to find how far they travel in a known time.

Open the interactive lesson, with simulation and quiz →

Frequently asked questions

Why is there a half in front of at squared?

Because the velocity increases steadily from u to v, so the average velocity is halfway between them over the time.

Can I use this if acceleration is not constant?

No. Use integration or a numerical method instead.

Can I use this equation if the object starts moving?

Yes, as long as the acceleration is constant. Use u for the initial velocity, which is zero if the object starts from rest.