s = ut + ½at²

The idea

For constant acceleration, the displacement in time t is s = ut + (1/2)at², obtained from the area under the velocity-time graph.

How far did it go?

A car is already moving at 5 m s⁻¹ and speeds up steadily. We want the distance it covers in 10 s.

Which shape on its velocity-time graph do you think helps us find the displacement?

• A rectangle and a triangle under the line
• The line itself
• The time axis alone

Split the area

Press Next step to cut the area under the line into two simple shapes.

Rectangle plus triangle

The rectangle came from the initial velocity acting for the whole time: ut. The triangle came from the extra velocity gained, at, growing steadily from 0: ½ × t × at = ½at². Together they gave 75 m.

For constant acceleration, the displacement in time t is s = ut + ½at², obtained from the area under the velocity-time graph: the rectangle ut plus the triangle ½at².

Notes

For constant acceleration, s = ut + ½at²: the rectangle ut plus the triangle ½at² under the velocity-time graph.

Check yourself

A car has u = 5 m s⁻¹ and a = 0.5 m s⁻². What is its displacement in 10 s?

Answer: 75 m

s = 5 × 10 + ½ × 0.5 × 10² = 50 + 25 = 75 m.

A car starts from rest (u = 0) with a = 2 m s⁻². What is its displacement in 5 s?

Answer: 25 m

s = 0 + ½ × 2 × 5² = 25 m.

In s = ut + ½at², which term comes from the rectangle under the velocity-time graph?

An object moves with constant velocity, so a = 0. What does s = ut + ½at² become?

• ½at². This term comes from the triangle.
• at. The quantity at is the gain in velocity, the height of the triangle, not an area.
• ut — correct. Yes!
• s = ut — correct. Yes!
• s = ½at². The term ½at² is zero when a is zero, so only ut remains.
• s = 0. The term ut remains, so the displacement is not zero for a moving object.