SUVAT Displacement — Speed Skater (Worked Example)

Easy

Problem

A speed skater starts from rest and accelerates uniformly at 1.5 m/s² for 6.0 s down the straight. How far does she travel in this time?

Given

u = 0 m/s
a = 1.5 m/s²
t = 6.0 s

Find

The displacement s.

Principle

Distance under constant acceleration is the area under the velocity-time graph — a triangle when starting from rest: s = ut + ½at².

Formula

s = ut + \frac{1}{2}at^2

s = ut + ½at²

Solution

Write the equation: $s = ut + \frac{1}{2}at^2$
The ut term is zero (starts from rest)
Substitute: $s = \frac{1}{2}(1.5)(6.0)^2$
Calculate: $s = 0.5 \times 1.5 \times 36 = 27\ \mathrm{m}$

27 m

Unit check

m/s² × s² = m. The result is a length — correct for displacement.

What this means physically

27 m is just over a quarter of a 100 m straight — a strong start, but the skater is still accelerating and will cover the next 27 m much faster.

Common mistake

Using s = vt with the final velocity instead of the SUVAT equation. Distance needs the average velocity here, which is exactly what ut + ½at² computes.