Displacement: s = ut + ½at²: formula, calculator and worked example

What the symbols mean

$s$: the displacement in metres: distance in a chosen positive direction (it can be negative) (m)
$u$: the initial velocity in m/s (m/s)
$a$: the acceleration in m/s², assumed constant; negative if it acts against the positive direction (g = 9.81 m/s² for free fall) (m/s²)
$t$: the time in seconds (s)

When to use it

Use it with constant acceleration when the final velocity is unknown and not wanted: how far something falls in a given time, how far a car travels while accelerating. Solving for t needs the quadratic formula, and there can be two answers: a ball thrown upwards passes a given height on the way up and again on the way down. Negative times are before the motion started and are discarded.

Worked example

A stone is dropped from rest (u = 0) down a well and hits the water after t = 3 s. With a = 9.81 m/s², how deep is the water surface?

Start from the formula.
$s = u t + \frac{1}{2} a t^{2}$
Substitute $u$ = $0$ , $a$ = $9.81$ , $t$ = $3$ and work it out.
$s = 0 \cdot 3 + \frac{1}{2} \cdot 9.81 \cdot 3^{2}$
$= 0 + 0.5 \cdot 9.81 \cdot 9$
$= 0 + 4.905 \cdot 9$
$= 0 + 44.145$
$= 44.145 m$

Answer s ≈ 44.15 m below the top

Why it works

It is the area under the velocity–time graph from 0 to t: a rectangle $u t$ under the starting velocity plus a triangle $1/2×t×at$ for the extra speed gained.

Common mistakes

Forgetting the ½ in front of $a t^{2}$ .
Squaring the whole $a t$ instead of only t.
Keeping both roots for t without checking which make sense (a negative time is before the start).

Displacement: s = ut + ½at²

Calculator

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What the symbols mean

When to use it

Worked example

Why it works

Common mistakes