Uniformly Accelerated Motion: Speed and Distance Formulas

For motion at constant acceleration, the speed after a time t is v = initial speed + acceleration × t, and the distance travelled is x = initial speed × t + ½ × acceleration × t².

Explanation

When an object moves in a straight line with a constant acceleration, its speed and the ground it covers are fixed by the starting speed, the acceleration and the time elapsed.

v = v_0 + a t

x = v_0 t + \dfrac{1}{2} a t^2

$v$ is the speed after time $t$ , in m/s
$v_0$ is the initial speed, in m/s
$a$ is the acceleration, in m/s²
$t$ is the elapsed time, in seconds (s)
$x$ is the distance travelled in that time, in metres (m)

Read the distance formula in two parts. The term $v_0 t$ is how far the object would have gone at its original speed, and $\frac{1}{2} a t^2$ is the extra distance the acceleration buys.

Example

The defaults are an initial speed of 5 m/s, an acceleration of 2 m/s² and a time of 6 s.

v = 5 + 2 \times 6 = 17\,\mathrm{m/s}

x = 5 \times 6 + \dfrac{1}{2} \times 2 \times 6^2 = 30 + 36 = 66\,\mathrm{m}

After 6 seconds the object is moving at 17 m/s and has travelled 66 m.

Notes

Both formulas require the acceleration to hold steady for the whole interval. They say nothing useful about motion whose acceleration changes along the way.

A negative acceleration models slowing down, but the formula keeps running after the object has stopped. Starting at 5 m/s with an acceleration of −2 m/s², the object halts at 2.5 s. Enter a longer time and you get the answer for something that turned around and drove back.

Free fall, and a ball thrown straight up, are this same motion with the acceleration set to gravity, 9.8 m/s².

The time cannot be negative.