Position–Time Equation

s = s_{0} + v_{0} t + \frac{1}{2} a t^{2}

Position under constant acceleration as an explicit function of time. The first two terms are the initial position and the uniform-velocity contribution; ½at² is the extra distance from speeding up or slowing down.

$s$ (m): Position at time t
$s_{0}$ (m): Initial position
$v_{0}$ (m/s): Initial velocity
$t$ (s): Elapsed time
$a$ (m/s²): Constant acceleration

Step-by-step solved example

s0 = 0, v0 = 8 m/s, a = 2 m/s². Position after 3 s?

s = s0 + v0 t + ½at² along a straight path.

1. Substitute
s = 0 + 8(3) + ½(2)(3)² = 24 + 9.
$s = 24 + 9$
2. Add
s = 33 m.
$s = 33 m$

Answer: 33 m

10 practice questions

0/10 correct

1.From rest, a = 4 m/s², t = 3 s. s =

2.s0 = 5 m, v0 = 8 m/s, a = 0, t = 4 s. s =

3.The ½at² term vanishes when

4.From rest, s = 20 m at a = 2.5 m/s². t =

5.v0 = 12 m/s, a = −4 m/s², t = 2 s, s0 = 0. s =

6.s(t) is quadratic in t when

7.Drop from rest, g = 9.81 m/s², t = 2 s. Distance ≈

8.Units of ½at² are the same as

9.s0 = 10 m, v0 = 0, a = 6 m/s², t = 2 s. s =

10.The coefficient of t² in s(t) is