Average Velocity Displacement Calculator (s = ½(u+v)t)

Solve s = ½(u + v)t for displacement, initial velocity, final velocity, or time using the average of the two velocities.

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Displacement from average velocity

When acceleration is constant, the average velocity over an interval is exactly the mean of the initial and final velocities. Multiplying by time gives the displacement:

s=12(u+v)ts = \frac{1}{2}(u + v)t

  • s — displacement, in metres (m)
  • u — initial velocity, in metres per second (m/s)
  • v — final velocity, in metres per second (m/s)
  • t — time elapsed, in seconds (s) — always positive

Enter any three values and leave the fourth blank; the calculator solves for it and shows the substitution step.

Worked example

A car speeds up from 0 m/s to 10 m/s over 5 seconds. How far does it travel?

  1. Formula: s = ½(u + v)t
  2. Substitute: s = ½ × (0 + 10) × 5
  3. Result: s = 25 m

Solving in reverse

The same formula rearranges to find initial velocity, final velocity, or time when the other three are known:

  • Initial velocity: u = 2s/t − v — enter s, v, and t, leave u blank.
  • Final velocity: v = 2s/t − u — enter s, u, and t, leave v blank.
  • Time: t = 2s / (u + v) — enter s, u, and v, leave t blank.

For example, a car covering 25 m while speeding up from 0 to 10 m/s takes t = 2 × 25 / (0 + 10) = 5 s.

Tip: This equation doesn't use acceleration at all — it's the fastest route to displacement or time when you already know both velocities. If u and v have opposite signs and don't match the direction of s, the solved time can come out negative; the calculator flags that instead of showing it.

Common mix-ups

  • Adding instead of averaging. The ½ in front of (u + v) is essential — forgetting it doubles the displacement.
  • Assuming acceleration is needed. This equation skips acceleration entirely — no need to compute it first if you already have both velocities and time.
  • Mixing this up with the acceleration form. When acceleration is known instead of the final velocity, use the displacement calculator (s = ut + ½at²) instead.

Where this shows up

s = ½(u + v)t is useful whenever a speedometer or radar gives you start and end speeds directly, without needing acceleration — average speed cameras and simple motion-sensor logs are common real-world sources. It cross-checks against final velocity (v = u + at) and velocity-squared (v² = u² + 2as) — all five SUVAT equations must agree for the same motion.

Frequently asked questions

What is the formula for displacement from average velocity?
The fourth SUVAT equation: s = ½(u + v)t, where s is displacement (m), u is initial velocity (m/s), v is final velocity (m/s), and t is elapsed time (s). It works because average velocity for constant acceleration is exactly the mean of the start and end speeds.
How do I find time if I know displacement and both velocities?
Rearrange the formula to t = 2s / (u + v). Enter the displacement, initial velocity, and final velocity, leave time blank, and the calculator solves it for you.
Why does the calculator reject some inputs when solving for time?
Time must be greater than 0. If the displacement, initial velocity, and final velocity you entered would only balance with a negative or zero elapsed time — for instance if the velocities point opposite to the displacement — the calculator flags the inputs instead of showing an impossible answer.
Can initial velocity or final velocity be negative?
Yes. A negative velocity means motion in the direction you called negative — the formula handles it correctly as long as displacement is measured on the same axis. Only time must stay positive.
What units does this calculator use?
SI units throughout — metres (m) for displacement, metres per second (m/s) for velocity, and seconds (s) for time.
How does this equation relate to the other SUVAT equations?
s = ½(u + v)t is one of two ways to get displacement without acceleration in the formula — the other, s = ut + ½at², needs acceleration instead. Combined with v = u + at and v² = u² + 2as, these five equations cover any constant-acceleration problem.

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