SUVAT Calculator

Solve constant-acceleration motion problems using SUVAT: displacement s, initial velocity u, final velocity v, acceleration a, and time t. Enter any 3 known values and we’ll compute the rest — with steps, unit support, and a mini v–t / s–t visual.

Background

SUVAT equations describe motion with constant acceleration (straight-line kinematics). They’re perfect for free-fall (ignoring air resistance), cars speeding up/slowing down, and uniform acceleration problems.

Enter any 3 known values

Unit system

Sign convention: use + for your positive direction. Example: if “up” is positive, gravity is negative.

Direction convention

Up / forward is positive (gravity is negative)

Down is positive (gravity is positive)

Solve mode

I know any 3 values (solve everything)

Solve for a target variable (tell me what inputs I need)

Variables (leave unknowns blank)

Displacement, s (m)

Initial velocity, u (m/s)

Final velocity, v (m/s)

Time, t (s)

Acceleration, a (m/s²)

Use Set a = ±g to apply gravity with your chosen sign convention.

Options

Round decimals

Prefer physical time (t ≥ 0)

Show step-by-step

Show mini visuals (v–t and s–t)

Quick picks:

Chips prefill common kinematics scenarios. You can tweak values after the fill.

Result:

No results yet. Enter any 3 known values and click Calculate.

How to use this calculator

Pick your sign convention (up/forward positive, or down positive).
Enter any 3 of the 5 SUVAT variables: s, u, v, a, t.
Click Calculate to solve the remaining variables.
If you see multiple valid solutions, choose the one you want (usually t ≥ 0).

Tip: In free fall (ignoring air resistance), use a = -9.81 m/s² if “up” is positive.

How this calculator works

It uses the 5 constant-acceleration equations and solves for missing variables, including quadratic cases.
If an equation yields two real solutions (common for time), you’ll be offered both, then filtered by t ≥ 0 when enabled.
Visuals show the implied motion: v–t is linear, s–t is parabolic when a ≠ 0.

Formula & Equation Used

1) v = u + at

2) s = ut + \u00bdat^2

3) v^2 = u^2 + 2as

4) s = \u00bd(u + v)t

5) s = vt - \u00bdat^2

Example Problem & Step-by-Step Solution

Example 1 — Free fall from rest (ignoring air resistance)

A ball is dropped from rest and falls for t = 3 s. Find v and s. Use u = 0 and a = -9.81 m/s² (up positive).

v = u + at = 0 + (-9.81)(3) = -29.43 m/s
s = ut + ½at² = 0(3) + ½(-9.81)(3²) = -44.145 m

Negative means “down” because we chose “up” as positive.

Answer: v = -29.43 m/s, s = -44.145 m

Example 2 — Braking to a stop (find stopping distance)

A car is moving at u = 25 m/s and brakes with constant acceleration a = -5 m/s² until it stops (v = 0). Find the stopping distance s.

Use v² = u² + 2as (no time needed).
Substitute: 0² = 25² + 2(-5)s
Solve: 0 = 625 - 10s → 10s = 625 → s = 62.5 m

The negative a means the car is slowing down in the positive direction of motion.

Answer: s = 62.5 m

Example 3 — Toss up & return to start (choose the physical time)

A ball is thrown upward with u = 12 m/s. Take “up” as positive, so gravity is a = -9.81 m/s². The ball returns to the launch point, so s = 0. Find the flight time t.

Use s = ut + ½at².
Substitute: 0 = (12)t + ½(-9.81)t²
Factor: 0 = t(12 - 4.905t)
Solutions: t = 0 or t = 12/4.905 ≈ 2.45 s
Pick the physical one: t ≈ 2.45 s (the non-zero flight time).

Getting t = 0 is normal — it’s the “starting instant.” The meaningful solution is the later return time.

Answer: t ≈ 2.45 s