Kinematics is the branch of mechanics that describes motion without worrying about forces — it tells you where an object will be, how fast it will be moving, and how long everything takes, given a constant acceleration. This calculator covers the two most important scenarios you will encounter in school, university, and engineering: uniform linear acceleration (the SUVAT equations) and projectile motion (two-dimensional launch under gravity).
The four SUVAT equations
All four equations relate the same five variables — s, u, v, a, t — in different combinations. You always need three knowns to find one unknown.
| Equation | Variables involved | Use when |
|---|---|---|
| v = u + a·t | v, u, a, t | You know t but not s |
| s = u·t + ½·a·t² | s, u, a, t | You know t and want displacement |
| v² = u² + 2·a·s | v, u, a, s | Time is not known |
| s = ½·(u + v)·t | s, u, v, t | You know both velocities |
The calculator rearranges each equation algebraically for the chosen unknown before substituting your numbers, so the working matches textbook form exactly.
Projectile motion
A projectile launched at speed v₀ and angle θ above the horizontal follows a parabolic arc. The horizontal and vertical components are independent:
- Horizontal: constant velocity vx = v₀·cos θ (no acceleration)
- Vertical: vy = v₀·sin θ − g·t (acceleration g = 9.81 m/s² downward)
From these two equations, the closed-form results are:
- Time of flight (landing at original height): T = 2·v₀·sin θ / g
- Range: R = v₀²·sin(2θ) / g
- Maximum height: H = v₀²·sin²θ / (2·g)
If the launch height h₀ is not zero, the calculator solves the quadratic h₀ + vy₀·t − ½·g·t² = 0 for the positive root, then uses that time to find range and max height.
Worked example — car braking
A car travelling at 30 m/s (about 108 km/h) brakes with deceleration 6 m/s². How far does it travel before stopping?
Known: u = 30 m/s, v = 0 m/s, a = −6 m/s². Unknown: s.
Use v² = u² + 2·a·s, rearranged to s = (v² − u²) / (2·a):
s = (0² − 30²) / (2 × −6) = −900 / −12 = 75 m
Set the calculator to equation “v² = u² + 2·a·s”, solve for s, enter u = 30, v = 0, a = −6, and you get 75 m with full working shown.
Worked example — projectile at 45°
A ball is kicked at 20 m/s at 45° from ground level (h₀ = 0):
- vx = 20·cos 45° ≈ 14.14 m/s
- vy₀ = 20·sin 45° ≈ 14.14 m/s
- Time of flight = 2 × 14.14 / 9.81 ≈ 2.88 s
- Range = 14.14 × 2.88 ≈ 40.8 m (also = 20²·sin 90° / 9.81 ≈ 40.77 m)
- Max height = 14.14² / (2 × 9.81) ≈ 10.2 m
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