Kinematic Equations Calculator (SUVAT)

The four kinematic equations

For motion with constant acceleration in one dimension, these four equations relate the five variables u (initial velocity), v (final velocity), a (acceleration), s (displacement), and t (time):

(1) v = u + a t (2) s = u t + ½ a t² (3) v² = u² + 2 a s (4) s = ½ (u + v) t

Each equation uses exactly four of the five variables; the variable not appearing is whichever you do not need for that particular calculation. If you know any three of u, v, a, s, t, you can solve for the other two using these equations.

Physicists call these the SUVAT equations after the variables (S for displacement, U for initial velocity, V for final velocity, A for acceleration, T for time). They are introduced early in mechanics courses and remain the workhorse for projectile motion, free fall, vehicle braking, and any other constant-acceleration problem.

When they apply

These equations require constant acceleration. If a varies with time, you need calculus instead (integrate to get v, integrate again to get s). Constant acceleration is the common case in introductory mechanics because gravity near the Earth surface, an idealized brake, or a steady rocket thrust all produce constant a.

Real-world problems usually have constant-enough acceleration that SUVAT is an excellent approximation. A car braking at “approximately 0.7 g” until it stops is a perfectly good SUVAT problem. Air resistance, banking on a turn, and engine variations make a vary in practice, but for the duration of most everyday problems the average a is close enough to constant.

Deriving the equations from each other

Equation (1), v = u + at, comes directly from the definition of acceleration: a = (v − u)/t, rearranged. Equation (4), s = ½(u + v)t, comes from the average velocity: when a is constant, the average is the arithmetic mean of initial and final velocities, and displacement = average velocity × time.

Equation (2) follows by substituting v = u + at into equation (4): s = ½(u + u + at)t = ut + ½at². Equation (3) follows by eliminating t between (1) and (4): from (1) t = (v−u)/a, plugging into (4): s = ½(u+v)(v−u)/a = (v² − u²)/(2a), rearranging gives v² = u² + 2as. So strictly speaking only two of the four equations are independent; the other two come for free.

A worked example: car braking

A car traveling at 27 m/s (about 100 km/h) brakes with constant deceleration of 6 m/s². How far does it travel before stopping, and how long does it take?

Known: u = 27 m/s, v = 0 (stops), a = −6 m/s² (deceleration). Find s and t.

Use (3) v² = u² + 2as to find s: 0 = 27² + 2(−6)s → s = 729/12 = 60.75 m. Use (1) v = u + at to find t: 0 = 27 + (−6)t → t = 4.5 s.

So at 100 km/h on dry pavement (a typical max deceleration), the stopping distance is about 60 meters and the time is about 4.5 seconds. Plus the driver reaction time (about 1 to 1.5 seconds adds 30 to 40 more meters at 100 km/h). This is why tailgating is dangerous.

Free-fall example: dropped object

A rock is dropped from a 45 m cliff. How fast is it going when it hits the ground, and how long does it take to fall?

Known: u = 0 (dropped, not thrown), a = 9.81 m/s² (free fall, gravity downward), s = 45 m. Find v and t.

Use (3) v² = u² + 2as: v² = 0 + 2(9.81)(45) = 882.9 → v ≈ 29.7 m/s. Use (1) v = u + at: 29.7 = 0 + 9.81 t → t ≈ 3.03 s.

A 45-meter drop takes about 3 seconds and the rock hits the ground at almost 30 m/s. That is roughly 67 mph, well into the lethal range for a falling object.

Projectile motion: independent horizontal and vertical

For 2D projectile motion (a ball thrown at an angle), the horizontal and vertical components are independent. The horizontal motion has zero acceleration (ignoring air drag), so x = u_x × t. The vertical motion has a = −g and uses the SUVAT equations directly with u_y as the initial vertical velocity. For a projectile launched at angle θ with initial speed v_0:

u_x = v_0 cos θ, u_y = v_0 sin θ

Range, max height, time of flight all follow from applying SUVAT to the vertical motion. This calculator handles the one-dimensional case; for trajectory work, decompose into x and y components and apply SUVAT twice.

Signs and convention

The biggest source of student error in SUVAT is sign conventions. Pick a positive direction (usually right or up) and stick with it. Velocities in the negative direction get negative signs. Acceleration in the negative direction (like deceleration when motion is positive, or gravity when up is positive) gets a negative sign.

A car traveling at +27 m/s and braking has a = −6 m/s², not +6. The deceleration “magnitude” is 6 m/s², but the signed acceleration is negative because it opposes the positive motion. Plugging in positive 6 into the equations gives wrong answers (car accelerates instead of slowing down).

Common pitfalls

Equation (3) gives v², not v. When solving, take the square root and pay attention to sign: the physical solution is usually positive (object moves in the original direction), but if motion reverses, v can be negative.

When a = 0 (constant velocity), equation (3) becomes 0 = 2(0)s = 0, which is satisfied trivially. You cannot solve for s from (3) alone in this case; use s = u × t (equation 2 with a = 0) or s = v × t (since u = v when a = 0).

When the unknown appears in both (2) and (1) (you know s and a but want t), you get a quadratic in t. Two solutions exist; the physically meaningful one is usually the smaller positive root.

Multi-stage motion

Many problems involve different accelerations during different phases (a car accelerating, then cruising, then braking). For each phase, apply SUVAT to that phase only, using the end conditions of one phase as the start conditions of the next. Total distance and time are sums of the phase distances and phase times.

This is also how you handle motion in changing-gravity environments (going from a planet surface to orbit) or staged rocket flight. SUVAT is the building block; chaining is the technique.

How this calculator works

Enter any three of the five SUVAT variables. Leave the other two blank. The calculator detects which three you provided, picks the right equation(s), and computes the missing two. It will tell you which equation it applied. If three knowns are inconsistent (no real solution exists), the result will say so rather than producing a misleading number.