Kinematics calculator

What are the kinematic equations?

For constant acceleration: v = u + at, s = ut + ½at², v² = u² + 2as and s = (u + v)t ÷ 2, where s is displacement, u the initial velocity, v the final velocity, a the acceleration and t the time.

How do I use the calculator?

Fill in any three values and leave the other two empty. A ball dropped from rest (u = 0) with a = 9.8 m/s² for t = 2 s falls s = 19.6 m and reaches v = 19.6 m/s.

How do I handle braking?

Use a negative acceleration. A car at 20 m/s that stops (v = 0) in 40 m decelerates at a = −5 m/s² and takes 4 s.

Which units should I use?

Any consistent set — meters, seconds, m/s and m/s² as labeled, or feet and ft/s² throughout.

What if there is no solution?

Some combinations are impossible, such as a car that must stop but is speeding up. The calculator tells you to check the signs.

How do I find the time when I know the distance, starting speed and acceleration?

Enter s, u and a and leave v and t empty. A car on a freeway ramp starts at 10 m/s, accelerates at 2 m/s² and covers 200 m: v² = u² + 2as gives v = 30.0 m/s, and t = (v − u) ÷ a = 10.0 s. The equation s = ut + ½at² is a quadratic in t and has a second, negative root; it is discarded as unphysical (OpenStax University Physics, Volume 1, Example 3.11).

How do I solve a ball thrown straight up?

Call up the positive direction, so the acceleration is a = −9.80665 m/s², and the speed at the top is v = 0. A ball thrown up at 20 m/s (u = 20, v = 0, a = −9.80665) takes 2.04 s to reach its peak, which is 20.4 m above the hand. With up positive, gravity stays −9.80665 whether the ball is rising or falling.

Does the calculator give distance traveled or displacement?

Displacement: the straight-line change in position, s = x − x₀ (OpenStax, section 3.1). A ball thrown up and caught at the same height has s = 0 even though it traveled twice the height of its flight.

The formulas and what they assume

All four equations hold only for motion in a straight line with constant acceleration (OpenStax University Physics, Volume 1, section 3.4, Eq. 3.10–3.14). They are not independent: five quantities (s, u, v, a, t) are tied by two equations, so any three known values determine the other two. Positive and negative follow your choice of direction, and s, u, v and a must all use it.

Worked example: braking from 30 m/s

OpenStax Example 3.10 stops a car from 30.0 m/s on dry concrete with a = −7.00 m/s². Enter u = 30, v = 0 and a = −7: the calculator returns s = 64.3 m and t = 4.29 s. On wet concrete (a = −5.00 m/s²) the distance grows to 90.0 m. A reaction time of 0.500 s adds 0.5 × 30 = 15.0 m, because the car keeps its speed until the driver brakes. Halve the speed to 15 m/s and the dry distance falls to 16.1 m, a quarter, since s depends on u².

Common mistakes

Braking needs a negative acceleration when the velocity is positive. A positive a with v = 0 for a moving car has no solution, and the calculator says so. Mixing units is the second trap: a speed in mph with a time in seconds does not give an acceleration in m/s², so convert to one system first. Third, when an object passes the same position twice, going up and then coming down, there are two valid times; the calculator reports the first pass, with the positive final velocity.

Sources: OpenStax University Physics, Volume 1, section 3.4 (motion with constant acceleration, Eq. 3.10–3.14; Examples 3.10 and 3.11) and section 3.1 (displacement); standard gravity 9.80665 m/s² is exact (NIST SP 811, Appendix B.8). Related: acceleration calculator, free fall calculator, projectile motion calculator and stopping distance calculator.