Calculator of Alexandria

SUVAT calculator (constant acceleration)

Enter any three of displacement, initial velocity, final velocity, acceleration and time to find the other two, with the five equations derived.

Use negative values for motion in the opposite direction.

Negative for slowing down. Gravity is −9.81 m/s² if up is positive.

Displacementss
Final velocity (m/s)vv
Initial velocity (m/s)uu
Acceleration (m/s²)aa
Time (s)tt

Formula

v=u+at,s=ut+12at2v = u + at, \quad s = ut + \tfrac{1}{2}at^2
How it's derived

The formula

Five quantities describe motion in a straight line with steady acceleration: displacement ss, initial velocity uu, final velocity vv, acceleration aa and time tt. Each of the five equations leaves one of them out:

v=u+atv = u + at s=12(u+v) ts = \tfrac{1}{2}(u + v)\,t s=ut+12at2s = ut + \tfrac{1}{2}at^2 s=vt−12at2s = vt - \tfrac{1}{2}at^2 v2=u2+2asv^2 = u^2 + 2as

Know any three, and one of these equations gives you a fourth. Then any other gives the fifth.

Where it comes from

Acceleration is the rate velocity changes. If it’s constant, velocity changes by the same amount every second, so after time tt:

v=u+atv = u + at

Velocity rises in a straight line from uu to vv, so the average velocity is exactly halfway between them. Displacement is average velocity times time:

s=u+v2 ts = \frac{u + v}{2}\,t

Substitute v=u+atv = u + at into the average-velocity equation to remove vv:

s=u+u+at2 t=ut+12at2s = \frac{u + u + at}{2}\,t = ut + \tfrac{1}{2}at^2

Substitute u=v−atu = v - at instead to remove uu, and you get s=vt−12at2s = vt - \tfrac{1}{2}at^2.

To remove time, rearrange v=u+atv = u + at as t=(v−u)/at = (v - u)/a and put it into the average-velocity equation:

s=u+v2⋅v−ua=v2−u22as = \frac{u + v}{2}\cdot\frac{v - u}{a} = \frac{v^2 - u^2}{2a}

Multiply both sides by 2a2a and you have v2=u2+2asv^2 = u^2 + 2as.

Worked example

A car pulls away from rest (u=0u = 0) at a=3 m/s2a = 3\ \text{m/s}^2 for t=10t = 10 s. The first equation gives the final velocity:

v=0+3×10=30 m/sv = 0 + 3 \times 10 = 30\ \text{m/s}

The third gives the distance covered:

s=0+12×3×102=150 ms = 0 + \tfrac{1}{2} \times 3 \times 10^2 = 150\ \text{m}

Check with the last equation: 302=0+2×3×150=90030^2 = 0 + 2 \times 3 \times 150 = 900. ✓

Watch out for

  • Signs matter. Pick one direction as positive. A ball thrown upward at 10 m/s has u=+10u = +10 and a=−9.81a = -9.81 m/s². If you leave out the minus sign, the ball never comes down.
  • Two answers. When you know ss, uu and aa, the time comes from a quadratic, and there can be two valid times. A ball thrown up passes 3 m on the way up and again on the way down. The calculator gives the first time.
  • Constant acceleration only. These equations don’t apply when acceleration changes, such as a car’s engine tapering off or a falling object reaching terminal velocity.

Sources

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