Any three of s, u, v, a and t give the other two, with the kinematic equation each step uses.
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The Coddy physics course teaches the method: you solve each problem in steps on an interactive board, and it tells you which step went wrong.
The kinematic equations describe motion in a straight line with a constant acceleration. They use five quantities: displacement s, initial velocity u, final velocity v, acceleration a and time t, which is why UK schools call them the SUVAT equations. There are five of them, and each one leaves out exactly one of the five letters.
That is the whole trick to choosing one. Write down the three values you know and the one you want; the letter left over is the one you neither have nor need, and the equation without it is the one to use. A car braking from 20 m/s to a stop in 50 m, with no time given and none asked for, calls for v² = u² + 2as, the equation without t.
The calculator does that choosing for you. Fill in three of s, u, v, a and t, leave the other two blank, and it finds both, one equation at a time, showing which equation each step uses and how it was rearranged. Where a square root leaves the sign open, it keeps the sign the other values allow; when both signs fit (a ball passes the same height going up and coming down), it shows both answers. If the three values contradict each other, it says which one does not fit. Values in km/h, mph, km or feet are converted to SI units first, as their own step.
Name each value by its letter. "From rest" means u = 0, "comes to a stop" means v = 0, and "dropped" means u = 0 with a = 9.81 m/s² downwards.
Give everything that points the other way a minus sign. A car braking while it moves forward has a negative a.
Of s, u, v, a and t, find the one you neither know nor want. Use the equation that leaves it out, from the table below.
Get the unknown on its own before you substitute, with every value in m, m/s, m/s² and s. In the calculator, leave the two unknowns blank and it shows each of these steps.
Each equation uses four of the five quantities. Pick the one that leaves out the quantity you neither know nor need.
| Equation | Leaves out | Use it when |
|---|---|---|
| v = u + at | s (displacement) | No displacement is given or asked for |
| s = ½(u + v)t | a (acceleration) | No acceleration is given or asked for |
| s = ut + ½at² | v (final velocity) | No final velocity is given or asked for |
| v² = u² + 2as | t (time) | No time is given or asked for |
| s = vt − ½at² | u (initial velocity) | No initial velocity is given or asked for |
u = 0 m/s, a = 3 m/s², t = 4 ss and v are missing. v = u + at, the equation without s, gives v = 0 + 3 · 4 = 12 m/s. Then s = ½(u + v)t = ½ · (0 + 12) · 4 = 24 m. Checking with s = ut + ½at² = 0 + ½ · 3 · 4² gives the same 24 m.
u = 20 m/s, v = 0 m/s, s = 50 ma and t are missing. s = ½(u + v)t rearranges to t = 2s/(u + v) = 2 · 50 / 20 = 5 s. Then a = (v − u)/t = (0 − 20)/5 = −4 m/s². The minus sign says the acceleration points against the motion: the car is slowing down.
s = 100 m, u = 0 m/s, a = 9.81 m/s²Take down as positive. With no time given, use v² = u² + 2as: v = √(0² + 2 · 9.81 · 100) = √1962 = 44.29 m/s. Then t = (v − u)/a = 44.29 / 9.81 = 4.52 s. Air resistance is ignored, as the equations assume.
u = 15 m/s, v = 0 m/s, a = −9.81 m/s²Take up as positive, so gravity is −9.81 m/s², and v = 0 at the top. t = (v − u)/a = (0 − 15)/(−9.81) = 1.53 s to reach the top. Then s = ½(u + v)t = ½ · 15 · 1.529 = 11.47 m, the highest point.
Five equations for motion with constant acceleration: v = u + at, s = ½(u + v)t, s = ut + ½at², v² = u² + 2as and s = vt − ½at². Here s is displacement, u initial velocity, v final velocity, a acceleration and t time.
The five quantities in the equations: s for displacement, u for initial velocity, v for final velocity, a for acceleration and t for time. SUVAT is the name used in UK schools; in the US the same five are usually called the kinematic equations or the equations of motion.
List the three values you know and the one you want. One of the five letters is left over, and the equation that does not contain it is the one to use. No time given or wanted means v² = u² + 2as.
Three. Each equation links four of the five quantities, so three known values leave one equation with a single unknown. Solving it gives a fourth value, and a second equation gives the fifth. With only two values, the calculator lists what is still missing.
Yes, as long as air resistance is small enough to ignore, because the acceleration is then constant at g = 9.81 m/s². Use a = 9.81 m/s² with down as positive, or a = −9.81 m/s² with up as positive.
A negative acceleration points in the negative direction. If the object is moving the positive way, it is slowing down; if it is moving the negative way, it is speeding up. A car braking from 20 m/s to a stop in 50 m has a = −4 m/s².
From v² = u² + 2as, both +v and −v square to the same number. The calculator gives the positive one; the negative one is the same speed the other way, which is the answer to take if the object has turned around, like a ball on its way back down.
