Using the Kinematics (SUVAT) Formulae

Picture a car waiting at a red light. The light turns green, it pulls away, and it is getting faster every second. Someone asks you how fast it is going five seconds later. You could measure how far it travelled in those five seconds and divide by five. That does give you a speed. It is just not the speed you were asked for. Distance over time hands you the average across those five seconds. The car was below that average early on and above it by the end, so its speed at the end is bigger than the number you calculated. Anything that speeds up has more than one speed.

So there is a second set of formulae for exactly this, where the speed changes at a steady rate. Five letters carry the whole thing.

The five letters are s, u, v, a and t, and together they get called the suvat formulae. Every question of this kind is built from those five quantities and nothing else. s is displacement, meaning how far the object has moved from where it started. Everything here travels one way in a straight line, so read it as the distance covered, in metres. u is the velocity it starts with, v is the velocity it ends with, a is the acceleration, which has to stay constant, and t is the time taken, in seconds. u and v are the pair people mix up, so here is your handle for this video: u comes before v in the alphabet, and u comes before v in the journey. u is the speed before, v is the speed after. Two bits of exam wording point at these letters without using a number. Starts from rest means u equals zero. Comes to a stop means v equals zero. Nobody writes those zeros down for you.

Try one now. A train is travelling at twelve metres per second when the driver brakes, and it stops eight seconds later. Which letter does that twelve metres per second belong to: u, v, or a?

Take your pick. I'll wait.

The answer is u, because twelve metres per second is the velocity the train starts with. And stops eight seconds later quietly fills in two more: v is zero, and t is eight.

Slowing down at a steady rate uses these same formulae, with a written as a negative number.

Before any of that helps, you have to notice you are in a kinematics question at all. That decision comes before a single number.

The check is one question asked of the wording: does the speed change? An object holding one steady speed the whole way needs distance equals speed times time, and none of the five letters. An object speeding up or slowing down at a steady rate needs a kinematics formula. Wording that means yes: accelerates, decelerates, starts from rest, comes to a stop, or an acceleration quoted in metres per second squared. Constant speed, steady speed, or the same speed throughout means no.

Here are two cyclists, ten seconds each. Cyclist A travels at a constant six metres per second. Cyclist B accelerates from two metres per second up to six metres per second over the same ten seconds.

One of those two needs a kinematics formula and one does not. Which is which?

Have a think. I'll wait.

Cyclist B is the one that needs it, because B's speed changes across the ten seconds and no single speed describes that ride. A holds six the whole way, so distance equals speed times time gives sixty metres.

B takes more work, because B needs a and a is nowhere in the question. It comes from v equals u plus at first. Six equals two plus ten a, so a is nought point four. Now the distance. s equals u t plus a half a t squared, which is two times ten, plus a half of nought point four times ten squared. That is twenty plus twenty, so B travels forty metres. Forty metres against sixty, over the same ten seconds. B was below six metres per second for almost the whole ride, so the shorter distance is what you should expect. Speed times time on B would have given sixty, and been twenty metres out.

Now the three formulae themselves, and a way of choosing between them that takes about two seconds.

The first is v equals u plus at. The second is s equals u t plus a half a t squared. The third is v squared equals u squared plus two a s. Look at what each one leaves out. The first has no s in it anywhere. The second has no v. The third has no t. That is why there are three of them. A kinematics question hands you three of the five letters and asks for a fourth. That leaves exactly one letter you neither know nor need, and the missing letter picks the formula.

Back to the car at the lights, with numbers this time. It starts from rest and accelerates at four metres per second squared for five seconds. Find the velocity it finishes at, writing your five letters down first.

Pause it there and work it out. I'll wait.

Starts from rest gives u equals zero, the acceleration is four, the time is five, and v is what you want. Three known letters and the unknown, so the spare one is s, and the formula with no s in it is v equals u plus at. Substituting: v equals zero plus four times five, so v is twenty metres per second. Metres and seconds went in, so metres per second comes out, and the unit gets written next to the answer every time.

Which puts one condition on all three: the units have to match. A speed arriving in kilometres per hour gets converted to metres per second first.

And here is the number from the opening. The distance covered is zero plus a half of four times twenty five, which is fifty metres. Fifty metres over five seconds is ten metres per second, a genuine average, and exactly half of what the car was doing at the end.

Two neighbouring skills sit just outside this. Making a different letter the subject is ordinary rearranging, covered in the video on using and rearranging a formula. Where these three come from, the gradient and area of a velocity-time graph, belongs to the video on graphs in real-world contexts.

Two things go wrong with these questions in real exams: not knowing the letters, and not spotting that the letters are needed. There is evidence on both.

The first comes from an examiner's report on an OCR paper, about a question that needed the kinematics formulae. Candidates need to learn the meaning of the variables in the kinematics formulae. These are clearly stated in Section six point zero two e of the specification. Learn the meanings, not just the shapes. A student who writes all three out perfectly and cannot say what u stands for never gets started. The following year, a report on a different OCR paper named the second failure. Some candidates did not recognise that, because the particle has constant acceleration, the given kinematic formulae need to be used. The speed equals distance divided by time formula is only to be used if there is no acceleration. Particle there just means the moving object. And that last sentence is the recognition check in an examiner's own words: speed equals distance over time is only for when nothing is accelerating. So here is the routine for the exam room. Settle whether anything is accelerating, then write the five letters down the margin and fill in every value the question gives you, zeros included. Then the letter still blank that nobody asked for picks your formula. Substitute, work it out, and put the unit on the answer.

Right, let's put the whole thing back together in one piece. s is displacement, u is the velocity you start with, v is the velocity you end with, a is constant acceleration, and t is time. u before v, in the alphabet and in the journey. One question decides whether you need them at all: does the speed change? A steady speed the whole way means distance equals speed times time. Speed changing at a steady rate means a kinematics formula. Then three formulae with one missing letter each. v equals u plus at has no s, s equals u t plus a half a t squared has no v, and v squared equals u squared plus two a s has no t. Spot the letter you neither know nor need, and the missing letter picks the formula.

That completes our chapter on Linear Equations, Formulae and Modelling. The next chapter is Quadratic and Simultaneous Equations.

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