The uniform acceleration equation: v^2 - u^2 = 2as

All that is left on the road is seventy-five metres of skid. A witness says the car was doing thirty metres per second, and it braked to a dead stop. What was its deceleration? Nobody timed it. Every method in this chapter so far needs a time somewhere, and a skid mark never records how long it took.

This is video five of five in Motion graphs and acceleration. If acceleration feels shaky, Acceleration and velocity-time graphs comes first, with the equation that does have a time in it.

When a question gives no time, and a distance is either given or asked for, there is one equation to reach for. That rule comes from the triangle on the line. Its base is a time, so with no time there is no triangle, and you need an equation that never mentions t. There is also a condition. The acceleration must be uniform, steady the whole way, and a journey with two different accelerations is done in two parts. Here are two questions. In one, a bus speeds up from five to fifteen metres per second in four seconds. In the other, it does the same over fifty metres. Both ask for the acceleration. Which of the two needs the equation with no time in it? The second. It gives a distance and no time, so the time-based equation cannot work. The equation has four quantities. v is the final velocity and u is the initial velocity, both in metres per second. a is the acceleration, in metres per second squared, and s is the distance, in metres. In symbols, v squared minus u squared equals two a s. In words, final velocity squared, minus initial velocity squared, equals two times acceleration times distance. It is on Foundation and Higher papers, and it is printed on the equation sheet, so nobody has to memorise it. The work is in choosing it, rearranging it and finishing it. Here is where it comes from. With steady acceleration, the distance is the average velocity, halfway between u and v, times the time. And the time is v minus u, over a. Put that time into the first line, and the t disappears: s equals v squared minus u squared, over two a.

Now the skid. Forwards counts as positive, so a slowing car has a negative acceleration. It starts at u equals thirty metres per second, ends at v equals zero, and travels s equals seventy-five metres. Which of the three rearrangements correctly gives a on its own? The first one: a equals v squared minus u squared, all over two s. Dividing both sides by two s leaves a alone. Now substitute. v squared is zero squared, which is zero. u squared is thirty squared, which is nine hundred. Two s is two times seventy-five, which is a hundred and fifty. Zero take away nine hundred is minus nine hundred. Minus nine hundred divided by a hundred and fifty is minus six metres per second squared. Why is the answer negative, and how do you write it as a deceleration? The car is slowing, which is what a negative acceleration means. Written as a deceleration it is six, in the same unit, and the word deceleration carries the sign.

Now a cyclist starts from rest and accelerates uniformly at one point two metres per second squared for forty metres. How fast is she going at the end? No time, and a distance given, so this is the equation. Add u squared to both sides, and v squared equals u squared plus two a s. She starts from rest, so u is zero. Write u squared, write zero, and cross it out, so you see the term go rather than quietly vanish. Two times one point two times forty is ninety-six. That makes v squared ninety-six metres squared per second squared. Here is a finish a student might hand in: v equals ninety-six metres per second. That finish is wrong. What is missing from that finish? The square root. That number is v squared, not v, and the question asked for a speed. Take the root on its own line. The square root of ninety-six is about nine point eight, so v is nine point eight metres per second, to two significant figures, meaning two digits that count. One examiner's report on a Foundation paper is about this equation. Most candidates managed to calculate the velocity squared by correctly substituting the numbers into the given equation, however, most did not go on to square root their answer to gain the final mark. The substitution is the part most got right, and the mark went on the last step. The fix is a habit: label the line v squared, then always write one more line. So: no time in the question? Take the square route, and when it wants a speed, finish on the root.

A stone is dropped from a bridge and falls twenty metres to the water. Over a fall this short, air resistance, the drag of the air pushing back, can be ignored. An object falling freely near the Earth's surface accelerates at about ten metres per second squared. Exam boards print it as nine point eight or ten, so use the value your paper gives you. What is the stone's speed at the water in metres per second, using ten? It hits the water at twenty metres per second. The stone is dropped, so u is zero. v squared is two times ten times twenty, which is four hundred metres squared per second squared. The square root of four hundred is twenty. With nine point eight instead, v squared is three hundred and ninety-two, and v is about nineteen point eight metres per second.

Back to the skid. Same car, same brakes, same road, but now it starts at sixty metres per second instead of thirty. How far does it skid now: a hundred and fifty, three hundred, or six hundred metres? Three hundred metres, which is four times as far. The speed is squared. Doubling u makes u squared four times bigger. With the same deceleration, s is u squared over two a, so s is four times bigger too. Check it: sixty squared is three thousand six hundred, and two times six is twelve. Three thousand six hundred over twelve is three hundred metres.

Three last checks on this equation, each question before its answer. A question gives u, v and a time, and asks for a. Which equation do you use? a equals v minus u, over t. There is a time, so the triangle equation works. A sprinter this time. From rest, she accelerates uniformly at two metres per second squared over twenty-five metres. What is the sprinter's speed at the end, in metres per second? Ten metres per second. v squared is a hundred, and the root of a hundred is ten. One more. Your last line reads v squared equals sixty-four. What is still left to do before that is an answer? Take the square root on a new line. v is eight metres per second. No clock in the question means the square route, and for a speed, the root is the last line. And the skid from the start: seventy-five metres, from thirty metres per second, meant a deceleration of six metres per second squared, with no clock anywhere.

If there is no need to see this one twice, a thumbs up says so, and your list stays honest. If it needs more time, save it. The rest of the course keeps using acceleration, so this comes back round.

That completes our chapter on Motion graphs and acceleration. Next chapter: Forces and their effects.

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