Distance from the area under a velocity-time graph

A car brakes from thirty metres per second to a dead stop in four seconds. How far does it travel while it stops? It is not thirty times four, because it was not doing thirty all the way. It is not zero times four either, because it was not stopped all the way. The distance is real. It is the length of skid left on the road, and the only place it is written down is the shape of the graph.

This is video four of five in Motion graphs and acceleration. It reads the same graph as Acceleration and velocity-time graphs, the other way round, so start there if that graph feels new.

On one exam board this is for every tier, and on the other two it is Higher only. Each board's row is shown here, and the video is for anyone who needs it.

Start with the plainest case. A tram holds twelve metres per second for twenty-five seconds, so its velocity-time line is flat. The shape under that line is a rectangle, twelve metres per second tall and twenty-five seconds wide. Its area is twelve times twenty-five, which is three hundred. Metres per second times seconds: what unit does this area come out in? Metres. The seconds cancel, so the rectangle's area is a distance: distance equals speed times time, drawn as a picture. When the line slopes, slice the area into thin upright strips. Each strip is a velocity times a short time, which is a short distance. Add up every strip, and you have every short distance the object travelled. That is why the whole area is the whole distance. That gives a third sentence for the stack: on a velocity-time graph, the area under the line is the distance travelled. All three together. On a distance-time graph, the gradient is the speed. On a velocity-time graph, the gradient is the acceleration, and the area under the line is how far it went. Back to the braking car. Its line falls straight from thirty metres per second to zero over four seconds, so the shape under it is a triangle. How far does it skid: a hundred and twenty, sixty, or zero metres? Sixty metres, which is the area of that triangle. The area of a triangle is a half times base times height. Here that is a half, times four seconds, times thirty metres per second, which is sixty metres. And the half has a reason. Thirty times four is the whole rectangle, and the triangle is half of it, because the car's speed fell steadily from thirty to nothing. One examiner's report on a Higher paper describes the wrong choice most of the unsuccessful students made. This was obtained by just multiplying the two numbers together. The fix is the shading. Shade the region first, name its shape, and only then pick the formula, so a triangle is never mistaken for a rectangle.

Real journeys come in parts. Put a start in front of that tram: it speeds up uniformly from rest to twelve metres per second over ten seconds, then holds twelve for twenty-five seconds. Shade the two parts differently: a triangle for the first ten seconds, and the rectangle from before for the next twenty-five. Pause and find both areas, then add them for the total distance. The total distance is three hundred and sixty metres. The triangle is a half, times ten seconds, times twelve metres per second, which is sixty metres. The rectangle is three hundred metres. Sixty plus three hundred is three hundred and sixty metres. Read which stretch the question asks about before you shade. If it asked only about the speeding-up part, you would shade the triangle alone, and the answer would be sixty metres. Now a coach brakes uniformly from twenty-four metres per second to six metres per second over six seconds. The shape under that line is a trapezium, a four-sided shape with two parallel sides. A trapezium's area is half the sum of the parallel sides, times the gap between them. Twenty-four plus six is thirty, and half of thirty is fifteen. Fifteen metres per second times six seconds is ninety metres. Now, can you split the trapezium into a rectangle and a triangle and add their areas? It comes to the same total, so either route works. The rectangle sits under six metres per second for six seconds, which is thirty-six metres. The triangle on top is a half, times six seconds, times eighteen metres per second, which is fifty-four metres. Together, ninety metres.

Some lines curve, and then there is no triangle or rectangle to use. A cyclist freewheels to a halt from ten metres per second, slowing quickly at first, then more gently. So count squares. Before counting a single one, decide what one square is worth from the two axis scales. Each square is one second wide and two metres per second tall. What distance is that? Two metres. Metres per second times seconds gives metres. Now count. There are eleven whole squares under the curve. Then count the part squares that are more than half full, and ignore the rest. There are seven. That makes eighteen squares, and eighteen times two metres is thirty-six metres. It is an estimate, because the part squares are judged by eye. A report on a Higher paper notes that some candidates did not know how to calculate the area under the graph from a non-linear line, a curve. Square counting is the fix. Value one square, count the whole ones, then count the part ones over half.

Here is one velocity-time graph with two questions. In its first section, a car goes from rest to twelve metres per second in four seconds. Question one asks for its acceleration in that section. Question two asks how far it travels in that section. Which tool does each question need: the gradient or the area? The gradient for the acceleration, because that is how fast the velocity changes. The area for the distance, because that is how far it goes. The gradient is twelve metres per second over four seconds, which is three metres per second squared. The area is a half, times four seconds, times twelve metres per second, which is twenty-four metres. So: the slope of the line for how fast it changes, and the area under the line for how far it goes. Another report, on a Higher paper, says this question required candidates to equate the distance travelled to the area under the graph, for the four seconds of braking. The majority of candidates did not realise this and did not gain any credit. Some candidates attempted to use an equation of motion or speed equals distance divided by time. The fix is to treat a distance off a velocity-time graph as an area, every time, whatever equation you know. An equation that does some of this without a graph has its own video, The uniform acceleration equation. On a different paper, it went the other way. The report on the multiplying slip records that the majority of candidates realised that the area under the graph was equal to the distance travelled by the object. If the strip idea has landed, you are with that majority.

Three to try before the end, each answer following its question. What does the area under a velocity-time graph give you? The distance travelled. Velocity times time is distance, strip by strip. Try a journey you have not seen. A bike speeds up uniformly from four to ten metres per second over six seconds. How many metres does the bike travel in those six seconds? Forty-two metres. Half of fourteen is seven, and seven times six is forty-two. Now the other tool. A car's acceleration from its velocity-time graph: gradient or area? The gradient. Acceleration is how fast the velocity changes, so it is the slope. Two tools, then: the slope for how fast, the area for how far. And the braking car from the start left sixty metres of skid: the area of its triangle, not thirty times four.

Is this one yours now? Then the thumbs up tells your list it is sorted. If not, save it, give it a day, and look again. The strips settle in once you have slept on them.

Next in the chapter: The uniform acceleration equation, v squared minus u squared equals two a s.

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