# Video: AQA GCSE Mathematics Higher Tier Pack 1 • Paper 1 • Question 23

AQA GCSE Mathematics Higher Tier Pack 1 • Paper 1 • Question 23

03:27

### Video Transcript

Michella was practising her sprinting. The graph shows the first 12 seconds of one particular sprint. Part a) Between which two times was her acceleration at its peak? Part b) Work out the distance travelled in her last two seconds.

For point a, we’re looking to find the time during which her acceleration is highest. Another way to think about acceleration is as change in a velocity over a given period of time.

On our graph, change in velocity is change in the 𝑦-direction. And the change in time is change in 𝑥-direction. We can say then that the acceleration is given by the gradient of the line. Remember gradient is change in 𝑦 over change in 𝑥.

This means we’re looking to find the time during which the gradient is highest or the graph is steepest. We can observe that that’s this part of the line. If we draw a vertical line down to the horizontal axis, we see that that finishes at two seconds.

And we can, therefore, say that her acceleration was at its peak between zero seconds and two seconds.

For part b, we’re looking to find the distance travelled in the last two seconds of her journey. The graph finishes at 12 seconds. So the last two seconds are between 10 and 12 seconds.

And just like the gradient of a velocity-time graph represents the acceleration, the area underneath it represents the distance travelled. We can join these two parallel lines and we see that we form a trapezium.

The formula for area of a trapezium is a half multiplied by 𝑎 plus 𝑏 multiplied by ℎ, where 𝑎 and 𝑏 are the lengths of the parallel sides and ℎ is the distance between them. We can spot quite easily that the distance between the two parallel sides is two units.

But before we can work out the lengths of the parallel sides themselves, we need to work out what the scale is on this 𝑦-axis. Five small squares on the 𝑦-axis represents two metres per second. And dividing through by five tells us that one small square is equal to two-fifths of a metre per second or 0.4.

The longer parallel side ends two squares above the number six. So that’s 6.8 units. And the shorter parallel side ends two squares about four. So that’s 4.8 units. So we substitute these values into our formula for the area of a trapezium. And we get one-half multiplied by 6.5 plus 4.8 multiplied by two.

Remember multiplication is commutative. That means it can be done in any order. If we do a half multiplied by two first, those cancel each other out since one-half multiplied by two or one-half of two is one. And this means the area of our trapezium is simply 6.8 plus 4.8.

We can use a column method to work this out. Eight plus eight is 16. So we put a six in the tenths column and carry the one. We add a decimal point in the answer directly below the decimal point in the question. Then, six plus four plus one is 11. This means the area of our trapezium is 11.6 square units.

We can see that the velocity is given to us in metres per second and the time is given to us in seconds. So the units for our distance are metres.

The distance travelled in the last two seconds of her journey was 11.6 metres.