Video: AQA GCSE Mathematics Higher Tier Pack 4 • Paper 2 • Question 23

The graph shows how a cyclist’s speed varies over time. a) Estimate the acceleration at 14 seconds. You must show your working. b)Estimate the average speed of the cyclist on their journey. You must show your working. Evaluate your answer to part b). Tick a box. Underestimate. Exact. Overestimate. And then make a comment.

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Video Transcript

The graph shows how a cyclist’s speed varies over time. And we’ve got a graph here which has got time in seconds as the 𝑥-axis and speed in metres per second as the 𝑦-axis. Part a) Estimate the acceleration at 14 seconds. You must show your working.

Now there’s also a part b) and a part c). But we’ll come back to those in a minute. Now, this is quite interesting. We’re told to estimate something. But we’re also told that we must show our working out. So we’re using the word estimate in the mathematical sense. We’re going to perform some calculations. But we’re going to use approximate values to put into our calculations to give an approximate answer to the question. Let’s just recap what we know about speed-time graphs.

The 𝑦-coordinate tells us the speed at a given time. But acceleration is the rate of change of speed. So the gradient of the curve or the line at any point tells us the acceleration of that time. And finally, the area under the graph tells us how much distance has been travelled. So since we’re looking for the acceleration when the time is equal to 14 seconds, we’re looking for the gradient of the curve or the line at that point. Now, if we look at the graph at the point where the time is equal to 14 seconds, we can see that the rate of change of speed isn’t constant. That’s not a straight line part of the graph.

So we’re going to need to put our pencil on that point, bring up the ruler, and swivel it round, and draw in a line that looks like it’s a decent tangent to the graph at that point. And this is where the estimation comes in. We could’ve drawn the line swiveled slightly that way. Or we could’ve drawn the line slightly swiveled that way. Either way, we’d have come up with slightly different answers. Now, we’ve got to work out the gradient of that line, that tangent to our speed-time graph when the time is equal to 14 seconds. And in order to do that, we just need to check that we understand the scale on the different axes first.

On the 𝑦-axis, the speed axis, four little squares is equal to two metres per second. So this means that two little squares must be equal to one metre per second. And looking up and down our tangent, that point at the top there seems to have an integer, a whole number, 𝑥-coordinate and an integer 𝑦-coordinate. Looking at our scale, we can see the 𝑦-coordinate is seven. Then, looking at the scale on the 𝑥-axis for time, two little squares is equal to two seconds. So each individual little square is worth one second. And we can see that the 𝑥-coordinate of that top point is six. And the tangent conveniently intersects the 𝑥-axis there at 17. So it’s got integer 𝑥-coordinate of 17 and an integer 𝑦-coordinate of zero.

This is gonna make our calculation slightly easier. We can see that our speed has gone from seven down to zero. So the change in speed is zero minus seven. And that was in the time between six seconds and 17 seconds. So the difference in time was 17 minus six. So that gives us an acceleration of negative seven elevenths of a metre per second squared. And if we put that into our calculator, we get negative 0.63 recurring, which is negative 0.636363 and so on, which we can round to two decimal places to get negative 0.64. So any of those answers would be correct. Now moving on to part b).

Estimate the average speed of the cyclist on their journey. You must show your working.

Now, the average speed is just the total distance travelled divided by the total time taken. And we can remember that the area under the graph is gonna tell us the total distance travelled. And we started at zero seconds. We finished at 18 seconds. So the total time taken is 18 seconds. And again, we’ve got to produce an estimate. But we’ve got to show our working out. So this is not just having a guess at the answer. This is doing a very specific calculation by using approximations for all or some of the component parts.

Now, as we said, from start to end of that journey, we read this off from the graph as being 18 seconds. So we know the total time taken is 18 seconds. Now, we can break the area that we’re looking for down into three regions. Area one is a triangle and we can calculate that accurately. Area two is a trapezium and we can calculate that area accurately. And area three is approximately a triangle. Okay, the area under the curve is a little bit smaller. But this is where our approximation, our estimation, is coming in.

So for area one, the area of a triangle is a half times base times height. Well, the base of our triangle here is nine. And the height goes up from zero to 3.5, so the height is 3.5. So the area is a half times nine times 3.5 which is 15.75. But what are the units? Well, the half was just a constant. And nine, remember, was a time in seconds. And the 3.5 is metres per second. So our calculation involves multiplying a constant by seconds by metres per second. Well, you can see that we can cancel out the seconds here. And so, we’ve got a constant times the number of metres. Our units are metres. And area one represents a distance of 15.75 metres on our cyclist’s journey.

Now, area two is a trapezium. And the area of a trapezium is the mean of the parallel sides, that’s a half of 𝑎 plus 𝑏, times the distance between those parallel sides, ℎ in this case. So working out the values for 𝑎 and 𝑏, reading off from the 𝑦-axis, I can see — well, I can see a few things. Firstly, I can see that 𝑎 comes out at 3.5. And then I can see that I’ve actually written something wrong in 𝑦-axis. That four that I’ve written in above the four should be a five. And then we can see that 𝑏 should be 5.5. And we can see that the width of the trapezium, ℎ, goes from nine to 11. So the difference between nine and 11 will tell us the value of ℎ. And when we work that out, we get an area of nine square units, which represents a distance of nine metres on the cyclist’s journey.

Then area three is another triangle. So the formula for the area of a triangle is a half times base times height. So the base goes from 11 to 18. So the difference between those two will be the base, 18 minus 11. And the height is 5.5 again. Multiplying those together gives us an area of 19.25 square units which represents 19.25 metres on our cyclist’s journey. And when we add those together, we see that the total distance that the cyclist travelled was 44 metres.

So the average speed is 44 divided by 18 which simplifies to 22 over nine. And remember, those units were metres per second. And if you want to put that into your calculator and convert it to decimal, that’s 2.4 recurring metres per second which is 2.444 and so on metre per second. Or, rounded to one decimal place, that’s 2.4 metres per second. Right, let’s move on to part c).

Evaluate your answer to part b). Tick a box: underestimate, exact, or overestimate. And then make a comment.

Well, in part b), we wanted to calculate area three, which was that curve. And we used a triangle to approximate that. And because we did that, we included all of this pink region in our approximation. So using a triangle to approximate area three gave us an overestimate of the distance travelled by the cyclist. This means that the cyclist didn’t travel as far as we said in the time. So we overestimated their speed. So the answer is: this is an overestimate.

So for part a), you’ll need a number somewhere near negative 0.64 metres per second squared. For part b), you’ll need a number close to 2.4 metres per second, And for part c), you need to say that it was an overestimate.

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