### Video Transcript

This is the distance-time graph for
a cyclist cycling 100 metres. Part a) Calculate the average speed
over the 100 metres. Give units in your answer. Part b) Estimate the distance at
which the instantaneous speed is equal to the average speed. You must show your working on the
graph.

Remember speed is a way of
measuring the change in distance over a given period of time. The formula for average speed is
distance divided by time. So to calculate the average speed
for our cyclist, we will need to work out the total distance they travelled and the
time it took. In fact, we’re told that the
cyclist travelled 100 metres. But we’ll need to read information
for the time it took from the graph.

We can see that the cyclist reached
100 metres at this point. So we can draw a line vertically
down to our 𝑥-axis. And we see that it took the cyclist
15 seconds to complete this distance. The speed is therefore found by
dividing 100 by 15. This isn’t a calculation we can
complete in our heads. So instead, we’ll simplify this
fraction as far as we can.

100 and 15 are multiples of
five. So we’re going to divide through by
five. And we see that this is equivalent
to twenty thirds. Now we could leave it in this form
or we could change it into a mixed number and therefore into a decimal. 20 divided by three is six
remainder two. So we can say that twenty thirds is
the same as six and two-thirds. Two-thirds is the same as .6
recurring or .67 correct to two decimal places. So the speed is 6.67 correct to two
decimal places.

But what are the units? Well, we divided a measurement in
metres by a measurement in seconds. So the units must be metres per
second. And the average speed over the 100
metres was 6.67 metres per second correct to two decimal places. Now let’s look at part b.

We now know the average speed for
the journey. It was 20 over three metres per
second. We now need to find a point on the
graph, where the gradient is 20 over three since the gradient of a distance-time
graph tells us the speed at that point. Since the graph is a curve, we’re
going to need to add a tangent with that gradient. Remember the gradient of a straight
line is change in 𝑦 over change in 𝑥, which is sometimes called rise over run.

So we need to find a tangent that
rises 20 units for every three units it runs. And we need to be a bit careful
with the scale when finding this. Notice that two squares on the
𝑦-axis represent 20 and three squares on the 𝑥-axis represent three. So let’s find a tangent which has
this slope. It’s here. The rise of this line is 20 and the
run is three. So the gradient is indeed 20 over
three. This is at the point where the
distance is approximately 68 metres.