Sherpa In this video, we take a look at how to use and interpret graphs. This is an AQA GCSE Maths topic, but the skills you learn here are applicable to many other subjects.

We start by looking at what a graph is and what different types of graphs there are. We then move on to looking at how to read data from a graph, including understanding scales and axes. Finally, we see how to interpret graphs in order to make deductions about trends or relationships by answering some exam-style questions and going through how to answer them.

Okay, we're looking at using and interpreting graphs. We'll first look at just reading graphs in general and what does it all mean? Then we'll go into displacement and velocity time graphs and then finally we'll look at instantaneous rates of change. So, yeah, first of all, what do we mean by reading graphs? What a graph?

What's the whole point of it? Let's just start with an example. We'll kind of go through and we'll create our example as we go. So I'm going to draw a graph here and let's say it's going to be a cost per hour graph. We're going to look at how much something costs per hour.

Let's say we're going to throw a party and we're going to hire Elton John to pair the party. Well, Elton is quite expensive. He charges £20,000 just to show up on the day. And then for every hour that we want him to play, he charges an additional £10,000. So it's going to cost us a fair bit of money.

After the first hour it's an additional £10,000. So that we'd then owe kind of £30,000 and same again. After the second hour it's another £10,000. And so in total we would go £40,000 and so on. You can kind of start to see the general rule here.

We're getting this nice straight line appearing we got online now and we've just constructed a graph. So if we were to kind of come to this graph completely blind and not really know what we're looking at here well, we know that the Y intercept always represents the starting amount here. The initial amount just show up. Or it could be the starting amount when you sit in a taxi and the metre is already running, is usually three £4 just to sit in a taxi. Or it could be the cover charge that a restaurant might charge for to sit down at a table.

So this is our wine intercept, is our kind of starting charge in a way. And if we look at what the gradient represents, well, we know that when we move along one in the X direction, we move up by whatever Elton John's hourly rate is, right? This is our hourly rate here. In this case it's £10,000. And we know that if we look at this, if we want to find the gradient of a graph, well, the gradient of the graph is just change in Y over change in X.

Well, in this case, the change in Y is the hourly rate. It's just the rate that charges. And the change in X is just one. Right? So it's just rate over one is just rate.

I'm actually going to put this on a new line. So we know that the gradient of this graph represents the hourly rate. Right, let's look at displacement time graphs now. So let's say Alison is paid for 3 hours and we actually don't want to pay him. We haven't got any money we completely made that whole thing up and we've got to sneak off from the party and avoid Alison finding out.

We start to sneak off from the party. So this is our displacement time graph that we're looking at here. This could be in metres, this could be in minutes or hours, it doesn't really matter. So for the first 20 minutes, we sneak off early and Nelson doesn't suspect a thing. Well, then he finds out we've got to pick up the pace, we've got to start running, and we've now got to increase the rate at which we escape.

So you've got to go from a gentle skulp into a run, and this could be at 40 minutes. After 40 minutes, we're able to hide from Elton and we can stop and take a break. So that'll be our displacement time graph. As we run away from Elton, we can now look at the velocity time graph. So what's the velocity?

What's happening with the velocity? Well, let's look at the same scenario, right? Our velocity is going to be in metres per second, and again, we'll look at time in minutes. And so for the first 20 minutes, we've got a fairly low velocity, right? We're just thinking of for the first 20 minutes, and obviously, as soon as we get found out, we've got to break into a run.

So our velocity drastically increases and we go from walking to running. Now, after 40 minutes, obviously we were able to take a break and hide somewhere and we can finally stop moving. And so that would be our velocity downgrade. And what we see is that the velocity actually represents the gradient of our displacement time graph. We end up with between zero and 20 minutes.

We have a small gradient, so our velocity is small. If we were to actually calculate the gradient, we could mathematically leave the two together. The gradient at that point there, or in general, at any point, the gradient is just again, it's the change in Y over the change in x. Well, in this case, that means the change in displacement divided by the change in time. Well, the amount, the distance we travel divided by the time it takes us is just equal to our velocity.

So the gradient of our displacement time graph is our velocity. And the same applies to velocity acceleration, right? It's the exact same analogue. The gradient above velocity time graph is our acceleration. Right, let's look at the instantaneous rate of change.

So let's say after you've had a break, after the first hour, we get to an hour, hop in a car and properly escape.

After an hour, again, this is going to be time. After the first hour, we can hop in a car and so our acceleration of the car, let's look at the velocity time graph and we capital acceleration from that. Let's look at what happens in the next 40 minutes. So we hop in our car and we are able to really put up the pace in the car, and then we start to reach the car's maximum speed. And so we kind of level off at maybe 80 miles an hour.

Well, if we wanted to look at our acceleration at any point on this graph, if we were to look at the acceleration at that point. Now, as I said, the acceleration is the gradient of a velocity time graph, right? It's just all we're doing is change in y divided by change in x, and in this case, it's change in t, because t is on our x axis. Well, we know the change in velocity of change in time is just acceleration, but it's not that simple to calculate the gradient at this point, right? What we can do is we can estimate it by drawing a tangent to the graph at that point, and the gradient of this gradient will therefore be a really good estimate of our instantaneous acceleration at that point in time.

At that point in time, at maybe, I don't know, 85 minutes into our journey, that's the exact acceleration at that point in time. It's the gradient of that tangent to our curve. All right, let's dive into a couple of example questions now. So, build a tax driver. We've got a graph for how much it costs per mile, in this case, for a taxi.

And three jenny, there's a fixed charge, plus the charge for the distance. So what's the fixed charge? Just like we had with Elton, right? It's what we charge initially as soon as we sit down, basically, as soon as something happens, as soon as we start off. In this case, it's what the metre will display as soon as we see in the taxi.

Well, at that point there, it's going to be £5 between zero and ten. So that's our fixed charge, it's £5. We want to work out the difference between the two journeys. So if one journey is 10 miles further than another journey, because it's a straight line, it doesn't really matter where we take it from, right? I could do the difference between going no miles and going 10 miles.

Well, if I travel 0 mile, I pay £5. If I travel 10 miles, I can see I have to pay £20. And so all I have to do is I work out the difference between those two values, it's going to be 20 minus five and I get £15. I could double cheque this, right? I could do the difference between ten and 20.

Well, if I travel 20 miles, I'm going to end up paying £35. And the difference between 20 and 35, again, is £15. Right, final question. We've got a velocity time graph for a car for the first 50 seconds, and we're going to work out the average acceleration during the 50 seconds. So I'm going to unline average here.

So we're really focusing on average. Well, why I underlined that is because it doesn't really matter for a velocity time graph. If we're looking at the average acceleration, it doesn't really matter what we're doing, right? We could be doing this as long as we end up at the same velocity at the end of it. After the full 50 seconds, our average acceleration is going to be the same as what we can do is we can actually take it to the other extreme.

We can basically say that I'm going to use the ruler, actually, for this to make it slightly neater. We can say that this is basically the path that we take. And if we're calculating the acceleration, when we know the acceleration is just the change in velocity divided by the change in time, so the change in velocity is going to be 30, the change in time is going to be 50. Well, that's just the same as three over five if I divide the top and bottom by ten, which is also the same as six over ten, which is just not .6. So I know my acceleration is zero six metres per second squared.

Finally, part B, estimate the time during the 50 seconds when the instantaneous acceleration equals the average acceleration. So you just calculated the average acceleration. We know it's. The average acceleration is equal to this gradient here. We calculated the change in change in Y over that change in x.

So if we're going to try and match the instantaneous acceleration to this acceleration, we want to find a point on the graph where our gradient on this curve is equal to the gradient of this line. Well, in that case, just eyeballing it for a second. It's probably going to be about that point there, right?

At least onto a square. It's going to be probably that point there and we could test this, right? So, again, I can draw the gradient to the curve at this point. It's going to be roughly that. I might move it down slightly and we end up drawing a line that parallel to my original line.

And I just realised I've just left it on the wrong tool. I'll put that back on there. So we know that these two lines here and here, these two lines, that line there and that line there are parallel to each other, right? And so therefore, that point there in time is going to be our corresponding time, where our instantaneous velocity is equal to our average velocity. So our answer would be roughly again, it's an estimate.

It's roughly going to be 11 seconds into our journey. That would be our answer. So that is the end of reading and using and interpreting graphs.

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