### Video Transcript

Michael made a 60-minute trip to town. The given graph shows the distance he was from his house throughout his trip. He stopped at two shops. At what times was that? At some point, he changed direction without stopping first. When was that? The third part of this question asks us at what times was he moving toward his house?

And there’s one further part to this question, which we’ll consider in a moment. We’ve been given a graph that represents the distance Michael was from his house at a given time. The graph doesn’t show the total distance traveled. It shows us, in other words, his displacement. It’s the distance he was from a fixed point, in this case, that was his house. And so, there are several things we can tell from a displacement–time graph.

Firstly, we’re looking to find the points at which he stopped at the shops. And so, we recall that horizontal sections of graph — that’s flat sections of graph — will represent a velocity of zero. In other words, Michael isn’t moving — his velocity is zero — when his displacement–time graph is given by a horizontal section. If we look carefully at our graph, we see that that is here and up here. By dropping in vertical dashed lines as shown, we’ll be able to read the values of 𝑡 off of our graph. We do, however, need to be a little bit careful with our scale.

We see on the horizontal scale that five small squares represent 10 time units. In fact, we’re told that the trip is 60 minutes. So, five small squares represents 10 minutes. If we divide through by five, we then see that one small square must represent two minutes. And so, this first dashed line, which sits three small squares above 10, must represent 16 minutes. This one sits three small squares above 20, so it represents 26 minutes. This third line is two small squares above 40, so it represents 44 minutes. And this final dashed line sits half a square back from 50, so it must represent 49 minutes.

Let’s use inequality notation to represent this. We see that he had stopped between and including 16 and 26 minutes and 44 and 49. So, in inequality notation, we see 𝑡 is greater than or equal to 16 and less than or equal to 26 and 𝑡 is greater than or equal to 44 and less than or equal to 49.

The second part of this question says that at some point, he changes direction without stopping first. Now, it follows that as he travels away from his home, the graph moves upwards essentially. And as he goes back towards his home, the graph must be sloping downwards. And so, a change of direction is shown on our graph by a change of sign of slope. Now, the slope of a curve is found by estimating the slope of the tangent at that point. And we can see that around here, the slope changes sign. It goes from being negative to positive, and so we drop a fifth vertical line in. We see that vertical line occurs at 𝑡 equals 32. Michael changes direction without stopping first at 32 minutes or 𝑡 equals 32.

Now, this will help us answer the third part of this question. We’re looking to find the times at which he was moving towards his house, in other words, the times at which the slope is negative. Let’s look at the points on the graph where this occurs. Well, we can see this would happen here and also on this section of the graph. All other sections of the graph either have a positive slope or those flat sections have a slope of zero.

Michael is traveling toward his house between and including 26 minutes and 32 minutes and 49 and 60 minutes. In inequality notation, we write 𝑡 is greater than or equal to 26 and less than or equal to 32. And 𝑡 is greater than or equal to 49 and less than or equal to 60.

Let’s clear some space and look at the final part of this question.

At which minute was he moving fastest?

Now, we’ve sort of inferred this so far, but now we quote that “the slope of a distance or displacement–time graph gives the velocity at that point.” And so, what we’re really asking ourself is “at which point is the tangent to the curve the most steep?” Well, if we do this by eye, we see it’s at one of these two places. We add some tangents in, and we see that we have fairly steep lines. In fact, we’re probably now able to observe which line is the most steep, but let’s perform some calculations to check.

We recall that the formula that helps us calculate the slope of a straight line is change in 𝑦 divided by change in 𝑥. In this case, that’s change in 𝑑 divided by change in 𝑡. Now, you might also have seen this as rise over run. Let’s begin by estimating the slope at 𝑡 equals eight. We add horizontal and vertical lines to our tangent. Now, we can add these at any point, but it makes sense to do them as spread out as possible. This will give us the most accurate result.

The scale on our distance axis is roughly the same. This time, though, it’s one square equals two distance units. Those could be kilometers or miles, for example. And we see that the change in distance, the length of this line, is 32 distance units. Change in 𝑡 is 16, or 16 minutes. This means the slope of this first tangent is 32 divided by 16, which is two. And the velocity which might be in, say, kilometers per minute or meters per minute is therefore two.

Let’s repeat this process for our second tangent. This tangent is drawn at 𝑡 equals 55, or 55 minutes. This time, change in 𝑑 is 30. Change in 𝑡 is only two. Of course, since this is sloping downwards, we can say that the change in its distance or displacement is negative. So, the slope is negative 30 divided by two, which is negative 15. Now, in fact, we were just looking to find which tangent was most steep. And so, we consider the magnitude of their slope, which, of course, tells us the magnitude of their velocity or their speed. We know that 15 is greater than two.

We can, therefore, say that the magnitude of his velocity, which is his speed, is greater at 𝑡 equals 55 than it is at 𝑡 equals eight. And so, we’ve answered the final part of this question. Michael was moving the most fast at minute 𝑡 equals 55.