Video: Reading and Interpreting Speed-Time Graphs

The speed-time graph shows the change of speed of a person walking during the time interval from 𝑡 = 0 seconds to 𝑡 = 6 seconds. What is the person’s speed at 𝑡 = 0 s? What is the person’s speed at 𝑡 = 3 s? What distance does the person walk between 𝑡 = 0 s and 𝑡 = 2 s? What distance does the person walk between 𝑡 = 2 s and 𝑡 = 4 s? What distance does the person walk between 𝑡 = 4 s and 𝑡 = 6 s? What distance does the person walk between 𝑡 = 0 s and 𝑡 = 6 s?

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Video Transcript

The speed–time graph shows the change of speed of a person walking during the time interval from 𝑡 is equal to zero seconds to 𝑡 is equal to six seconds.

So, this here is the speed–time graph in question, and we can see that on the vertical axis we’ve been given the speed of the person walking in meters per second. And on the horizontal axis, we’ve got the time in seconds.

Now, the first part of the question asks us, what is the person’s speed at 𝑡 is equal to zero seconds?

Now, as we’ve already mentioned, the horizontal axis of the graph shows us the time when the person is walking. And from this graph, we can find the speed at which the person is walking at any given time. So, we’ve been asked to find the speed of the person at 𝑡 is equal to zero seconds. Now, 𝑡 is equal to zero seconds is the time at the very beginning of the graph. Specifically, this vertical line here represents 𝑡 is equal to zero seconds.

And so, what we can do is to draw a vertical line upwards at 𝑡 is equal to zero seconds until it meets the graph representing the person’s speed. And at this point, we can read off on the vertical axis that the person speed is five meters per second. Hence, our answer to this part of the question is that the person’s speed at 𝑡 is equal to zero seconds is five meters per second. Let’s move on to the next part of the question then.

This part asks us, what is the person’s speed at 𝑡 is equal to three seconds?

So, now we want to go along the horizontal axis, the time axis, until we arrive at 𝑡 is equal to three seconds. At this point, we want to draw a vertical line upwards at 𝑡 is equal to three seconds until we meet the blue line because the blue line is representing the speed of the person walking as time progresses. And once we meet the blue line, we can go across to the vertical axis, which tells us that the speed of the person at the time 𝑡 is equal to three seconds is six meters per second. And hence, six meters per second is our answer to the second part of the question.

Moving on to the next part of the question then, this part asks us, what distance does the person walk between 𝑡 is equal to zero seconds and 𝑡 is equal to two seconds?

Now, as we’ve already seen, 𝑡 is equal to zero seconds is represented by this vertical line here. And we can draw in this vertical line to represent 𝑡 is equal to two seconds. Now, we’ve been asked to find the distance walked by the person between 𝑡 is equal to zero seconds and 𝑡 is equal to two seconds. To do this, we can recall the equation that links speed, distance, and time. Two quantities which we have on our graph, speed and time, and the third quantity which we’re trying to find, distance. We can recall that the speed of an object is equal to the distance moved by the object divided by the time taken for that object to move that distance. And so, we can take this equation and multiply both sides by the time 𝑡 in order to solve for the distance 𝑑.

When we do this, we find that the speed of the object 𝑠, in this case the person walking, multiplied by the time 𝑡 for which the person walks at that speed is equal to the distance moved by that person, 𝑑. Now, in our particular graph, we’ve got the speed of the person on the vertical axis and the time on the horizontal axis. And for the first two seconds of time given to us on the graph, we’ve been told that the person walks at a constant speed of five meters per second. That’s what the blue line is telling us. This means that for the first two seconds, the person walks at a constant speed of five meters per second, and so the distance moved by the person is going to be the constant speed, five meters per second, multiplied by the time, two seconds.

But interestingly, this also happens to be equal to the area underneath our blue line, in this case now drawn in pink. In other words, that’s this area here because that area is equal to five meters per second multiplied by two seconds, which is exactly what’s going to give us the distance moved by the person. And so, we can say that the distance moved by the person between 𝑡 is equal to zero seconds and 𝑡 is equal to two seconds, we’ll call this distance 𝑑, is equal to the speed of the person, five meters per second, multiplied by the amount of time for which the person is walking at that speed, two seconds.

Now, quickly looking at the units, we can see that we’ve got meters per second multiplied by seconds which means that we’ve got units of seconds in the numerator and the denominator. And those cancel, leaving us with just the unit of meters. And this works perfectly because on the left-hand side, we’ve got the distance moved by the person. This needs to have a unit of meters. And hence, the distance moved by the person is going to be five multiplied by two meters. In other words, that’s 10 meters. And hence, we can say that the distance the person walks between 𝑡 is equal to zero seconds and 𝑡 is equal to two seconds is 10 meters.

Now, moving on to the next part of the question, we are asked, what distance does the person walk between 𝑡 is equal to two seconds and 𝑡 is equal to four seconds?

In other words now, what distance does the person walk between this point in time and this point in time? Well, drawing vertical dotted lines from 𝑡 is equal to two seconds and 𝑡 is equal to four seconds all the way up to the blue line shows us that the area that we now need to calculate is this area here. Because even though between 𝑡 is equal to two seconds and four seconds, the person is no longer walking at a constant speed because we can see that the speed increases between these two points in time. The fact of the matter is that the area under a speed–time graph still gives us the distance moved by that particular object.

So, to find the area underneath the speed-time graph, this time between 𝑡 is equal to two seconds and 𝑡 is equal to four seconds, we need to find this shaded area as we’ve already mentioned. Now, to make life easier for ourselves, we can split up this area into two smaller ones. Firstly, this rectangle here. And secondly, this blue triangle. We can call this area one, and we can call this area two. And we can say that the total area that we’re trying to find is area one plus area two, the area of the triangle plus the area of the rectangle.

Now, with area two, the area of the rectangle, we can notice something. We’ve actually already calculated the size of that rectangle because the area of that rectangle is equal to this time interval here, between two seconds and four seconds, multiplied by this speed interval which is five meters per second minus zero meters per second or five meters per second overall. Now, the time interval is four seconds minus two seconds, which happens to be an interval of two seconds. And so, we can say that area two is equal to five meters per second multiplied by two seconds, which we’ve already seen from earlier is 10 meters. In other words, area two is the same size as this rectangle we saw earlier. Which means all we’ve got to do now is to calculate area one, the area of the triangle.

To do this, we can recall that the area of a triangle, we’ll call it 𝐴 subscript triangle, is equal to half multiplied by the length that is the base of the triangle multiplied by the height of the triangle. In this case, the base of the triangle is the same as the width of the rectangle from two seconds to four seconds. Which, actually of course, is a time interval since on the horizontal axis we’ve got the time. And the height of the triangle is this distance here between seven meters per second and five meters per second. In other words, this interval here is seven meters per second minus five meters per second, which is two meters per second. And hence, the area of the triangle is equal to half multiplied by the base, which we’ve said is two seconds, multiplied by the height, which we’ve said is two meters per second.

Once again, we see that the unit of seconds in the numerator and denominator cancel to give us an overall unit of meters which is once again good. We are calculating a distance after all. And the numerical value is half multiplied by two multiplied by two. Overall then, we find that the area of triangle one is two meters. And hence, as we’ve said earlier, the total area of triangle one plus rectangle two is equal to two meters plus 10 meters. And hence, we can say that the total distance moved by the person between 𝑡 is equal to two seconds and 𝑡 is equal to four seconds is 12 meters, which is our answer to this part of the question.

Moving on to the next part of the question then, we are now asked, what distance does the person walk between 𝑡 is equal to four seconds and 𝑡 is equal to six seconds?

So, we can draw vertical lines from the time axis for 𝑡 is equal to four seconds and 𝑡 is equal to six seconds, as we’ve seen already. We need to find the area beneath the blue line between these two times: in other words, the area underneath this straight line. And so, that’s this shaded area here. That area happens to have a width that’s equivalent to this time interval and a height that’s equivalent to this speed. It helps us, by the way, that the person is walking once again at a constant speed because now the speed–time graph is a flat line in this time interval. Which means the area that we’re trying to find is once again a rectangle.

And to find the area of that rectangle, we firstly see that six seconds minus four seconds gives us a width for the rectangle of two seconds. And the height is equal to seven meters per second, which is the highest point of that rectangle, minus zero seconds, which is the lowest point. Or, in other words, we’ve got a total height of seven meters per second. Hence, we can say that the distance moved in this time interval between 𝑡 is equal to four seconds and six seconds is equal to seven meters per second, that’s the height of the rectangle, multiplied by the width, which is two seconds. Calculating this area then, we find that the distance moved by the person in this time interval is 14 meters. And that is our final answer to this part of the question.

Now, the last part of the question asks us, what distance does the person walk between 𝑡 is equal to zero seconds and 𝑡 is equal to six seconds?

So, that’s the distance walked between this time and this time. In other words, we need to find the area underneath the line for the entire graph now between 𝑡 is equal to zero seconds and 𝑡 is equal to six seconds. However, as we’ve seen from previous parts of the question already, the distance walked by the person between 𝑡 is equal to zero seconds and two seconds was 10 meters. That’s the area of this rectangle. The distance walked by the person between 𝑡 is equal to two seconds and four seconds was 12 meters. And the distance walked between 𝑡 is equal to four seconds and six seconds was 14 meters.

So, the total distance walked between 𝑡 is equal to zero seconds and 𝑡 is equal to six seconds is just going to be 10 meters plus 12 meters plus 14 meters. We’ll call this total distance 𝑑 subscript tot or 𝑑 subscript total. And as we’ve already said, it’s the sum of all the three distances we’ve calculated so far. This distance ends up being 36 meters, and hence that’s our final answer to the final part of the question.

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