Video Transcript
Do the speeds corresponding to the lines shown on the following distance–time graph change value in the same ratio for any two adjacent lines?
In this question, we’re shown three lines — a blue line, a red line, and a green line — drawn on a distance–time graph. That’s a graph that plots distance on the vertical or 𝑦-axis against time on the horizontal or 𝑥-axis. We are being asked if the speeds that correspond to these three lines change value in the same ratio for any two adjacent lines. Let’s label the speeds corresponding to the blue, red, and green lines as 𝑉 subscript b, 𝑉 subscript r, and 𝑉 subscript g, respectively. We can see from the graph that the blue line and the red line are adjacent to each other and that the red line and the green line are also adjacent to each other.
By asking us if the speeds corresponding to any two adjacent lines change value in the same ratio, the question is effectively asking us whether this mathematical equality is true. On the left-hand side of this, 𝑉 subscript b divided by 𝑉 subscript r is the ratio of the speed corresponding to the blue line to the speed corresponding to the red line. Then, on the right, we’ve got 𝑉 subscript r divided by 𝑉 subscript g. That’s the ratio of the speed corresponding to the red line to the speed corresponding to the green line. This equality then is asserting that the ratio of the blue line speed to the red line speed is the same as the ratio of the red line speed to the green line speed.
So what we need to do then is to work out the left-hand side and the right-hand side and see if they are indeed equal. If we find that they are equal, then our answer to the question is yes, the speeds corresponding to adjacent lines do change value in the same ratio. By the same token, if we find that they’re not equal, then our answer is no, the speeds corresponding to adjacent lines do not change value in the same ratio. In order to work out these two ratios, let’s begin by calculating the individual speeds 𝑉 subscript b, 𝑉 subscript r, and 𝑉 subscript g.
We can recall that the speed of an object is defined as the rate of change of the distance moved by that object with time. This definition means that if an object moves a distance of Δ𝑑 and it takes a time of Δ𝑡 in order to do this, then the speed of that object, which we’ll label as 𝑉, is equal to Δ𝑑 divided by Δ𝑡. We can also write this fraction in another way. If between a time of 𝑡 one and a time of 𝑡 two the distance moved by an object changes from a value of 𝑑 one to a value of 𝑑 two, then for this period of motion, Δ𝑑 is equal to 𝑑 two minus 𝑑 one and Δ𝑡 is equal to 𝑡 two minus 𝑡 one. We have then that the speed 𝑉 is equal to 𝑑 two minus 𝑑 one divided by 𝑡 two minus 𝑡 one.
Since the distance–time graph plots distance on the vertical axis against time on the horizontal axis, then if 𝑡 one, 𝑑 one and 𝑡 two, 𝑑 two are the coordinates of two points on a straight line drawn on a distance–time graph, that means that this expression for the speed 𝑉 is equal to the change in the vertical coordinate between these two points divided by the change in the horizontal coordinate between the same two points. In other words, if we have a straight line drawn on a distance–time graph, then this expression is calculating the slope of this line. We can say then that the speed of an object is equal to the slope of the corresponding line on a distance–time graph.
Now, we said earlier that we want to calculate these three speeds: 𝑉 subscript b, 𝑉 subscript r, and 𝑉 subscript g. In each case, we can do that by using this expression to calculate the slope of the corresponding line on the graph. And we know that that slope will give us the speed. Let’s clear some space on the board and make a start on this. Let’s begin with the blue line on the graph, so that’s calculating the speed 𝑉 subscript b. To calculate the slope of this blue line, we need to pick two points on that line to calculate the slope between. And we’ll label those coordinates as 𝑡 one, 𝑑 one and 𝑡 two, 𝑑 two. We’ll choose our first point on this line as the origin of the graph, so that’s a time value of zero seconds and a distance value of zero meters. Then, for the second point, we’ll choose this one here.
If we trace vertically down from this point until we get to the time axis, we can see that this point occurs at a time value of four seconds. So that’s our value for the quantity 𝑡 two. If we then trace across horizontally to the distance axis, we can see that at this point, the object has moved a distance of eight meters, so that’s our value for the quantity 𝑑 two. We can now take these four values and substitute them into this equation in order to calculate the speed corresponding to the blue line 𝑉 subscript b. When we do that, we get this expression here. In the numerator, we’ve got eight meters, that’s the distance 𝑑 two, minus zero meters, that’s our value for 𝑑 one. And then, in the denominator, we have four seconds, that’s the time value 𝑡 two, minus zero seconds, which is 𝑡 one.
We can simplify this expression by noticing that eight meters minus zero meters is simply equal to eight meters, and likewise four seconds minus zero seconds is just four seconds. So we have then that the speed 𝑉 subscript b is equal to eight meters divided by four seconds. This works out as a speed of two meters per second.
Now that we found the speed corresponding to the blue line, let’s move on and do the same thing for the red line. Just as with the blue line, to find the speed 𝑉 subscript r corresponding to the red line, we need to begin by picking two points on that line. Again, for the first point, we’ll choose the origin of the graph, which means that we have 𝑡 one is equal to zero seconds and 𝑑 one is equal to zero meters. For the second point, we’ll choose this one here. Tracing down to the time axis, we can see that this point occurs at a time value of four seconds, and that’s our value for the quantity 𝑡 two. Then, tracing across to the distance axis, we can see that the distance value at this point is equal to four meters, so that’s our value for 𝑑 two.
If we now substitute these four values into this equation, we get this expression for the speed 𝑉 subscript r. In the numerator, four meters minus zero meters is just four meters. And similarly in the denominator, four seconds minus zero seconds is equal to four seconds. So then we have that 𝑉 subscript r is equal to four meters divided by four seconds, which works out as a speed of one meter per second. The final speed value left to find is the speed 𝑉 subscript g corresponding to the green line. Again, we’ll choose the origin for the first point on this line, and that gives us 𝑡 one equals zero seconds and 𝑑 one equals zero meters. For the second point on the green line, let’s choose this point here. We can see that this point occurs at a time of four seconds, so that’s our value for the quantity 𝑡 two.
We can also see that the distance moved at this point is equal to two meters, so that’s our value for 𝑑 two. Using these values in this equation gives us this expression for the speed 𝑉 subscript g. Evaluating this gives a speed of 0.5 meters per second. Now that we found the values of the three speeds corresponding to each of the three lines on this distance–time graph, we’re ready to calculate these two ratios and see whether or not they are equal. Let’s now clear some space on the board to do this.
Let’s begin by calculating 𝑉 subscript b over 𝑉 subscript r. That’s the ratio of the speed represented by the blue line to the speed represented by the red line. Substituting in that 𝑉 subscript b is equal to two meters per second and 𝑉 subscript r is one meter per second, we have that this ratio is equal to two meters per second divided by one meter per second. In this expression, we can see that the units of meters per second will cancel from the numerator and denominator, leaving us with a dimensionless quantity. Then, evaluating two divided by one, we find that this ratio of 𝑉 subscript b to 𝑉 subscript r is equal to two.
Now, let’s calculate 𝑉 subscript r divided by 𝑉 subscript g. That’s the ratio of the speed represented by the red line to the speed represented by the green line. Substituting in that 𝑉 subscript r is one meter per second and 𝑉 subscript g is 0.5 meters per second, we get this expression here. Like we had before, the units of meters per second cancel from the numerator and denominator. That leaves us with one divided by 0.5, which works out as a ratio of two.
Since we’ve calculated the same ratio of two in each case, then we can say that this statement is indeed true. So then, our answer to this question is yes. The speeds corresponding to the lines on this distance–time graph do change value in the same ratio for any two adjacent lines.
