# Question Video: Comparing Speeds from a Distance-Time Graph Physics

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?

08:48

### 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?

This question is asking us about lines 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’re asked to compare the speeds for each of these four lines that are drawn on this graph and to work out whether for any two adjacent lines those speeds change value in the same ratio. 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 of speed means that if an object travels 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 Δ𝑑 is equal to 𝑑 two minus 𝑑 one and Δ𝑡 is equal to 𝑡 two minus 𝑡 one. So we have that the speed 𝑉 is equal to 𝑑 two minus 𝑑 one divided by 𝑡 two minus 𝑡 one.

Now, 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 different points on a line that’s drawn on a distance–time graph, that means that this expression is the change in the vertical coordinate distance between those two points divided by the change in the horizontal coordinate time between those same two points. In other words, if we’ve got a straight line drawn on a distance–time graph, then this expression will give us the slope of that line. We can say then that if we’ve got an object whose motion is being represented by a straight line on a distance–time graph, that the speed of that object is equal to the slope of that line.

In this question, we’ve got a distance–time graph with four straight lines shown on it. There’s an orange line, a green line, a blue line, and a red line. We’ve labeled the speeds corresponding to each of these lines as 𝑉 subscript o, 𝑉 subscript g, 𝑉 subscript b, and 𝑉 subscript r, respectively. We can use this expression here in order to calculate the slope of each of these four lines, and we know that that gives us the speed that each line corresponds to. Let’s clear some space on the board and make a start on this.

To find the speed of each line, we need to pick two points on that line which we’ll label with coordinates 𝑡 one, 𝑑 one and 𝑡 two, 𝑑 two and then use this expression in order to calculate the slope between those two points. Let’s begin with the orange line on the graph, so that’s calculating the speed 𝑉 subscript o. We’ll choose our first point on this line as the origin of the graph. So that’s the time value of zero seconds and a distance value of zero meters. For the second point on our orange line, we’ll choose this point here.

Tracing vertically downward from this point until we get to the time axis, we can see that this point has a time value which lies halfway between the zero-second mark and the two-second mark on the axis. The value that’s halfway between zero and two is one. And so, the time value of this point 𝑡 two must be equal to one second. Then tracing horizontally across from the second point to the distance axis, we see that the distance value at this point is equal to eight meters. So that’s our value for the quantity 𝑑 two.

We can now take these four values and substitute them into this expression in order to calculate the speed correspondent to the orange line, 𝑉 subscript o. When we do that, we get this expression here. In the numerator, we have eight meters, that’s our value for 𝑑 two, minus zero meters, our value for 𝑑 one. Then, in the denominator, we have one second, that’s our time value 𝑡 two, minus zero seconds, which is 𝑡 one. Now, eight meters minus zero meters is simply equal to eight meters. And likewise, one second minus zero seconds is just equal to one second. So, we have that 𝑉 subscript o is equal to eight meters divided by one second. This works out as a speed of eight meters per second.

Now that we found the speed corresponding to the orange line, let’s move on and do the same thing for the green line. Let’s again choose the origin of the graph as the first point, which gives us 𝑡 one equals zero seconds and 𝑑 one equals zero meters. Then, for the second point on this green line, we’ll choose this point here. Tracing down to the time axis, we can see that this point occurs at a time value of two seconds and that’s our value for the quantity 𝑡 two. Then tracing across to the distance axis, we can see that the object has moved a distance of eight meters at this point, so that’s our value for the quantity 𝑑 two.

We can notice that our values for 𝑡 one, 𝑑 one, and 𝑑 two are the same for the green line as we had for the orange line. The only difference is the value of 𝑡 two, which is now two seconds rather than the one second that we had for the orange line. In other words, twice as much time has passed; at the moment the object represented by the green line has moved a distance of eight meters as compared to the object represented by the orange line. If we now substitute these four values into this equation, we get this expression for the speed 𝑉 subscript g. In the numerator, eight meters minus zero meters is just eight meters. Then, in the denominator, two seconds minus zero seconds is just two seconds. 𝑉 subscript g then is equal to eight meters divided by two seconds. And this works out as a speed of four meters per second.

Now, let’s move on and do the same thing for 𝑉 subscript b, the speed corresponding to the blue line. Again, we’ll choose the origin as our first point. So, we have that 𝑡 one is zero seconds and 𝑑 one is zero meters. For our second point on this line, we’ll choose this one here, we can see that just as with the orange line and the green line, the distance value at this second point is equal to eight meters. So again, we’ve got the same three values for the quantities 𝑡 one, 𝑑 one, and 𝑑 two. Then for the time value 𝑡 two, when we trace down from this point on the graph to the time axis, we read off a value of four seconds.

Let’s now substitute these four values into this equation to calculate the speed 𝑉 subscript b. When we do this, we get this expression here, which works out as a speed of two meters per second. The last speed value left to find is 𝑉 subscript r, the speed corresponding to the red line on the graph. Again, we’ll choose the origin for the first point on this red line, which gives us 𝑡 one and 𝑑 one as zero seconds and zero meters, respectively. For the second point, which is this one here, which we can see has a time value of eight seconds and a distance value of eight meters. Using these values in this equation gives us this expression for the speed 𝑉 subscript r, which works out as one meter per second.

Okay, so we’ve now found the four speeds corresponding to the four different lines on the distance–time graph. The question is asking us whether these speeds change value in the same ratio for any two adjacent lines on the graph. We can see that these lines are ordered orange, then green, then blue, and then red. Let’s recall that the ratio of two speeds 𝑉 one and 𝑉 two is equal to the speed 𝑉 one divided by the speed 𝑉 two. We’re being asked whether the ratio of 𝑉 subscript o to 𝑉 subscript g is equal to the ratio of the speeds 𝑉 subscript g and 𝑉 subscript b and equal to the ratio 𝑉 subscript b to 𝑉 subscript r. We can calculate each of these three ratios using our values for the four speeds.

Starting with the ratio 𝑉 subscript o over 𝑉 subscript g then substituting in that 𝑉 subscript o is eight meters per second and 𝑉 subscript g is four meters per second, we get this expression here. The units of meters per second cancel from the numerator and denominator, leaving us with a dimensionless quantity. Then evaluating eight divided by four gives a result for this ratio of two. If we now consider this second ratio and substitute in our values for the speeds 𝑉 subscript g and 𝑉 subscript b, then again the units cancel, leaving us with four divided by two, which works out as a ratio of two.

For the final ratio 𝑉 subscript b over 𝑉 subscript r, our values for these two speeds give us an expression of two meters per second divided by one meter per second. Canceling the units between the numerator and denominator and doing the division two divided by one, we find that this final ratio is also equal to two. Since we’ve calculated the same ratio of two in each of these three cases, then we can say that this expression about the speeds of the lines on the graph is indeed true.

So then, our answer to this question is yes, the speeds corresponding to the lines shown on this distance–time graph do indeed change value in the same ratio for any two adjacent lines.