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?

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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?

We can see that in this question, we’re given 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. There are four different lines drawn on this graph, and we’re being asked to compare the speeds corresponding to each of these lines. Let’s label the speed corresponding to the red line as 𝑉 subscript 𝑟, the speed corresponding to the blue line as 𝑉 subscript 𝑏, the green line 𝑉 subscript 𝑔, and the orange line 𝑉 subscript 𝑜. We need to work out if these speeds change value in the same ratio for any two adjacent lines on the graph.

From the graph, we can see that the red line is adjacent to the blue line and the ratio of the corresponding speeds is 𝑉 subscript 𝑟, the speed for the red line, divided by 𝑉 subscript 𝑏, the speed for the blue line. Now the blue line is also adjacent to the green line and the ratio of these speeds is 𝑉 subscript 𝑏 divided by 𝑉 subscript 𝑔. Lastly, we can see that the green line is also adjacent to the orange line, and these speeds have a ratio of 𝑉 subscript 𝑔 divided by 𝑉 subscript 𝑜. These three ratios of speeds are the ratios for each pair of adjacent lines on the graph. So when the question asks us if the speeds change value in the same ratio for any two adjacent lines, that’s the same as asking us whether or not it’s true that these three ratios all have the same value.

So then to answer this question, we need to work out each of these three ratios to see whether or not they’re equal. If we find that all of these ratios do have the same value, that is, if we find that this mathematical equality is true, then we know that the speeds do change value in the same ratio for any pair of adjacent lines. Conversely, if we find that the statement isn’t true, then we know that the speeds don’t change value in the same ratio between adjacent lines. To work out each of these three ratios, let’s begin by finding the values of the four individual speeds.

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 means that if an object moves a distance of Δ𝑑 and it takes a time of Δ𝑡 in order to do this, then the average speed of the object during this time, 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 object moves from a distance of 𝑑 one to a distance of 𝑑 two, then its average 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 at the coordinates of two points along a straight line on a distance–time graph, that means that this expression 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 then, this expression gives the slope of a straight line drawn on a distance–time graph. We can say then that the speed of an object is equal to the slope of the corresponding line on a distance–time graph.

A straight line is a line that has a constant slope. The slope has the same value at all points along the line. So on a distance–time graph, a straight line represents a constant speed. We can see that all four of the lines on the graph are straight lines, and so they all represent motion at a constant speed. For motion at a constant speed, the average speed is the same as the speed at any point during the motion. This means that we can safely use this expression for the average speed 𝑉 with the coordinates of any two points along each line to work out each of the four speeds. Let’s now clear some space on the board so that we can do this.

We can notice that all four of the lines on the graph pass through the origin. That’s a time value of zero seconds and a distance value of zero meters. This means that we can use the origin as the first point on each of the four lines, which gives us 𝑡 one equals zero seconds and 𝑑 one equals zero meters in all four cases. Let’s begin by finding the value of 𝑉 subscript 𝑟. That’s the speed corresponding to the red line.

For the second point on this red line, we’ll choose this one here because at this point the line intersects with both a vertical and a horizontal grid line, which will make it easier to read off a time and a distance value. By tracing down along the vertical grid line until we get to the time axis, we can see that this point occurs at a time of four seconds. So for this red line, that’s our value for 𝑡 two. Then tracing across from the point to the distance access along the horizontal grid line, we can see that at this point the object has moved a distance of eight meters. This gives us our value for the quantity 𝑑 two.

We can now take these four values for the quantities 𝑡 one, 𝑑 one, 𝑡 two, and 𝑑 two and substitute them into this equation in order to calculate the speed 𝑉 subscript 𝑟. We find that 𝑉 subscript 𝑟 is equal to eight meters, which is our value for the distance 𝑑 two, minus zero meters, that’s the distance 𝑑 one, divided by four seconds, which is the time 𝑡 two, minus zero seconds. That’s the time 𝑡 one. In the numerator, eight meters minus zero meters is simply equal to eight meters. And similarly in the denominator, four seconds minus zero seconds is just four seconds. We have then that 𝑉 subscript 𝑟 is equal to eight meters divided by four seconds. This works out as a speed of two meters per second.

Now, let’s move on and do the same thing for the blue line. So that’s finding the value of 𝑉 subscript 𝑏. As before, we’re using the origin as the first point. So we’ve got 𝑡 one is zero seconds and 𝑑 one is zero meters. For the second point on the blue line, we’ll choose this one here. We can see that this has the same time of four seconds as we had on the red line. So just like before, we’ve got 𝑡 two is equal to four seconds. Then tracing across horizontally to the distance axis, we can see that at the second point on the blue line, the distance moved is equal to four meters. So that’s our value for the quantity 𝑑 two.

If we now take these four values and substitute them into this equation, we get this expression for the speed 𝑉 subscript 𝑏. In the numerator, we’ve got four meters minus zero meters, which is just equal to four meters. And then in the denominator, we have four seconds minus zero seconds, which is simply four seconds. So then, the speed 𝑉 subscript 𝑏 is equal to four meters divided by four seconds. And this works out as one meter per second.

Now, let’s move on to the green line and find the speed 𝑉 subscript 𝑔. Taking the origin as the first point on this line gives us 𝑡 one is zero seconds and 𝑑 one is zero meters, just as we had before. Then for the second point on the green line, let’s choose this one here. We can see that just as with these other two points, this second point on the green line has a time value of four seconds. So again, we’ve got 𝑡 two is equal to four seconds.

Tracing horizontally across from this point, we find that we meet the distance axis at a height of two meters. So this gives us our value for 𝑑 two. Then, substituting these four values into this equation, we get this expression for 𝑉 subscript 𝑔. Two meters minus zero meters divided by four seconds minus zero seconds is just the same as two meters divided by four seconds. When we evaluate this, we find that 𝑉 subscript 𝑔 is equal to 0.5 meters per second.

Lastly, we need to work out the value of 𝑉 subscript 𝑜. That’s the speed corresponding to the orange line. Since we’re using the origin as the first point, we’ve still got 𝑡 one as zero seconds and 𝑑 one as zero meters. For the second point on the orange line, we could choose this one here which has the same time value of four seconds as we had for the points on the other three lines. However, we’re not going to do this because at this point the orange line doesn’t intersect with a horizontal grid line. Instead, we’ll choose this point over here which intersects with both a vertical and a horizontal grid line making it easier to read the values from the axes. We find that the time value is eight seconds and the distance value is two meters. So that’s our values for the quantities 𝑡 two and 𝑑 two, respectively.

Then, using these four values in this equation, we find that 𝑉 subscript 𝑜 is equal to two meters minus zero meters divided by eight seconds minus zero seconds. And this is just the same as two meters divided by eight seconds. This works out as a speed 𝑉 subscript 𝑜 equal to 0.25 meters per second.

Now that we found the values for each of the four speeds corresponding to the four lines on the graph, we’re ready to calculate these three ratios between the speeds of adjacent lines. First, though, we need to clear some space on the board.

Let’s begin with this first ratio 𝑉 subscript 𝑟 divided by 𝑉 subscript 𝑏. Substituting in that 𝑉 subscript 𝑟 is two meters per second and 𝑉 subscript 𝑏 is one meter per second, this ratio becomes two meters per second divided by one meter per second. We can notice that the units will cancel between the numerator and the denominator, which leaves us with a dimensionless quantity. Evaluating two divided by one, we find that this first ratio is equal to two.

We can now do the same thing for these other two ratios. When we substitute in our values for the speeds, we get these two expressions. Again, in both cases, the units cancel out between the numerator and the denominator. So in this expression, we’ve got one divided by 0.5, and this works out as two. Then in this expression, we have 0.5 divided by 0.25, which again comes out as two. We have found then that all three of these ratios have the same value of two. That means that the statement is indeed true. So our answer to this question is yes, the speeds corresponding to the lines shown on the distance time graph do change value in the same ratio for any two adjacent lines.

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