Video Transcript
Do the speeds corresponding to the lines shown on the following distance–time graph change value equally for any two adjacent lines?
Okay, so in this question, we’re given a distance time–graph. That’s a graph that puts distance on the vertical or 𝑦-axis against time on the horizontal or 𝑥-axis. We’re being asked to compare the speeds corresponding to these four different lines that are drawn on the graph. Let’s recall then 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 moving with a constant speed travels a distance of Δ𝑑 and it takes a time of Δ𝑡 in order to do this, then the speed of this object, which we’ll label as 𝑣, is equal to Δ𝑑 divided by Δ𝑡.
A distance–time graph is showing us how much distance an object has moved at each instant in time. This expression for the speed of an object is the change in distance. So that’s the change in the vertical coordinate on a distance–time graph divided by the change in time. So that’s the change in the horizontal coordinate on this graph. The change in the vertical coordinate over some section of a line on a graph divided by the change in the horizontal coordinate over that same section is defined as the slope of that 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.
For a straight line, the slope of that line has a constant value. And so a straight line on a distance–time graph must represent an object that’s moving with a constant speed. We can see that all four of the lines on this distance–time graphs are indeed straight lines, so all of them must represent motion at a constant speed. We can also write this expression for the speed 𝑣 in another way.
Let’s consider two points along a line on a distance–time graphs with coordinates 𝑡 one, 𝑑 one and 𝑡 two, 𝑑 two, respectively. This is equivalent to an object moving from a distance of 𝑑 one to a distance of 𝑑 two between a time of 𝑡 one and a time of 𝑡 two. In other words, the change in distance moved by the object Δ𝑑 is equal to 𝑑 two minus 𝑑 one. And the change in time Δ𝑡 over which this distance change occurs is equal to 𝑡 two minus 𝑡 one.
With this in mind, let’s use this expression now to calculate the slope, and therefore the speed, corresponding to each of these four lines on the distance–time graph. We’ve labeled the speed of the orange line as 𝑣 subscript o, the speed of the green line as 𝑣 subscript g, the blue line 𝑣 subscript b, and the red line 𝑣 subscript r. To work out these values of speed, we need to pick two points on each line with coordinates of 𝑡 one, 𝑑 one and 𝑡 two, 𝑑 two, respectively. Using the values of these four quantities for each line in this equation will then allow us to calculate the speed corresponding to that line.
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 a time value of zero seconds and a distance value of zero meters. This gives us our values for the quantities 𝑡 one and 𝑑 one. For the second point on this orange line, let’s choose this one here because at this point the orange line conveniently intersects with both a horizontal grid line and a vertical grid line.
Tracing straight down from this point along the vertical grid line until we get to the time axis, we see that this point has a time value of two seconds. So that’s our value for the quantity 𝑡 two. Then, tracing across along the horizontal grid line until we get to the distance axis, we can see that at this point the distance moved by the object is equal to eight meters. So that gives us our value for 𝑑 two.
We can now take these four values and substitute them into this expression in order to calculate the value of 𝑣 subscript o. When we do that, it gives us this expression here. In the numerator, we’ve got eight meters, that’s our value for the distance 𝑑 two, minus zero meters, that’s 𝑑 one. And then this gets divided by two seconds, that’s the time 𝑡 two, minus zero seconds, the time 𝑡 one. Now eight meters minus zero meters is simply equal to eight meters. And likewise, two seconds minus zero seconds is just equal to two seconds. We have then that 𝑣 subscript o is equal to eight meters divided by two seconds. This works out as a speed of four 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. So that’s finding the speed 𝑣 subscript g. Let’s again choose the origin of the graph as the first point on this green line. This gives us that 𝑡 one is equal to zero seconds and 𝑑 one is equal to zero meters just like we had before. If we then choose this point on the green line as our second point, we can see that it occurs at the same time value of two seconds as we had for the second point on the orange line. So then just as we had before, our value for 𝑡 two is two seconds. So then the only thing that differs between the sections of the orange line and the green line that we’ve chosen is the value for the quantity 𝑑 two.
Tracing horizontally across from this second point on the green line, we meet the distance axis at a value of six meters. And this is our value for 𝑑 two. What we have then is that while between a time of zero seconds and a time of two seconds, the object represented by the orange line moved from a distance of zero meters to a distance of eight meters, in that same interval between zero and two seconds, the object represented by this green line moved from the same starting distance of zero meters to an end distance of six meters. If we now substitute these four values into this equation, we get this expression for the speed 𝑣 subscript g, which looks almost exactly like the one we had before for 𝑣 subscript o, except now we’ve got the six meters in the numerator where previously we had eight meters.
We can see that six meters minus zero meters will just be equal to six meters. And likewise, two seconds minus zero seconds will just be two seconds. So we’ve got that 𝑣 subscript g is equal to six meters divided by two seconds, which works out as three meters per second. Let’s now move on and do the same thing for the speed 𝑣 subscript b corresponding to the blue line.
Again, we’ll choose the origin as the first point, giving us 𝑡 one equals zero seconds and 𝑑 one equals zero meters. For the second point on this line, we’ll choose this one here, which, we can see, has a time value of two seconds and a distance value of four meters. This gives us our values for the quantities 𝑡 two and 𝑑 two. So again, the values for 𝑡 one, 𝑑 one, and 𝑡 two haven’t changed. The only value that has changed is the value of 𝑑 two. Using these four values in this equation gives us this expression for the speed 𝑣 subscript b. We can then evaluate this to get a result of two meters per second.
The last speed value left to find is 𝑣 subscript r, the speed corresponding to the red line. As before, we’ll choose the origin as the first point on this line, giving us zero seconds and zero meters for the values of 𝑡 one and 𝑑 one, respectively. As the second point, we’ll choose this one here, which we can see has a time value of two seconds and a distance value of two meters. That gives us our values for the quantities 𝑡 two and 𝑑 two. Using these values in this equation gives us this expression for 𝑣 subscript r. This works out as a speed of one meter per second.
Okay, so we’ve now found the four speeds corresponding to the four different lines on this distance–time graph. The question is asking us whether these speeds change value equally for any two adjacent lines. We can see that these lines are ordered orange, then green, then blue, and then red. So the question is asking us whether the difference between the orange line speed and the green line speed is the same as the difference between the green line speed and the blue line speed and the same as the difference between the blue line speed and the red line speed. In other words, we’re being asked whether or not this statement is true.
On the left, we’ve got 𝑣 subscript o minus 𝑣 subscript g. That’s the difference between the speeds represented by the orange line and the green line. In the middle, we’ve got 𝑣 subscript g minus 𝑣 subscript b, which is the difference between the speeds of the green line and the blue line. Lastly on the right, 𝑣 subscript b minus 𝑣 subscript r is the difference between the blue line and the red line speeds. We can work out each of these three differences using our values for the four speeds.
When we do that, we end up with these three expressions for those differences. We can see that in all three cases, this difference is the same and is equal to one meter per second. Since we have found the same difference of one meter per second between the speeds corresponding to each pair of adjacent lines on the graph, then our answer to this question is yes, the speeds corresponding to the lines on this distance–time graph do change value equally for any two adjacent lines.
