Video Transcript
The graph shows how the stopping distances of two cars change depending on the speeds at which the cars are moving at the moment they start to decelerate. What is the difference in the stopping distances of the cars when they both stop at the speed of 15 meters per second?
So we can see that we’ve been given a graph which plots stopping distance in units of meters on the vertical or 𝑦-axis against speed in units of meters per second on the horizontal or 𝑥-axis. There are two lines plotted on the graph; there’s this blue line and there’s this orange one. And each of these lines corresponds to the motion of a different car. We’ll refer to these cars as the blue car and the orange car, respectively.
What’s going on here is that both of these cars are traveling at some initial speed, and then they each decelerate until they stop. The graph is plotting the stopping distance of each car, that’s the distance that it takes the car to come to a stop, for a range of different initial speeds. In this first bit of the question, we’re asked to work out the difference in the cars’ stopping distances when each of them is initially moving with a speed of 15 meters per second.
We can see from the graph that the blue line is always vertically above the orange line for any given value of speed. Now, the two cars may have different stopping distances because they’ve got different-quality brakes or because they were traveling on different surfaces or in different weather conditions, for example. Whatever the reason, the graph shows that the blue car will always have a greater value of stopping distance than the orange car and that this is true no matter which of these initial speeds we’re considering.
Let’s label the stopping distance of the blue car as 𝑑 subscript b and the stopping distance of the orange car as 𝑑 subscript o. Since we know that the orange car is the one that’s going to have the smaller stopping distance, then the difference in their stopping distances will be 𝑑 subscript b minus 𝑑 subscript o. So that’s subtracting the smaller stopping distance from the larger stopping distance. We can read off each of these values 𝑑 subscript b and 𝑑 subscript o from the graph that we’re given. To do this, we find the value of 15 meters per second on the horizontal speed axis.
For 𝑑 subscript o, the stopping distance of the orange car, we trace vertically upward from 15 meters per second until we get to the orange line on the graph. We then trace across from this point horizontally until we get to the vertical stopping distance axis. We find that our line meets the vertical axis at a height of 12. And since the stopping distance axis is measured in units of meters, that means that the stopping distance of this orange car when the speed was 15 meters per second is equal to 12 meters. So, 12 meters is our value for the quantity 𝑑 subscript o.
Then, to find 𝑑 subscript b, we need to do the same thing, but with the blue line on the graph. Tracing vertically upward from 15 meters per second on the speed axis until we get to the blue line and then tracing across horizontally from this point on the graph until we get to the vertical stopping distance axis, we read off a value of 30, which again we know is in units of meters. So when the blue car is initially traveling at 15 meters per second, it has a stopping distance 𝑑 subscript b equal to 30 meters.
Now, we just need to calculate the difference 𝑑 subscript b minus 𝑑 subscript o. Substituting in our values that we’ve read from the graph, we find that this difference is equal to 30 meters minus 12 meters, which works out as 18 meters. So then our answer to this first bit of the question is that the difference in the stopping distances of the cars when they both stop at the speed of 15 meters per second is equal to 18 meters.
Now, let’s clear some space and look at the second part of the question.
What is the difference in the stopping distances of the cars when they both stop at the speed of 20 meters per second?
This second part of the question is asking us to do exactly the same thing we did for the first part, but now for a speed of 20 meters per second. Once again, we’ll refer to the stopping distance of the blue car as 𝑑 subscript b and the stopping distance of the orange cars as 𝑑 subscript o. And also like before, we know that the difference in the stopping distances will be 𝑑 subscript b minus 𝑑 subscript o. Let’s start with 𝑑 subscript o, the stopping distance of the orange car.
We find the value of 20 meters per second on the speed axis. We then trace vertically up from this point until we meet the orange line on the graph and then horizontally across from this point until we get to the vertical stopping distance axis. We see that this line hits the axis at a value of 20, which means that a speed of 20 meters per second for the orange car corresponds to a stopping distance of 20 meters. So then 20 meters is our value for 𝑑 subscript o.
Let’s now do the same thing for 𝑑 subscript b, the stopping distance of the blue car. Tracing up from a speed of 20 meters per second until we get to the blue line and then tracing across to the stopping distance axis, we read off a value of 50. So for an initial speed of 20 meters per second, the blue car has a stopping distance of 50 meters. And this is our value for 𝑑 subscript b. The difference in stopping distances 𝑑 subscript b minus 𝑑 subscript o is then equal to 50 meters minus 20 meters, and this works out as 30 meters.
So, our answer for this second part is that when both cars stop from the speed of 20 meters per second, the difference in their stopping distances is equal to 30 meters.
