The figure shows a velocity–time graph for a particle moving in a straight line. Find the total distance the particle traveled.
Remember, given a function for velocity, we can integrate it with respect to time to find a function for displacement. However, we also know that integration is a way to evaluate the area between a curve and the 𝑥-axis. So, we can use our graph to find the area and hence find the total distance traveled.
We do need to be a little bit careful, though. We’re working with distance, which is a scalar quantity. It’s the magnitude of displacement. Usually, if the area between the curve and the 𝑥-axis sits below the 𝑥-axis, we end up with a negative value on integration. Here though, we don’t need to worry about that since we’re just looking for distance. We’re going to find the total area between the graph and the 𝑥-axis.
We have three distinct areas. I’ve marked them up as 𝐴 one, 𝐴 two, and 𝐴 three. And each area is enclosed by a polygon. 𝐴 one and 𝐴 three are made up of triangles. And we know that the area of a triangle is a half times the base times the vertical height. The base of the triangle enclosing 𝐴 one is four units wide, and its vertical height is also four units. So, 𝐴 one is equal to a half times four times four, which is eight square units. The triangle around 𝐴 three is three units wide and four units high. So, its area is a half times three times four, which is six square units.
𝐴 two is enclosed by a trapezium. And the area of a trapezium is a half times 𝑎 plus 𝑏 times ℎ, where 𝑎 and 𝑏 are the length of the parallel sides and ℎ is the distance between them. 𝐴 two is a half times three plus one times two, which is four square units. The total distance traveled by the particle is the sum of these three values. It’s eight plus six plus four, which is 18. The total distance the particle traveled is 18 meters.
Now, it’s also useful to know that we could have worked out the displacement of the particle. This time, that would’ve been eight plus six minus four, which is 10 meters.