Given the velocity–time graph of a particle moving in a straight line, determine the distance covered by the particle in the time interval zero to eight.
The motion of this particle can be split into four parts. Firstly, it decelerates from five meters per second to negative five meters per second. This takes two seconds. Secondly, it travels at a constant speed from 𝑡 equals two seconds to 𝑡 equals six seconds. It then accelerates from negative five meters per second to five meters per second by 10 seconds. Finally, it decelerates to rest at 𝑡 equals 14 seconds. In this question, we’re interested in the time interval zero to eight seconds.
In any velocity–time graph, we can calculate the displacement or distance by looking at the area under the graph. In this question, we will split this into two sections. The section between 𝑡 equals zero and 𝑡 equals one, where the area is above the 𝑥-axis, and the area between 𝑡 equals one and 𝑡 equals eight, where it is below the 𝑥-axis. Our value for B, if we were considering displacement, would be negative. But as we’re dealing with distance and this must be positive, we take the absolute value of the displacement.
Our first shape, A, is a triangle, and we can calculate the area of this by multiplying the base by the height and then dividing by two. The distance travelled between 𝑡 equals zero and 𝑡 equals one second is one multiplied by five divided by two. This is equal to 2.5 meters. Shape B is a trapezium. And we can calculate this area by adding 𝑎 and 𝑏, dividing by two, and then multiplying by the height, where 𝑎 and 𝑏 are the parallel sides.
The parallel sides of the trapezium have lengths seven and four. And the distance between them or height is five. 11 divided by two is 5.5, and multiplying this by five gives us a distance of 27.5 meters. We can then calculate the total distance by adding 2.5 to 27.5. This is equal to 30. So, the distance covered by the particle in the time interval zero to eight is 30 meters.