Given the velocity–time graph of a particle moving in a straight line, determine the displacement of the particle within the time interval zero to nine.
Before we begin finding the displacement, let’s have a look at the diagram. We can say that our particle started at a velocity of four meters per second. It moved for one second until it reached a velocity of zero. At two seconds, the velocity was negative four meters per second. And it held at this constant velocity until seven seconds. Its velocity then decreased until it reached zero again at nine seconds. We’re asked to find the displacement of this particle within this time interval. To do this, we can use the fact that, to find the displacement in a velocity–time graph, we find the area under the graph.
So we need to work out the area of the triangle in the section labeled one and the area of the trapezium in the section labeled two. So to find the area of the triangle, we can say that the base is one and the height is four. And using the formula that the area of the triangle is half times the base times the height, then we calculate half times one times four giving us an area of two. To find the area of section two, we use the formula for the area of a trapezium that’s 𝑎 plus 𝑏 over two times ℎ, where 𝑎 and 𝑏 are the basis and ℎ is the height. So our first base will be a length of eight. The lower base is length five. And the height is four.
So our formula is ℎ plus five divided by two times four. And evaluating this will give us an area of 26. It’s important to note that in a velocity–time graph, the area under the 𝑥-axis will have a negative displacement. So the displacement in the first section will be two and the displacement in the second section will be negative 26. So our total displacement will be two plus negative 26 which is negative 24 meters. And since we were only asked for the displacement in the time interval naught to nine, then we don’t need to worry about the displacement in the remainder of the graph.