0.2 Motion in one dimension  (Page 13/16)

We are asked to calculate the distance and displacement of the car. All we need to remember here is that we can use the area between the velocity vs. time graph and the time axis to determine the distance and displacement.

Break the motion up: 0 – 5 seconds, 5 – 12 seconds, 12 – 14 seconds and 14 – 15 seconds.

For 0 – 5 seconds: The displacement is equal to the area of the triangle on the left:

For 12 – 14 seconds the displacement is equal to the area of the triangle above the time axis on the right:

Now the total distance of the car is the sum of all of these areas:

Now the total displacement of the car is just the sum of all of these areas. HOWEVER, because in the last second (from $t=14$ s to $t=15$ s) the velocity of the car is negative, it means that the car was going in the opposite direction, i.e. back where it came from! So, to find the total displacement, we have to add the first 3 areas (those with positive displacements) and subtract the last one (because it is a displacement in the opposite direction).

The velocity is given by the gradient of a position vs. time graph. During the first 4 seconds, this is

For the last 3 seconds we can see that the displacement stays constant. The graph shows a horisontal line and therefore the gradient is zero. Thus $v=0\mathrm{m}·{\mathrm{s}}^{-1}$ .

The motion of the car can be divided into three time sections: 0 – 2 seconds; 2 – 4 seconds and 4 – 6 seconds. To be able to draw the velocity vs. time graph, the velocity for each time section needs to be calculated. The velocity is equal to the area of the square under the graph:

For 0 – 2 seconds:

The velocity vs. time graph looks like this: