The graph below is meant to represent an a-t graph for the following situation: A point moves away from origin at a slow constant velocity for 5s Then it moves away at a medium-fast constant velocity for 5s Then it stands still for 5s Then it moves toward the origin at a slow constant velocity for 5s Finally it stands still for 5s
My problem is that I am getting the last two bars as being of equal length. This is because in the third to last stage, it is simply standing still. Therefore, the acceleration it takes to move toward the origin at a slow constant velocity must be equal and opposite to the one it requires to go back to rest. This just seems to be common sense to me although it can be proved as follows.
When it is at rest, v=0. Let the acceleration required to move again be -a (-ve since towards origin), so because this is instantaneous (since the velocity is constant after), the constant velocity we get at the last stretch is 0-a, or just -a. Finally, to go back from this velocity to 0, we need an acceleration of +a. So the two bars at the end should be equal, one is -a the other is +a.
I obtain this time-graph for acceleration and velocity: