Okay. You understand that, as a vector quantity, displacement is a number value representing distance combined with an angle, measured from a certain point. This problem takes away the need for an angle, because the car can only move in two directions - forward and backward. This makes it easy. We call forward positive and backward negative.
So, if I move forward 1m from point X, my displacement is 1m.
If I move backward 1m from point X, my displacement is -1m.
Knowing this, and knowing that we have a velocity-time graph, we need to look at how velocity relates to displacement. Velocity is also a vector, and so - in this case - must either be positive or negative. You can think of it as speed with a direction. Again forward is positive and backward is negative. So. Given that velocity = distance / time:
If I move 1m forward and it takes me 1 second, my velocity is 1 metre per second (m/s)
If I move 1m backward and it takes me 1 second, my velocity is -1m/s.
If I move for 2 seconds at 1m/s, my displacement is 2m.
If I move for 2 seconds at -1m/s, my displacement is -2m.
Knowing all this, we can look at the graph. Clearly you know that the area under the graph gives displacement. If you don't know why:
We're multiplying the x-axis by the y-axis.
The x-axis is time and the y-axis is velocity.
Therefore, x = t, and y = d / t.
x*y = (d / t) * t = d.
d = displacement.
So looking at the graph we see that there are two distinct sections. One where velocity is negative (backwards) and one where velocity is positive (forwards). This translates into the car reversing from Point A and then moving forwards towards point A and continuing.
Clearly, because (v / t) * t = d, if we take a negative velocity and multiply it by time we'll get a negative displacement. Accordingly we will get a positive displacement from a positive velocity.