Physics question displacement

Hi, please could I have help on this question? I don’t get how it’s any of the answers becauase the total displacement would be the two areas subtracted since they are in opposite directions so surely the total displacement would just be 0?
Here is the question: https://app.gemoo.com/share/image-annotation/583292309293592576?codeId=DW2Ne9l2ZA6zW&origin=imageurlgenerator
Thank you!

You are mixing 2 "different" concepts - instantaneous and average quantity.

Indeed, total displacement for a time interval of T is zero but the instantaneous displacement from the origin need not be zero.

This is in fact a SHM.

Thank you for your reply, please could you explain the difference between instantaneous and average quantity. We have only covered displacement as the difference between the two areas of the graphs?

Its probably easier to think of it as time sliding from 0 to T, and at a point in time t, the dispalcement is the area under the velocity curve up to that point in time. Pick an arbitrary time and shade the relevant area to the left of that value.

So at time t=T, the displacment is zero, but t=T/2 would correspond to the maximum positive displacement as all the area to the left is positive and the velocity is zero at this point in time and becomes negative. Thats enough to deterime the answer, but at t=T/4 the velocity is largest so the displacement is increasing the fastest, and similarly at t=3T/4 the velocity is minimum (negative) so the displacement is decreasing the fastest etc ...

Alternatively, get your ruler out and run it over each time-displacement curve. So your ruler forms the local tangent to the curve at each point in time and the gradient of your ruler is the corresponding velocity. Which would match the given velocity curve? B and C could be discounted using t=0 as the initial velocity would be non-zero. D goes positve-negative-positive-negative whereas the actual velocity is simply positive-negative.