Simple Kinematics (v-t graph)

Question

A particle moves in a straight line. Its velocity–time graph is shown.

a. After 165 seconds the particle returns to the starting point. Find the value of T.

b. At what time does the particle have maximum speed?

c. Draw the displacement–time graph for this particle.

I'm struggling with A - how do I use the given information to work out the value T?

TIA.

Answer 1

The area below the $t$-axis is $\frac{1}{2} 8 \cdot (165-75)=360$ m.

The area above the $t$-axis is $6T+\frac{1}{2}\cdot 6\cdot (75-T)$ m.

We need $360=6T+3(75-T)$ and we get the correct solution $T=45$ s.