Kinematics of motion in a straight line — practice question

The diagram presents the velocity-time graph for a particle $P$ moving along the straight line $BAC$. It leaves $A$ and reaches $B$ in $5\,\text{s}$. It then reverses direction and goes from $B$ to $C$ in $10\,\text{s}$. During the first $3\,\text{s}$ of $P$’s motion, the acceleration is constant. For the next $12\,\text{s}$, the velocity of $P$ at time $t\,\text{s}$ after leaving $A$ is $v\,\text{m s}^{-1}$, where $v = -0.2t^2 + 4t - 15$ for $3 \le t \le 15$.