The velocity-time graph of a truck is plotted below.
$(a)$ Calculate the magnitude of displacement of the truck in $15$ seconds.
$(b)$ During which part of the journey was the truck decelerating?
$(c)$ Calculate the magnitude of average velocity of the truck.

(N/A) Displacement is equal to the area under the velocity-time graph.
Displacement $= \text{Area of triangle} + \text{Area of rectangle} + \text{Area of triangle}$
$= (\frac{1}{2} \times 5 \times 4) + ((12 - 5) \times 4) + (\frac{1}{2} \times (15 - 12) \times 4)$
$= 10 + 28 + 6 = 44 \text{ m}$
$(b)$ The truck is decelerating when the velocity decreases with time,which occurs between $12 \text{ s}$ and $15 \text{ s}$.
$(c)$ Average velocity $= \frac{\text{Total displacement}}{\text{Total time}}$
$= \frac{44 \text{ m}}{15 \text{ s}} \approx 2.93 \text{ m s}^{-1}$