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.
The graph consists of a triangle from $t = 0$ to $t = 5$ s, a rectangle from $t = 5$ to $t = 12$ s, and a triangle from $t = 12$ to $t = 15$ s.
Area $= (1/2 \times \text{base} \times \text{height}) + (\text{length} \times \text{breadth}) + (1/2 \times \text{base} \times \text{height})$
Area $= (1/2 \times 5 \times 4) + (7 \times 4) + (1/2 \times 3 \times 4) = 10 + 28 + 6 = 44 \text{ m}$.
$(b)$ Deceleration occurs when the velocity decreases with time. This corresponds to the part of the graph where the slope is negative, which is from $t = 12$ s to $t = 15$ s.
$(c)$ Average velocity $= \text{Total displacement} / \text{Total time} = 44 \text{ m} / 15 \text{ s} = 2.93 \text{ m s}^{-1}$.