Explanation

(b) From the give graph. From 0s to 8s, particle is accelerated, then from 8 to 12 s, particle moves with constant acceleration. The form 12 to 16s, the particle is in the condition of deceleration. Hence, maximum velocity will be during 8s to 12s. During 0 to 4s, the acceleration will be function of time. The equation of straight lines is
$a = \frac{5}{4}t$
$\therefore\ \ \frac{dv}{dt} = \frac{5}{4}t$
$\therefore\ v = \int_{0}^{t}{\frac{5}{4}t\ dt} = \frac{5}{4}\frac{t^{2}}{2} = \frac{5}{8}t^{2}$
The velocity at $t = 4$s is $u = 10ms^{- 1}$
The distance travelled during 4 to 8s is
$s_{2} = ut + \frac{1}{2}at^{2}$
$= 10 \times 4 + \frac{1}{2} \times 5 \times 4^{2}$
$= 40 + 40 = 80$m
The velocity at $t = 8s$ is
$v = 10 + 5 \times 4 = 30ms^{- 1}$
This is the maximum velocity.
Tricky approach: The area of $a - t$ graph gives change in velocity. The area of the graph from 0 to 8s
$= v - u = \frac{1}{2} \times 4 \times 5 + 4 \times 5 = 3$0
But $u = 0$
$v = 30{ms}^{- 1}$