A) 60 m/s
B) 30 m/s
C) 120 m/s
D) 90 m/s
Correct Answer: B
Solution :
From the given graph, the particle accelerates from Os to 8 s, then from 8 to 12 s, particle moves with constant acceleration. From 12 to 16 s, the particle is in the condition of deceleration. Hence, maximum speed will be during 8 to 12 s. The area of\[a-t\]graph gives change in velocity. The area of the graph from 0 to 8 s \[=v-u=\frac{1}{2}\times 4\times 5+4\times 5=30\] But \[u=0\] \[\therefore \] \[v=30\,m/s\] Hence, the correction option is [b].
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