Help
Q). A particle is initially at rest, it is subjected to a linear acceleration a, as shown in the figure. The maximum speed attained by the particle is

(A) 605 m/s
(B) 110 m/s
(C) 55 m/s
(D) 550 m/s

From the graph you can see that acceleration is decreasing but it is always positive and hence its direction is same throughout from t=0 to t=11 and velocity will keep on increasing and will be in the direction of acceleration. hence magnitude of velocity can be taken as speed. and maximum speed will be at t=11 sec a=dv/dt dv= adt integrating from t=0 to t=11 change in velocity=v2-v1 area under a-t graph=1/2 x 11 x 10 =55m/s since v1 is zero then v2=55 m/s (B) answer