Question

31. A particle starts form rest. Its acceleration (a)
versus time (t) is as shown in the figure. The
maximum speed of the particle will be :
10m/s? 1 B
11s
(a) 110 m/s
(c) 550 m/s
(b) 55 m/s
(d) 660 m/s

what is the solution​

Answer 1

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