Question

please answer Q11 please give proper explanation

Answer 1

first body accelerated , and at time , velocity is maximum , then velocity constant and then deaccelerated

now,
for , vmax = u + at
Vmax = 5t

distance travel at t sec
X1= ut + 1/2at²
= 0×t + 1/2(5)t²
X1=(5/2)t²

here acceleration and deceleration both are same . use symmetric concept here .
e.g time taken to covered distance by acceleration = time taken to covered distance by deacceleration .

we also know,
area of velocity - time graph = displacement

so, displacement = 1/2( 25 + 25 -2t )(5t)
where 5t is maximum height ( maximum velocity )

displacement = 1/2(50 -2t)(5t)=(25-t)(5t)

average velocity = displacement/total time = (25 -t)(5t)/25 = (25-t)t/5

20 = (25 -t )t/5

100 = 25t -t²

t² -25t +100 =0

t= ( 25±15)/2 = 5, 20

here you see two time out comes
one t = 5 sec is for acceleration and t = 20 sec is for retardation .

here question ask ,
time for acceleration
so, t = 5 sec