Question

An object starting from rest move on a straight road for time 't' and comes to rest finally. The distance is converted in two parts. In the first part it is accelerated at constant acceleration 'A' and then decelerate at rate 'B'. Then maximum velocity is
(1)At

(2)Bt

(3) ($\frac{A+B}{2}$)(t)

(4) ($\frac{AB}{A+B}$)(t)

Answer 1

Answer:

Maximum velocity of the object will be At m/sec

So option (a) is correct option  

Explanation:

We have that firstly the object starts from rest so its initial velocity will u = 0 m /sec

Its acceleration is $Am/sec^2$ and deceleration is $Bm/sec^2$'

As it is given that from rest its first accelerate and then decelerate so just before deceleration object will be at maximum speed

From first equation of motion we know that v = u+at

So $v=0+At=Atm/sec$

So maximum velocity of the object will be At m/sec

So option (a) is correct option