The displacement `x` of a body varies with time `t` as `x= -2/3 t^(2)+16 t+2`. The body will come to rest after :

If displacements of a particle varies with time t as s = 1/t^(2) , then.

At t=0 , a body starts from origin with some initial velocity. The displacement x(m) of the body varies with time t(s) as x=-(2//3)t^2+16t+2 . Find the initial velocity of the body and also find how long does the body take to come to rest? What is the acceleration of the body when it comes to rest?

At t=0 , a body starts from origin with some initial velocity. The displacement x(m) of the body varies with time t(s) as x=-(2//3)t^2+16t+2 . Find the initial velocity of the body and also find how long does the body take to come to rest? What is the acceleration of the body when it comes to rest?

The displacement of a body of mass 2 kg varies with time.t. as S =t^(2) + 2t , where S is in metres and .t.is in seconds. The work done by all the foces acting on the body during the time interval t = 2s to t = 4s is

If the displacement of a particle varies with time as x = 5t^2+ 7t . Then find its velocity.

The displacement X (in meters) of a body varies with time (in sec) as x=- (4)/(5)t^(2) + 16+ + 20 . The velocity of body will be zero at time: