If a particle moves according to the law `s=6t^(2)-(t^(3))/(2)`, then the time at which it is momentarily at rest is

A particle moving according to the law s=6t-1/2t^3 . At what time its velocity vanishes ?

A particle moves according to the law s=t^3-6t^2+9t+ 15 . Find the velocity when t=0 .

A particle moves according to law s=t^(3)-6t^(2)+9t+15 . Find the velocity when t = 0.

A particle moves along a line according to the law s=t^4-5t^2+8 . The intial velocity is

If a particle moves alonga straight line by S = 4t^(2) - 8t + 3 and the time at which the particle comes of rest is 't' seconds then '2t' is

A particle moves according to the law x=t^(3)-6t^(2)+9t+5 . The displacement of the particle at the time when its acceleration is zero, is

A particle moves according to the law s=t^(3)-9t^(2)+24t . The distance covered by the particle before it first comes to rest is -

A particle moves along x-axis according to the law x=(t^(3)-3t^(2)-9t+5)m . Then :-

A particle moves according to the law s= t^3-6t^2+9t +5 . The displacement of the particle at the time when its acceleration is zero is........ A) 9 B) -7 C) 7 D) 0

A particle moves according to the law x = rcos (pi)t/(2) The distance covered by it in the time interval between t=0 to t=3s is