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

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^3-3t^2+3t+12 . The velocity when the acceleration is zero 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 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

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

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

A particle moves so that the distance moved is according to the law s(t)=(t^(3))/(3)-t^(2)+3 . At what time the velocity and acceleration are zero respectively ?

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