If `s=2t^(3)+3t^(2)+2t+8` then find time at which acceleration is zero.

The displacement s of a particle at time t is is given by s=2 t^3 - 4 t^2 -5t. Then find the time when the acceleration is 14ft/ sec^2

If S=2+3t-t^(2)+t^(3) , then velocity when acceleration is zero is

The distance moved by the particle in time t is given by s=t^3-12t^2+6t+8 . At the instant when its acceleration is zero, the velocity is

if position vector is vecr=t^3−6t^2+3t+4m find the time when acceleration is zero.

A particle moves along a line according to the law s(t)=2t^(3)-9t^(2)+12t-4 , where tge0 . Find the particle's acceleration each time the velocity is zero.

The displacement of a particle at time t is given by s= 2 t^3 - 5 t^2 + 4t -3. Find the time when the acceleration is 14 ft/ sec^2 , find the velocity and the displacement at that time.

If the distance from the initial point at a time t is given by x = t - 6t^2 + t^3 : then at what time the acceleration will be zero?