The position of a body with respect to time is given by `x = 4t^(3) – 6t^(2) + 20 t + 12` . Acceleration at t = 0 will be-

Completion type Question : The position of a body with respect to time is given by x = 2t^(3) - 6t^(2) + 92t + 6 . If x is in metres and t in seconds, then acceleration at t = 0 is

The position of a body wrt time is given by x = 3t^(3) -6t^(2) + 12t + 6 . If t = 0, the acceleration is___________

The position of a point in time 't' is given by x = 1 + 2t + 3t^(2) and y = 2 - 3t + 4t^(2) . Then its acceleration at time 't' is

The position of a point in time t is given by x=a+bt-ct^(2),y=at+bt^(2) . Its acceleration at time t is

The position of a point in time t is given by x=a+bt-ct^(2),y=at+bt^(2) . Its acceleration at time t is

The displacement of a body is given by x = 2t^(3) - 6t^(2) + 12t + 6 . The acceleration of body is zero at time ti is equal to :

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?