Given `t=2+root(3)4+root(3)2,` determine the value of `t^3-6t^2+6t-2`.

The distance in meters described by a car in time t seconds is given by s = 4t^(3) - 3t^(2) + 6t -1 . Determine the velocity and acceleration when (i) t = 0, (ii) t =2 seconds.

If s=(t^(3))/(3)-(1)/(2)t^(2)-6t+5 , Find the value of (d^(2)s)/(dt^(2)) at t=1

If s=(t^(3))/(3)-(1)/(2)t^(2)-6t+5 , Find the value of (d^(2)s)/(dt^(2)) at t=1

(root(6)3 sqrt2 + (1)/(root (3) 3))^(n) , (t _(7) "fron the " 1^(st))/( t _(7) "from the last")=1/6, then the value of n is

Distance covered by a particle moving in a straight line from origin in time t is given by x = t- 6t^(2)+t^(3) . For what value of t, will the particle's acceleration be zero?

if position vector is vecr =t^3−6t^2+3t+4m find the value of velocity and acceleration.

The instantaneous angular position of a point on a rotating wheel is given by the equation theta (t) = 2t^3 - 6t^2 . The torque on the wheel becomes zero at : .......... .

If s=2t^(3)-6t^(2)+a t+5 is the distance travelled by a particle at time t and if the velocity is -3 when its acceleration is zero, then the value of a is

If x = (t^(3))/(3) - (5)/(2)t^(2) + 6t + 1 , then the value (d^(2)x)/(d t^(2)) , when (dx)/(d t) is zero is :