`t^3+t^2+t+6=0`

g(t)={(max(t^3-6t^2+9t-3,0), , t in[0,3]),(4-t,,tin(3,4)):} , find the point of non-differentiability

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

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

x=f(t) satisfies (d^2x)/dt^2=2t+3 and for t=0, x=0, dx/dt=0 , then f(t) is given by (A) t^3+t^2/2+t (B) (2t^3)/3+(3t^2)/2+t (C) t^3/3+(3t^2)/2 (D) none of these

x=f(t) satisfies (d^2x)/dt^2=2t+3 and for t=0, x=0, dx/dt=0 , then f(t) is given by (A) t^3+t^2/2+t (B) (2t^3)/3+(3t^2)/2+t (C) t^3/3+(3t^2)/2 (D) none of these

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-

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

Let a function g : [0,4] to R be defined as g(t)={(max(t^3-6t^2+9t-3,0), , t in[0,3]),(4-t,,tin(3,4)):} then the number of points in the interval (0,4) where g(x) is NOT differentiable, is _____________.

The value of p(t) = 2 + t + 2t^2 -t^3 for p(0) is :

The positon of an object moving along x-axis is given by x(t) = (4.2 t ^(2) + 2.6) m, then find the velocity of particle at t =0s and t =3s, then find the average velocity of particle at t =0 s to t =3s.