The position x of a particle varies with time t, as `x = at^(2)-b t^(3)`. The acceleration of the particle will be zero at time t equals to

Options

A. `(2a)/(3b)`
B. `(a)/(b)`
C. `(a)/(3b)`
D. zero

Answer

c

Explanation

`dx//dt = 2at -3bt^(2),d^(2)x//dt^(2) = 2a-6bt`
`t = 2a//6b = a//3b`