The position x of a particle varies with time t as `x = at^(2) -bt^(3)`. For what value of time acceleration is zero?

Options

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

Answer

a

Explanation

`x=at^(2)-bt^(3)`.Differentiating `(dx)/(dt)=2at-3bt^(2)`
Again diff.`(d^(2)x)/(dt^2)=a=2a-6bt`
Now a=0 if 2a -6bt=0
or 2a=6bt or `t=(2a)/(6b)=(a)/(3b)`