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

Answer

c

Explanation

Here `x=at^(2)-bt^(3)`
`v=(dx)/(dt)=2at-3bt^(2)`
Acceleration =`f=(dv)/(dt)=2a-6bt`
Now when f=0
2a-6bt=0
`2a=6bt` or `t=(a)/(3b)`