The position `x` of a particle with respect to time `t` along the x-axis is given by `x=9t^(2)-t^(3)` where `x` is in meter and `t` in second. What will be the position of this particle when it achieves maximum speed along the positive `x` direction

Answer

54

Explanation

Let us first find speed.
Speed, `v = (dx)/(dt) =d/(dt)(9t^(2) -t^(3)) = 18t - 3t^(2)`
For maximum speed, using `(dv)/(dt) = 0`
`(dv)/(dt) = 0 rArr 18 - 6t = 0 rArr t = 3`
Now finding displacement at that time, we have :
`rArr x_("max") = 81 - 27 = 54 m`