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

Options

A. 54m
B. 81m
C. 24m
D. 22m

Answer

a

Explanation

Given, x is the position with respect to time t
`x=9t^(2)-t^(3)`
`v-(dx)/(dt)=18t-3t^(2)`
`v=18t-3t^(2)`
`(dv)/(dt)=18-6t`
For maxima or minima, `(dv)/(dt)=0`, the time
`18-6t=0`
t=3
The distance covered in time t=3 seconds
`x=9(3)^(2)-(3)^(2)=54m`