The position x of a particle w.r.t. time t along x-axis is given by a `x = 9 t^(2)-t^(3)`, where x is in metre and t in sec. What will be the position of this particle when it achieves maximum speed along the +x direction ?

Options

A. 24 m
B. 32 m
C. 54 m
D. 81 m

Answer

C

Explanation

The position x of a particle w.r.t. to time t along X-axis
`x=9t^(2)-t^(3)` ….(i)
Differentiating Eq. (i) w.r.t. time, we get speed, i.e.,
`upsilon = (dx)/(dt)=(d)/(dt)(9t^(2)-t^(3)) rArr upsilon = 18t-3t^(2)` …(ii)
Again differntiating Eq. (ii) w.r.t. time, we get acceleration, i.e.,
`a=(d upsilon)/(dt)=(d)/(dt)(18t-3t^(2)) rArr a=18 -6t` ...(iii)
Now, when speed of particle is maximum, its acceleration is zero.
`a= 0 rArr 18-6t=0 rArr t=3s`
Putting in Eq. (i), we obtain position of particle at that time
`x=9(3)^(2)-(3)^(3)=9(9)-27=81-27=54m`