The displacement 'x' and time of travel 't' for a particle moving an a straight line are related as `t = sqrt((x+1)(x-1))`. Its acceleration at a time `t` is

Options

A. `(1)/(x)-(1)/(x^(2))`
B. `(1)/(x^(3))`
C. `(-t^(2))/(x^(3))`
D. `(-t)/(x^(2))`

Answer

B

Explanation

`t_(2) =x_(2) -1`
`x_(2) =1 + t_(2)`
`x(dx)/(dt) =t`
`rArr ((dx)/(dt))^(2)+x(d^(2)x)/(dt)=1`
`rArr (xd^(2)x)/(dt^(2))=1-((dx)/(dt))^(2)=1-((t)/(x))^(2)`
`=(x^(2)-t^(2))/(x^(2))=(1)/(x^(2))rArr a=(d^(2)x)/(dt^(2))=(1)/(x^(3))`.