A point moves in a straight line so its displacement `x` metre at time `t` second is given by `x^(2)=1+t^(2)`. Its acceleration in `ms^(-2)` at time `t` second is .

Options

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

Answer

C

Explanation

`x^(2) =1 +t^(2)` or `x=(1+t^(2))^(1//2)`
`(dx)/(dt) =(1)/(2) (1+t^(2)^(-1//2)2tt =t (1+t^(2))^(-1//2)`
`a=(d^(2)x)/(dt^(2)) =(1+t^((2)))^(-1//2)+t(-(1)/(2))(1+t^(2))(1+t^(2))^(-3//2)2t`
`=(1)/(x) -(t^(2))/(x^(3))`.