A point moves in a straight line so its displacement `x` meter 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//2))`
B. `(-t)/(x^(3))`
C. `(1)/(x)-(t^(2))/(x^(3))`
D. `(1)/(x)-(1)/(x^(2))`

Answer

C

Explanation

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