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

Answer

B

Explanation

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