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

Answer

C

Explanation

` 2x (dx)/(dt) = 2t rArr xv =t ` differential w.r.t. ` rArr v^(2) +xa =1 rArr a= (1-v^(2))/(x) `
`rArr a= ( 1-(t//x)^(2))/(x) =(x^(2) -t^(2))/(x^(3)) ." " `(since `x^(2) -t^(2) =1` Ans . could be `1//x^(2) too) `