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A point moves in a straight line so its ...

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 .

A

` (t)/(x^(3)) `

B

` (-t)/(x^(3))`

C

` (1)/(x) - ( t^(2))/(x^(3)) `

D

`(1)/(x) -(1)/(x^(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
` 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) `
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