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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

`(1)/(x^3)`

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
a,c
`x^2=t^2+1`
2x(dx/dt)=2t
`impliesv=(dx)/(dt)=(t)/(x)`
`impliesa=(dv)/(dt)=(1)/(x)-(t^(2))/(x^(3))=(1)/(x)(1-(t^(2))/(x^(2)))=1/x^3`
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