InterviewSolution
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InterviewSolution
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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//2))`B. `(-t)/(x^(3))`C. `(1)/(x)-(t^(2))/(x^(3))`D. `(1)/(x)-(1)/(x^(2))` |
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Answer» Correct Answer - C 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))` |
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