InterviewSolution
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InterviewSolution
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A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to `v(x) = beta x^(-2 n)` where `beta` and `n` are constant and `x` is the position of the particle. The acceleration of the particle as a function of `x` is given by.A. `- 2 n beta^2 x^(-2 n - 1)`B. `- 2n beta^2 x ^(- 4n - 1)`C. `-2 n beta^2 x^(-2 n + 1)`D. `-2 b beta^2 e^(-4 n + 1)` |
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Answer» Correct Answer - B We are given velocity of the particle `v(x) = beta x^(- 2n)` We know acceleration `a = v (dv)/(dx)` `a = beta x^(- 2n) (d)/(dx) (beta x^(- 2n))` =`beta^2 x^(- 2 n) (- 2 n)x^(- 2 n - 1) = -2 n beta^2 x^(-2 n - 1 - 2n)` `a = - 2n beta^2 x^(-4 n - 1)`. |
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