InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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 nb^(2) x^(-4n -1)`B. `- 2 nb^(2) x^(-2n +1)`C. `- 2 nb^(2) x^(-4n + 1)`D. `- 2 nb^(2) x^(-2n - 1)` |
|
Answer» `v = bx^(-2xn)` `(dv)/(dx) = (b) (-2n) x^(-2n - 1)` `a = v (dv)/(dx) = bx^(-2n) (b) (-2n) x^(-2n - 1)` `= - 2nb^(2) x^(-4n - 1)` |
|