Calculus Based Questions: Practice Problem & Solution
A particle of unit mass undergoes one dimensional motion such that its velocity varies according to \(v(x) = \beta x^{-2n}\) where \(\beta\) and (n) are constants and (x) is the position of the particle. The acceleration of the particle as a function of (x), is given by: (2015)
Solution Explained:
To solve this problem, we apply the core principles of Calculus Based Questions. Understanding the underlying formula is key to arriving at the correct answer below:
Given \(v = \beta x^{-2n}\). Acceleration \(a = v \frac{dv}{dx}\). First find \(\frac{dv}{dx} = \beta (-2n)x^{-2n-1}\). Then \(a = (\beta x^{-2n})(-2n\beta x^{-2n-1}\) = \(-2n\beta^2 x^{-4n-1}\).
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