Physical Applications of Calculus Q (help needed pls) (1 Viewer)

littlespartan · Apr 9, 2017

A particle attached to an elastic string is dropped from an origin with a velocity of 4m/s. After it has travelled x metres it has an acceleration given by (2-2x)m/s^2. Find the maximum velocity of the particle in exact form.

InteGrand · Apr 9, 2017

littlespartan said:
A particle attached to an elastic string is dropped from an origin with a velocity of 4m/s. After it has travelled x metres it has an acceleration given by (2-2x)m/s^2. Find the maximum velocity of the particle in exact form.

$\noindent Assume downwards is positive direction. We have $\ddot{x} = 2-2x$. Since $\ddot{x} = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{2}v^{2}\right)$, we get$

$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{2}v^{2}\right) &= 2 - 2x \\ \Rightarrow \frac{1}{2}v^{2} &= 2x -x^{2} + \frac{1}{2}c \quad \text{for some constant }c \\ \Rightarrow v^{2} &= 4x -2x^{2} + c.\end{align*}$

$\noindent Since $v^{2} = 16$ at $x = 0$, we have $c = 16$. So $v^{2} = -2x^{2} + 4x + 16$. Now, velocity is maximised when $\ddot{x} = 0 \iff 2-2x = 0 \iff x = 1$. So the maximum value of $v^{2}$ is $-2 + 4 + 16 = 18$. Hence the maximum velocity is $\sqrt{18} = 3\sqrt{2}$ (in metres per second).$

Physical Applications of Calculus Q (help needed pls) (1 Viewer)

littlespartan

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InteGrand

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