Solved Problem on Kinematics
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A body moves with acceleration given by
\[
\begin{gather}
a=\alpha x
\end{gather}
\]
where α is a positive real constant that makes the expression dimensionally consistent. The
initial velocity of the body is equal to v0 at position x0. Determine
the expression for velocity as a function of position.
Solution
The acceleration is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{a=\frac{dv}{dt}}
\end{gather}
\]
multiplying and dividing this expression by dx
\[
\begin{gather}
a=\frac{dv}{dt}\frac{dx}{dx}
\end{gather}
\]
changing the order of terms
\[
\begin{gather}
a=\frac{dv}{dx}\frac{dx}{dt}
\end{gather}
\]
we have to
\[
\begin{gather}
\bbox[#99CCFF,10px]
{v=\frac{dx}{dt}}
\end{gather}
\]
is the definition of velocity, substituting in the expression above
\[
\begin{gather}
a=v\frac{dv}{dx}
\end{gather}
\]
substituting the acceleration given in the problem
\[
\begin{gather}
\alpha x=v\frac{dv}{dx}
\end{gather}
\]
separating the variables and integrating both sides
\[
\begin{gather}
\int_{v_{0}}^{v}{}v\;dv=\int_{x_{0}}^{x}\alpha x\;dx
\end{gather}
\]
the limits of integration for v are v0, the initial velocity, up to v, the
velocity at any instant t, and for x, they are x0, the initial position, up
to x, any position.
Integration of
\( \displaystyle \int_{v_{0}}^{v}{}v\;dv \)
\[
\begin{gather}
\int_{v_{0}}^{v}v\;dv=\left.\frac{v^{2}}{2}\;\right|_{\;v_{0}}^{\;v}=\frac{v^{2}}{2}-\frac{v_{0}^{2}}{2}
\end{gather}
\]
Integration of
\( \displaystyle \int_{x_{0}}^{x}{}\alpha x\;dx \)
\[
\begin{gather}
\alpha \;\int_{x_{0}}^{x}x\;dx=\left.\frac{x^{2}}{2}\;\right|_{\;x_{0}}^{\;x}=\alpha\left(\frac{x^{2}}{2}-\frac{x_{0}^{2}}{2}\right)
\end{gather}
\]
\[
\begin{gather}
\frac{v^{2}}{2}-\frac{v_{0}^{2}}{2}=\alpha\left(\frac{x^{2}}{2}-\frac{x_{0}^{2}}{2}\right)\\[5pt]
\frac{v^{2}}{2}-\frac{v_{0}^{2}}{2}=\frac{\alpha}{2}\left(x^{2}-x_{0}^{2}\right)\\[5pt]
v^{2}-v_{0}^{2}=\alpha\left(x^{2}-x_{0}^{2}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v^{2}=v_{0}^{2}+\alpha \left(x^{2}-x_{0}^{2}\right)}
\end{gather}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .