Solved Problem on Kinematics
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A body moves with acceleration given by
\[
a=\alpha -\beta v
\]
α and β are real positive constants that make the expression dimensionally consistent. Determine
the expressions for the velocity and displacement as a function of time.
Solution
The instantaneous acceleration is given by
\[ \bbox[#99CCFF,10px]
{a=\frac{dv}{dt}}
\]
substituting this value in the given expression
\[
\frac{dv}{dt}=\alpha -\beta v
\]
separating the variables and integrating both sides
\[
\int \frac{dv}{\alpha -\beta v}=\int dt
\]
The limits of integration are v0, initial speed, and v(t), a speed at an
instant t for dv, and 0, initial time, and t, for dt
\[
\int_{v_{0}}^{v(t)}\frac{dv}{\alpha -\beta v}=\int_{0}^{t}dt
\]
Integration of \(\displaystyle \int_{v_{0}}^{{v(t)}}\frac{dv}{\alpha -\beta v} \)
Changing the variable
for v = v0, we have \( u=\alpha -\beta v_{0} \)
for v = v(t), we have \( u=\alpha -\beta v(t) \)
substituting in the integral
Changing the variable
\[
\begin{array}{l}
u=\alpha -\beta v\\[10pt]
\dfrac{du}{dv}=-\beta v\Rightarrow dv=-{\dfrac{1}{\beta }}du
\end{array}
\]
changing the limits of integration
for v = v0, we have \( u=\alpha -\beta v_{0} \)
for v = v(t), we have \( u=\alpha -\beta v(t) \)
substituting in the integral
\[
\begin{align}
\int_{{\alpha -\beta v_{0}}}^{{\alpha -\beta v(t)}}\frac{1}{u}\left(-{\frac{1}{\beta}}\;du\right)&=-{\frac{1}{\beta }}\int _{{\alpha -\beta v_{0}}}^{{\alpha -\beta v(t)}}\frac{du}{u}=-{\frac{1}{\beta}}\;\left.\ln u\;\right|_{\;\alpha -\beta v_{0}}^{\;\alpha -\beta v(t)}=\\[5pt]
&=-\frac{1}{\beta}\;\left[\ln (\alpha -\beta v(t))-\ln (\alpha -\beta v_{0})\right]=\\[5pt]
&=-{\frac{1}{\beta }}\;\ln \left(\frac{\alpha -\beta v(t)}{\alpha -\beta v_{0}}\right)
\end{align}
\]
Integration of \( \displaystyle \int_{0}^{t}dt \)
\[
\int_{0}^{t}dt=\left.t\;\right|_{\;0}^{\;t}=(t-0)=t
\]
\[
\begin{gather}
-{\frac{1}{\beta }}\;\ln \left(\frac{\alpha -\beta v(t)}{\alpha -\beta v_{0}}\right)=t\\[5pt]
\ln \left(\frac{\alpha -\beta v(t)}{\alpha -\beta v_{0}}\right)=-\beta t\\[5pt]
\frac{\alpha -\beta v(t)}{\alpha -\beta v_{0}}=\operatorname{e}^{-\beta t}\\[5pt]
\alpha -\beta v(t)=\operatorname{e}^{-\beta t}(\alpha -\beta v_{0})\\[5pt]
\beta v(t)=\alpha -\operatorname{e}^{-\beta t}(\alpha -\beta v_{0})
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{v(t)=\frac{\alpha }{\beta}-\frac{\operatorname{e}^{-\beta t}}{\beta}(\alpha -\beta v_{0})}
\]
The instantaneous speed is given by
\[ \bbox[#99CCFF,10px]
{v=\frac{dx}{dt}}
\]
substituting in the speed expression above
\[
\frac{dx}{dt}=\frac{\alpha }{\beta}-\frac{\operatorname{e}^{-\beta t}}{\beta}(\alpha -\beta v_{0})
\]
we integrate this expression in dt on both sides
\[
\begin{gather}
\int \frac{dx}{dt}dt=\int \left[\frac{\alpha }{\beta}-\frac{\operatorname{e}^{-\beta t}}{\beta}(\alpha -\beta v_{0})\right] dt\\
\int \frac{dx}{dt}dt=\int\left[\frac{\alpha }{\beta}-\frac{\alpha -\beta v_{0}}{\beta}\operatorname{e}^{-\beta t}\right]dt
\end{gather}
\]
in the integral on the left-hand side
\( \dfrac{dx}{dt}dt=dx \),
and on the right-hand side of the equation the integral of the difference is the difference of the integral
\[
\int dx=\int \frac{\alpha }{\beta}\;dt-\int\frac{\alpha -\beta v_{0}}{\beta}\operatorname{e}^{-\beta t}dt
\]
as
\( \dfrac{\alpha }{\beta} \)
and
\( \dfrac{\alpha -\beta v_{0}}{\beta} \)
are constants, they are carried outside of the integral sign.The limits of integration are x0, initial position, and x(t), a position at an instant t for dx, and 0, initial time, and t, for dt
\[
\int_{x_{0}}^{{x(t)}}dx=\frac{\alpha }{\beta}\int_{0}^{{t}}dt-\frac{\alpha -\beta v_{0}}{\beta}\int_{0}^{{t}}\operatorname{e}^{-\beta t}dt
\]
Integration of \( \displaystyle \int_{x_{0}}^{x(t)}dx \)
\[
\int_{{x_{0}}}^{{x(t)}}dx=\left.x\;\right|_{\;x_{0}}^{\;x(t)}=x(t)-x_{0}
\]
The integral in dt has already been calculated above.
Integration of \( \displaystyle \int_{0}^{t}\operatorname{e}^{-\beta t}dt \)
Changing the variable
for t = 0, we have \( u=0 \)
for t = t, we have \( u=-\beta t \)
substituting in the integral
Changing the variable
\[
\begin{array}{l}
u=-\beta t\\[10pt]
\dfrac{du}{dt}=-\beta \Rightarrow dt=-{\dfrac{1}{\beta}}du
\end{array}
\]
changing the limits of integration
for t = 0, we have \( u=0 \)
for t = t, we have \( u=-\beta t \)
substituting in the integral
\[
\begin{align}
\int_{0}^{{-\beta t}}\operatorname{e}^{u}\left(-{\frac{1}{\beta}}\right)du&=-{\frac{1}{\beta}}\int_{0}^{{\beta t}}\operatorname{e}^{u}du=-{\frac{1}{\beta}}\;\left.\operatorname{e}^{u}\;\right|_{\;0}^{\;-\beta t}=\\[5pt]
&=-{\frac{1}{\beta}}\left(\operatorname{e}^{-\beta t}-\operatorname{e}^{0}\right)=-{\frac{1}{\beta}}\left(\operatorname{e}^{-\beta t}-1\right)
\end{align}
\]
\[
x(t)-x_{0}=\frac{\alpha }{\beta}t-\frac{\alpha -\beta v_{0}}{\beta}\left[-{\frac{1}{\beta}}\left(\operatorname{e}^{-\beta t}-1\right)\right]
\]
\[ \bbox[#FFCCCC,10px]
{x(t)=x_{0}+\frac{\alpha }{\beta}t+\frac{\alpha -\beta v_{0}}{\beta^{2}}\left(\operatorname{e}^{-\beta t}-1\right)}
\]
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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 .