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{\bf Question}

A particle of mass $m$ slides down an inclined plane under the
influence of gravity.  If the motion is resisted by a force
$f=kmv^2$, show that the time required to move a distance $d$
starting from rest is $$ t = \frac{\cosh^{-1}(e^{kd})}{\sqrt{kg
\sin \theta}},$$ where $\theta$ is the inclination of the plane to
the horizontal.


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{\bf Answer}

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\put(2.25,.8){\makebox(0,0){$m$}}

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Use the {\bf i} and {\bf j} aligned with the slope and the
reaction.

Using Newton's 2nd law gives:
\begin{eqnarray*} m\ddot x{\bf i}  & = & -kmv^2 {\bf
i} + R {\bf j} \\ & & + mg ( {\bf i} \sin \theta - {\bf j} \cos
\theta)
\\ m \ddot x & = & -k m \dot x^2 + mg \sin \theta \\ \ddot x & = &
-k \dot x^2 + g \sin \theta \\ {\rm put\ \ }  \dot x = v
\hspace{.3in} v \frac{dv}{dx} & = & -kv^2 + s \sin \theta \\ \int
\frac{v \, dv}{g \sin \theta - kv^2} & = & x + A \\ -\frac{1}{2k}
\ln|g \sin \theta - kv^2| & = & x + A
\end{eqnarray*}

Substituting the initial conditions of $v = 0$ when $x = 0$ to
find $A$:

\begin{eqnarray*} A & = & -\frac{1}{2k} \ln |g\sin \theta| \\
\Rightarrow \hspace{.2in} -\frac{1}{2k} \ln|g \sin \theta - k v^2|
& = & x - \frac{1}{2k} \ln |g \sin \theta| \\ \ln|g\sin \theta| &
= & -2kx + \ln|g \sin \theta| \\ g \sin \theta - kv^2 & = &
e^{-2kx+ ln|g\sin \theta|} \\ -kv^2 & = & g \sin \theta
(e^{-2kx}-1) \\ v^2 & = & \frac{g}{k} \sin \theta (1 - e^{-2kx})
\\ \Rightarrow v & = & \sqrt{\frac{g\sin \theta}{k}}\sqrt{(1 -
e^{-2kx})} \\ {\rm substituting\  back\  for} \hspace{.2in} \dot x
= v \\ \Rightarrow \int \frac{dx}{\sqrt{1 - e^{-2kx}}} & = &
t\sqrt{\frac{g \sin \theta}{k}} + const. \\ {\rm substitute\ } u =
e^{kx} &  & \Rightarrow du = ku \, dx \\ {\rm integral\ becomes\ }
\int \frac{k^{-1}u^{-1}}{\sqrt{1 - u^{-2}}} & = & \frac{1}{k} \int
\frac{du}{\sqrt{u^2 - 1}} = \frac{1}{k} \cosh{-1}u \\ {\rm Hence\
} \frac{1}{k} \cosh^{-1} (e^{kx}) & = & t \sqrt{\frac{g \sin
\theta}{k}} + const \\ {\rm Hence\  on\ rearranging\ this\ result}
\\ t & = & \frac{\cosh^{-1}(e^{kd})}{\sqrt{kg \sin \theta}}
\end{eqnarray*}





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