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 after starting from rest is $t = \frac{ \cosh^{-1}(e^{kd}) }{ \sqrt{kg \sin \theta} }$ where $\theta$ is the anlge of inclination of the plane.

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 after starting from rest is $t = \frac{ \cosh^{-1}(e^{kd}) }{ \sqrt{kg \sin \theta} }$  where $\theta$ is the anlge of inclination of the plane.

ganeshie8 · Answer

$$m\ddot{x} = mg\cos\theta - km {\dot{x}}^2$$

astrophysics · Answer

cos?

ganeshie8 · Answer

$$m\ddot{x} = mg\sin\theta - km {\dot{x}}^2$$

astrophysics · Answer

Too easy for you :p

ganeshie8 · Answer

How are we gonna solve it ?

astrophysics · Answer

Try rewriting it in a different form

ganeshie8 · Answer

https://www.wolframalpha.com/input/?i=solve+x%27%27%28t%29+%3D+g*sin%28theta%29+-+k*%28x%27%28t%29%29^2,+x%280%29%3D0,+x%27%280%29%3D0

ganeshie8 · Answer

lumping the constants, i get

$$\ddot{x} + k\dot{x}^2 = l$$

$k, l$ are constants

astrophysics · Answer

Well I sort of did it a messy way I guess, got a separable, if you write it as $$\frac{ 1 }{ k } \frac{ dv }{ \frac{ g }{ k }\sin \theta -v^2 } = dt$$

astrophysics · Answer

Yeah looking up the integral is probably best I think

ganeshie8 · Answer

Ahh reduction of order !

dan815 · Answer

since we simply have to show it after writing the DE , showing that equation satisfies the ODE is good enough

dan815 · Answer

so rewrite d=f(t) and show that the ODE is satisfied

ganeshie8 · Answer

yeahh that will do
also solving it isn't that hard... only the algebra is a pain..