Newtonian Mechanics - Particle in Motion with Air Resistance

[derravaragh]

Homework Statement

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 = [arccosh(e^(kd))]/√(kgsin(θ)

where θ is the angle of inclination of the plane.

Homework Equations

F = ma
F_g = mgsin(θ)
Resistance = -kmv^2
Motion of particle => ma = mgsin(θ) - kmv^2

The Attempt at a Solution

My attempt was to set up the equation for the motion (ma = mgsin(θ) - kmv^2) and use differential equations to solve. After dividing by mass, I had:
dv/dt = gsin(θ) - kv^2
which I then divided by k, and substituted (g/k)sinθ for C^2 giving

dv/kdt = C^2 - v^2

After collecting terms and integrating I came to

t = arctan(v/√((g/k)sinθ))/√(kgsinθ)

I thought I was on the right track as I have the numerator correct, but I do not know where to go from here, or if this is even correct so far. Any help would be much appreciated. Also, I know this question has been asked before, but the answer given didn't make sense to me so it doesn't help.

 

[TSny]

Try writing the acceleration as $a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v \frac{dv}{dx}$ where $x$ is distance along plane. Separate variables and integrate to find $v$ as a function of $x$. Then, writing $v = \frac{dx}{dt}$ you can separate variables again and integrate to find $t$ as a function of $x$.

[EDIT: Actually, your way works too. However, I think there is a typo in your final expression for t. Did you mean to write arctanh rather than arctan? If you solve your (corrected) expression for $v$ and then let $v = \frac{dx}{dt}$ you can separate variables and integrate to find $t$ as a function of $x$]

 

[derravaragh]

Yes, it was supposed to be arctanh (and also denominator was correct not numerator). I just finished this, thank you very much for your assistance.