Find time elapsed, given Force as a function f(velocity,displacment)

[Saracen Rue]

Homework Statement

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A particle of mass $2 kg$ is initially traveling with a constant velocity of $\frac{\pi }{2}\ ms^{-1}$. The particle is acted upon by a resistance force, $F=-2v\tan \left(v\right)e^{2x}$, where $v$ is the velocity in $ms^{-1}$ and $x$ is the displacement of the particle in meters at any time $t$. Let $t$ be the time elapsed, in seconds, after the force is applied. Find the time taken for the particle to travel a distance of $1.55 m$, expressing your answer in minutes correct to $2$ decimal places.

Homework Equations

Using calculus to convert between displacement, velocity and acceleration

$v=\frac{dx}{dt}$

$a=\frac{dv}{dt}=\frac{d^2x}{dt^2}=v\cdot \frac{dv}{dx}=\frac{d\left(\frac{1}{2}v^2\right)}{dx}$

The Attempt at a Solution

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$F=-2v\tan \left(v\right)e^{2x}$

$a=-v\tan \left(v\right)e^{2x}$

$v\cdot \frac{dv}{dx}=-v\tan \left(v\right)e^{2x}$

$\frac{v}{-v\tan \left(v\right)}dv=e^{2x}dx$

$\int _{ }^{ }-\cot \left(v\right)dv=\int _{ }^{ }e^{2x}dx$

$-\ln \left(\sin \left(v\right)\right)=\frac{1}{2}e^{2x}+c$

"initially traveling with a constant velocity of $\frac{\pi }{2}\ ms^{-1}$"
Therefore at $t=0$, $v=\frac{\pi }{2}$. The question also states that $x$ is the displacement after the force has been applied, therefore at $t=0$, $x=0$.

$-\ln \left(\sin \left(v\right)\right)=\frac{1}{2}e^{2x}+c$

$-\ln \left(\sin \left(\frac{\pi }{2}\right)\right)=\frac{1}{2}e^{2\left(0\right)}+c$

$-\ln \left(1\right)=\frac{1}{2}+c$

Therefore, $c=-\frac{1}{2}$

$-\ln \left(\sin \left(v\right)\right)=\frac{1}{2}e^{2x}-\frac{1}{2}$

$\sin \left(v\right)=e^{-\frac{1}{2}\left(e^{2x}-1\right)}$

$v=\arcsin \left(e^{-\frac{1}{2}\left(e^{2x}-1\right)}\right)$

$\frac{dx}{dt}=\arcsin \left(e^{-\frac{1}{2}\left(e^{2x}-1\right)}\right)$

$\frac{dt}{dx}=\frac{1}{\arcsin \left(e^{-\frac{1}{2}\left(e^{2x}-1\right)}\right)}$

$t=\int _{ }^{ }\frac{1}{\arcsin \left(e^{-\frac{1}{2}\left(e^{2x}-1\right)}\right)}dx$

This is where I get stuck. There's no way to get the integral of $\frac{1}{\arcsin \left(e^{-\frac{1}{2}\left(e^{2x}-1\right)}\right)}$ that I know of. I tried putting it into Wolfram Alpha but it said no result found in terms of standard mathematical functions. Have I done some of the intermediate steps wrong to arrive at this function which can't be integrated? Or is there another way of doing this? Any help is much appreciated.

 

[MarkFL]

Your solution looks good to me...I would suggest using W|A to approximate the definite integral:

$\displaystyle t=\int_0^{1.55}\frac{1}{\arcsin\left(e^{-\frac{1}{2}\left(e^{2x}-1\right)}\right)}\,dx$

 

[Saracen Rue]

Your solution looks good to me...I would suggest using W|A to approximate the definite integral:

$\displaystyle t=\int_0^{1.55}\frac{1}{\arcsin\left(e^{-\frac{1}{2}\left(e^{2x}-1\right)}\right)}\,dx$

Sorry for late reply, I ended up realising to use a definite integral instead of an indefinite one on my own and forgot about this thread. I'm not sure what I was thinking trying to find the indefinite integral first - I guess I just needed a couple hour break before seeing how easy the final step actual was haha.

Anyway, thank you for your help regardless :)