- #1

Saracen Rue

- 150

- 10

## 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}##

*##x##*

*and**##t##*

*is the displacement of the particle in meters at any time**##t##*

*. Let**##1.55 m##*

*be the time elapsed, in seconds, after the force is applied. Find the time taken for the particle to travel a distance of**##2##*

*, expressing your answer in minutes correct to*

*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.