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A car is standing still (\(x(0) = 0, v(0) = 0\)). It accelerates at a rate of

 

\(a(t) = v'(t) = \frac{dv}{dt} = k_1 - k_2*v^2\)

 

Given \(k_1\) and \(k_2\) and \(v(t_f) = v_f\) solve for \(x(t_f) = x_f\) .  (\(k_1, k_2 \) and \(v_f\) are known values)

 

 

 

My first idea was to solve the differential equation for \(v(t)\) but I got a really ugly answer(which I had to look up on Wolframalpha) so I assume I'm doing something wrong.

 

The idea was that to solve for \(x_f\) but if you find \(v(t)\) , solve for \(t_f\) you could integrate to find \(x(t)\) and then input \(t_f\) but I'm stuck at the \(v(t)\) part. Thanks for help

 Apr 24, 2019
 #1
avatar+28363 
+4

I think the solutions can be written as:

 

\(v=\sqrt{\frac{k_1}{k_2}}\tanh{(\sqrt{k_1k_2}t)}\text{ and }x=\ln{(\cosh{(\sqrt{k_1k_2}t)})}/k_2\)

 

which I obtained by guessing a few forms for the velocity and manipulating them until they matched numerical solutions for all the values of k1 and k2 that I tried!!   I've no doubt there is a more straightforward way of doing it!!

 Apr 25, 2019
 #2
avatar+267 
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Thank you this helped me get the correct answer when plugging in values for \(k_1, k_2\) and \(v_f\)

I still struggle to understand how did you solve to find these x(t) and v(t)?

Quazars  Apr 25, 2019
 #3
avatar+78 
+4

If you've been there before, so to speak, and you know what the solution is to look like, then Alan's ' trial solution ' method is quickest.

Assume that

 \(\displaystyle v=A\tanh(Bt), \text{ then}\\\frac{dv}{dt}=AB\text{ sech} ^{2}(Bt)=AB(1-\tanh^{2}(Bt))=AB -\frac{B}{A}v^{2}\)

Equate coefficients with the given equation and solve for A and B.

 

If you have to get there from scratch, with one eye on the upcoming algebra, write \(\displaystyle \frac{dv}{dt}=k_{2}(K^{2}-v^{2}),\text{ where, for convenience, } K^{2}=k_{1}/k_{2}.\)

then  \(\displaystyle \int \frac{dv}{K^{2}-v^{2}}=\int k_{2}dt\)

\(\displaystyle \frac{1}{2K}\int \frac{1}{K+v}+\frac{1}{K-v}dv=\frac{1}{2K}\ln\left(\frac{K+v}{K-v}\right)=k_{2}t + C\),   \(\displaystyle v=0 \text{ when } t=0, \text{ so } C=0.\)

\(\displaystyle \frac{K+v}{K-v}=e^{2Kk_{2}t} \text{ from which }\quad v= K\left(\frac{e^{2Kk_{2}t}-1}{e^{2Kk_{2}t}+1}\right)=K\left(\frac{e^{Kk_{2}t}-e^{-Kk_{2}t}}{e^{Kk_{2}t}+e^{Kk_{2}t}}\right)=K\left(\frac{\sinh (Kk_{2}t)}{\cosh(Kk_{2}t)}\right)\\=K\tanh(Kk_{2}t)\).

Substitution for K now gets us the earlier mentioned solution (Alan's solution).

 

To find x, integrate wrt t,

 

\(\displaystyle \int v\,dt= x =K\int\tanh(Kk_{2}t)\,dt=\frac{K}{Kk_{2}}\ln(\cosh(Kk_{2}t))+D\\\text{ and since } x=0\text{ when }t=0, D=0.\)

Substituting for K,

\(\displaystyle x=\frac{1}{k_{2}}\ln(\cosh(\sqrt{k_{1}k_{2}}t)\)

.
 Apr 25, 2019
edited by Tiggsy  Apr 25, 2019
 #4
avatar+28363 
+4

Here is a brief version of how I approached it - pretty much as Tiggsy explained, though I did a numerical solution first to get a feel for the shape of the velocity as a function of time:

 

 

Note: Tiggsy approach from scratch is the better way to do this.

 Apr 25, 2019
edited by Alan  Apr 26, 2019

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