We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
110
4
avatar+267 

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+28065 
+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 
0

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+28 
+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+28065 
+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

4 Online Users

avatar
avatar