+0

# Solve for distance(and time) it takes to accelerate from v=0 to v=v_f

0
201
4

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
+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
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
+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
+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