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I'm trying to solve this problem, in particular part a. As the hint suggests I make the substitution $v=\frac{ds}{dt}$. Carrying out the computation I get $ln\frac{v}{V_0}=-KHe^{s/H}$, provided I haven't made any mistakes. Where do I go from here? How do I calculate the maximum deceleration?

EDIT: here is how I got $v$. By making the substitution $v=\frac{ds}{dt}$ we get $v\frac{dv}{ds}=-ke^{s/H}(v)^2$, hence $\frac{dv}{ds}=-ke^{s/H}v$, separating the variables $\frac{1}{v}dv=-ke^{s/H}ds$. Integrating $ln\;v=-kHe^{s/H}+c$, evaluating at $-\infty$ we get $c=ln\;V_0$.

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    @Aretino : I just wrote how I got $v$. Could you point out the mistake, please? – 2017-02-17
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    You are right, sorry: your derivation is correct. You just have to insert your solution $v$ into the equation for $dv/dt$ and find the maximum of the resulting function of $s$. – 2017-02-17

2 Answers 2

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You correctly integrated the differential equation and found $$ \ln{ds/dt\over V_0}=-KHe^{s/H}, \quad \hbox{that is:}\quad {ds\over dt}=V_0\exp\big(-KHe^{s/H}\big). $$ Plugging that into the equation for the acceleration $a=d^2s/dt^2$ one then finds: $$ a=-Ke^{s/H}V_0^2\exp\big(-2KHe^{s/H}\big) =-KV_0^2\exp\left(-2KHe^{s/H}+{s\over H}\right). $$ Notice that $a\to0$ for $s\to\pm\infty$, so there must be a maximum value for $-a$, which can be found by solving the equation $da/ds=0$.This yields $$ -2Ke^{s/H}+{1\over H}=0, \quad\hbox{whence}\quad e^{s/H}={1\over 2HK}. $$ Now substitute that into the above expression for $-a$ to find the maximum: $$ (-a)_\max=K{1\over 2HK}V_0^2\exp\big(-2KH{1\over 2HK}\big)={V_0^2\over 2H e}. $$

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Well, we have that:

$$\text{s}''\left(t\right)=-\text{K}\cdot\exp\left(\frac{\text{s}\left(t\right)}{\text{H}}\right)\cdot\text{s}'\left(t\right)^2\tag1$$

Treating $\text{s}$ as the independent variable, let $\rho\left(\text{s}\right)=\text{s}'\left(t\right)$:

$$\rho'\left(\text{s}\right)\cdot\rho\left(\text{s}\right)=-\text{K}\cdot\exp\left(\frac{\text{s}}{\text{H}}\right)\cdot\rho\left(\text{s}\right)^2\tag2$$

Now, solve the separable equation:

$$\frac{\rho'\left(\text{s}\right)}{\rho\left(\text{s}\right)}=-\text{K}\cdot\exp\left(\frac{\text{s}}{\text{H}}\right)\space\Longleftrightarrow\space\int\frac{\rho'\left(\text{s}\right)}{\rho\left(\text{s}\right)}\space\text{d}\text{s}=\int-\text{K}\cdot\exp\left(\frac{\text{s}}{\text{H}}\right)\space\text{d}\text{s}\tag3$$

So, we get that:

$$\ln\left|\rho\left(\text{s}\right)\right|=\text{C}-\text{H}\cdot\text{K}\cdot\exp\left(\frac{\text{s}}{\text{H}}\right)\tag4$$

So, we got two solutions:

  1. $$\text{s}'\left(t\right)=0\space\Longleftrightarrow\space\text{s}\left(t\right)=\text{C}_1\tag5$$
  2. $$\ln\left|\text{s}'\left(t\right)\right|=\text{C}_2-\text{H}\cdot\text{K}\cdot\exp\left(\frac{\text{s}\left(t\right)}{\text{H}}\right)\tag6$$