<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

A particle starts from rest and moves in a straight line with
acceleration

$$\ds\frac{dv}{dt}=a-kv^2$$

where $v$ is the velocity and $a$ and $k$ are constants. Find the
times taken to acquire a velocity $V$ and show the distance
travelled in this time is

$$x=\ds\frac{1}{2k}\ln\left(\ds\frac{a}{a-kV^2}\right)$$

(Hint: Note that if $v=\ds\frac{dx}{dt}$,
then$\ds\frac{d^2x}{dt^2}=\ds\frac{dv}{dt}=\ds\frac{dv}{dx}\ds\frac{dx}{dt}=v\ds\frac{dv}{dx}$.
This will enable you to change the differential equation above to
one in terms of $x$ and $v$ only)

\medskip

{\bf Answer}

$\ds\frac{dv}{dt}=a-kv^2$ variables separable

$\Rightarrow \ds\int\ds\frac{dv}{a-kv^2}=\ds\int dt$

$\Rightarrow
\ds\int\ds\frac{dv}{(\sqrt{a}+\sqrt{k}v)(\sqrt{a}-\sqrt{k}v)}=t+c$

Partial fractions

$\Rightarrow
\ds\frac{1}{2\sqrt{a}}\ds\int\ds\frac{dv}{(\sqrt{a}=\sqrt{k}v)}+
\ds\frac{1}{2\sqrt{a}}\ds\int\ds\frac{dv}{(\sqrt{a}=\sqrt{k}v)}=t+c$

$\Rightarrow \ds\frac{1}{2\sqrt{ak}}\ln(\sqrt{a}+\sqrt{k}v)-
\ds\frac{1}{2\sqrt{ak}}\ln(\sqrt{a}+\sqrt{k}v)=t+c$

$\Rightarrow
\ln\left(\ds\frac{\sqrt{a}+\sqrt{k}v}{\sqrt{a}-\sqrt{k}v}\right)=2\sqrt{ak}(t+c)$

$\Rightarrow
\ds\frac{\sqrt{a}+\sqrt{k}v}{\sqrt{a}-\sqrt{k}v}=e^{2\sqrt{ak}(t+c)}$

$\sqrt{a}+\sqrt{k}v=e^{2\sqrt{ak}(t+c)}(\sqrt{a}-\sqrt{k}v)$

$\sqrt{k}v(1+e^{2\sqrt{ak}(t+c)})=\sqrt{a}(e^{2\sqrt{ak}(t+c)}-1)
\Rightarrow
\un{v=\sqrt{\ds\frac{a}{k}}\left(\ds\frac{e^{2\sqrt{ak}(t+c)}-1}
{e^{2\sqrt{ak}(t+c)}+1}\right)}$

Now let $v=0$ when $t=0$ and $v=V$ when$t=T$

Thus

$$0=\sqrt{\ds\frac{a}{k}}\left(\ds\frac
{e^{2\sqrt{ak}c}-1}{e^{2\sqrt{ak}c}+1}\right)$$

So $e^{2\sqrt{ak}c}=1$

$\Rightarrow 2\sqrt{ak}c=\ln 1=0$

If $a$ and $k \ne 0$ we have $c=0$.

Thus $v=\sqrt{\ds\frac{a}{k}}\left(\ds\frac{e^{2\sqrt{ak}t}-1}
{e^{2\sqrt{ak}t}+1}\right)$

Thus $v=\sqrt{\ds\frac{a}{k}}\left(\ds\frac{e^{2\sqrt{ak}T}-1}
{e^{2\sqrt{ak}T}+1}\right)$

or, after some algebra

$$\un{T=\ds\frac{1}{2\sqrt{ak}}\ln
\left(\ds\frac{\sqrt{a}+\sqrt{k}V}{\sqrt{a}-\sqrt{k}V}\right)}$$

Use hint and return to original equation

$\ds\frac{dv}{dt} = \ds\frac{d^2x}{dt^2} =
v\ds\frac{dv}{dx}=a-kv^2$

$\Rightarrow v\ds\frac{dv}{dx}=a-kv^2$

Also variables separable

$$\ds\int\ds\frac{v dv}{a-kv^2}=\ds\int\,dx$$

Standard integral

$$\Rightarrow \ds\frac{1}{-2k}\ln(a-kv^2)=x+c'$$

where $c'$ is a new constant

$$\Rightarrow
x=-c'+\ds\frac{1}{2k}\ln\left(\ds\frac{1}{a-kv^2}\right)$$

What are the boundary conditions?

Well $v=0$ when $x=0$

$\Rightarrow 0=-c'=\ds\frac{1}{2k}\ln\left(\ds\frac{1}{a}\right)$

$\Rightarrow c'=\ds\frac{1}{2k}\ln\left(\ds\frac{1}{a}\right)$

$\Rightarrow
x=-\ds\frac{1}{2k}\ln\left(\ds\frac{1}{a}\right)+\ds\frac{1}{2k}
\ln\left(\ds\frac{1}{a-kv^2}\right)$

$\Rightarrow x=\ds\frac{1}{2k}\ln\left(\ds\frac{a}{a-kv^2}\right)$

Thus the distances travelled to when $v=V$ is

$$\un{\ds\frac{1}{2k}\ln\left(\ds\frac{a}{a-kV^2}\right)}\ \rm{as\
required}$$
\end{document}
</pre></body></html>