One may assume that as a car moves that there is a force resisting the movement that is proportional to the car’s velocity $v$. Suppose that the car has a mass of $m$ we get then that the force resisting the car’s forward motion is given by
$F=mass$ x $acceleration=m\frac{dv}{dt}=-kv$ where $k>0$
Suppose that a car that has an initial velocity $v_0$ begins to coast. Find its position $s(t)$ at time $t$ given that its initial position is given by $s(0) = 0$. How far does the carcoast before stopping?
Can someone help me with this please? I'm not sure where to start. I tried to differentiate this but I wasn't able to get anything solid as we don't know $m$, or $v.
Best Answer
Rewrite $$ m\cdot\frac{dv}{dt}=-kv $$ as $$ \frac{dv}{v}=-\frac km\ dt. $$ Integrating both sides yields $$\eqalign { \int_{v_o}^{v_t}\frac{dv}{v}&=-\int_{0}^t\frac km\ dt\\ \ln v_t-\ln v_o&=-\frac kmt + C_1\\ \ln\frac{v_t}{v_o}&=-\frac kmt + C_1\\ v_t&=C_2\ e^{\Large-\frac kmt}\quad\Rightarrow\quad C_2=v_o\ e^{C_1}\\ \frac{ds}{dt}&=C_2\ e^{\Large-\frac kmt}\\ ds&=C_2\ e^{\Large-\frac kmt}\ dt\\ \int_{s_o}^{s_t}\ ds&=C_2\int_{0}^t e^{\Large-\frac kmt}\ dt\\ s_t-s_o&=C_3\left(1-e^{\Large-\frac kmt}\right)\quad\Rightarrow\quad C_3=\frac{mC_2}{k}=\frac{mv_o\ e^{C_1}}{k}\\ \color{blue}{s(t)}&\color{blue}{=C_3\left(1-e^{\Large-\frac kmt}\right)}.\quad\Rightarrow\quad s_o=0 }$$