A car is travelling at $100$ km/h on a level road when it runs out of fuel. Its speed $v$ (in km/h) starts
to decrease according to the formula $$\frac{dv}{dt}=-kv \quad (1)$$ where $k$ is constant. One kilometre after running out of fuel its speed has fallen to $50$ km/h. Use the chain rule substitution $$\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=\frac{dv}{ds}v$$ to solve the differential equation.
(Note: Although I haven't solved it yet, the answers say that this isn't a reasonable model as the velocity is always positive; I didn't make a typo in the question.)
What I'm trying to do is solve velocity as a function of displacement ($s$, in km), velocity as a function of time ($t$, in hours), and displacement as a function of time (I need these functions for later parts of the question).
So far I've found velocity as a function of displacement ($v(s)$):
$$\frac{dv}{dt}=-k\frac{ds}{dt} \quad \text{(from (1))}$$
$$\int \frac{dv}{dt} dt = -k \int \frac{ds}{dt} dt $$
$$v(s) = -ks+C$$
$$v(0) = 100 \implies C=100, \ v(1)=50 \implies k=50$$
$$v(s) = -50s+100$$
Then I've tried to find velocity as a function of time ($v(t)$), but I've got stuck. I can't find any differential equation I can use to get this, or to get displacement as a function of time ($s(t)$).
The answer key says $v(t)=100e^{-50t}$ and $s(t)=2(1-e^{-50t})$.
I've solved such questions many times before, but it's been a while so I'm a bit rusty. So, even a hint might be enough for me to realise what to do.
Thanks
Best Answer
We can find $v(t)$ using amsmath's comment, and that $k=50$: $$\frac{dv}{dt}=-kt=-50t$$ $$v(t)=Ae^{-50t},\ \text{for some constant $A$}$$ $$v(0)=100\implies A=100$$ $$v(t)=100e^{-50t}$$ We can find $s(t)$ using $\frac{ds}{dt}$ and the chain rule, however I'm not sure if this is the most efficient method. We know from $v(s)$ that $\frac{dv}{ds}=-50$, so $\frac{ds}{dv}=-\frac{1}{50}$. We can also see from $v(t)$ that $\frac{dv}{dt}=-5000e^{-50t}$. $$\frac{ds}{dt}=\frac{ds}{dv}\frac{dv}{dt}=-\frac{1}{50}\times-5000e^{-50t}=100e^{-50t}$$ $$\int \frac{ds}{dt} dt = 100\int e^{-50t}dt$$ $$s(t) =100\times-\frac{1}{50}e^{-50t}+C$$ $$s(t)=-2e^{-50t}+C$$ $$s(0)=0 \implies C=2$$ $$s(t)=-2e^{-50t}+2=2(1-e^{-50t})$$