Kyle Sutton

2022-07-14

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

$\frac{dv}{dt}=-kv\phantom{\rule{1em}{0ex}}(1)$

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)):

$\frac{dv}{dt}=-k\frac{ds}{dt}\phantom{\rule{1em}{0ex}}\text{(from (1))}$

$\int \frac{dv}{dt}dt=-k\int \frac{ds}{dt}dt$

$v(s)=-ks+C$

$v(0)=100\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=100,\text{}v(1)=50\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}k=50$

$v(s)=-50s+100$

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 $v(t)=100{e}^{-50t}$ 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.

$\frac{dv}{dt}=-kv\phantom{\rule{1em}{0ex}}(1)$

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)):

$\frac{dv}{dt}=-k\frac{ds}{dt}\phantom{\rule{1em}{0ex}}\text{(from (1))}$

$\int \frac{dv}{dt}dt=-k\int \frac{ds}{dt}dt$

$v(s)=-ks+C$

$v(0)=100\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=100,\text{}v(1)=50\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}k=50$

$v(s)=-50s+100$

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 $v(t)=100{e}^{-50t}$ 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.

poquetahr

Beginner2022-07-15Added 18 answers

We can find v(t), taht k=50.

$\frac{dv}{dt}=-kt=-50t$

$v(t)=A{e}^{-50t},\text{}\text{for some constant}A$

$v(0)=100\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A=100$

$v(t)=100{e}^{-50t}$

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 $\frac{dv}{dt}=-5000{e}^{-50t}$.

$\frac{ds}{dt}=\frac{ds}{dv}\frac{dv}{dt}=-\frac{1}{50}\times -5000{e}^{-50t}=100{e}^{-50t}$

$\int \frac{ds}{dt}dt=100\int {e}^{-50t}dt$

$s(t)=100\times -\frac{1}{50}{e}^{-50t}+C$

$s(t)=-2{e}^{-50t}+C$

$s(0)=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=2$

$s(t)=-2{e}^{-50t}+2=2(1-{e}^{-50t})$

$\frac{dv}{dt}=-kt=-50t$

$v(t)=A{e}^{-50t},\text{}\text{for some constant}A$

$v(0)=100\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A=100$

$v(t)=100{e}^{-50t}$

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 $\frac{dv}{dt}=-5000{e}^{-50t}$.

$\frac{ds}{dt}=\frac{ds}{dv}\frac{dv}{dt}=-\frac{1}{50}\times -5000{e}^{-50t}=100{e}^{-50t}$

$\int \frac{ds}{dt}dt=100\int {e}^{-50t}dt$

$s(t)=100\times -\frac{1}{50}{e}^{-50t}+C$

$s(t)=-2{e}^{-50t}+C$

$s(0)=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=2$

$s(t)=-2{e}^{-50t}+2=2(1-{e}^{-50t})$

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