# Modelling with differential equation

• Sep 8th 2010, 03:05 AM
acevipa
Modelling with differential equation
A cyclist freewheeling on a level road experiences a negative acceleration which is proportional to the square of his speed. His speed is reduced from $20m/s$ to $10m/s$ in a distance of $100m$. Find the average speed (w.r.t. time) during this period.

Would the equation be

$\dfrac{dv}{dt}=-kv^2$ where $k$ is a constant of proportionality
• Sep 8th 2010, 03:57 AM
Quote:

Originally Posted by acevipa
A cyclist freewheeling on a level road experiences a negative acceleration which is proportional to the square of his speed. His speed is reduced from $20m/s$ to $10m/s$ in a distance of $100m$. Find the average speed (w.r.t. time) during this period.

Would the equation be

$\dfrac{dv}{dt}=-kv^2$ where $k$ is a constant of proportionality

yes.
• Sep 8th 2010, 04:49 AM
acevipa
Quote:

yes.

Thanks. So I'm a little unsure of what to do next.

$\dfrac{dv}{dt}=-kv^2$

$-\dfrac{dv}{v^2}=k \ dt$

$\displaystyle\int -\dfrac{dv}{v^2}=\int k \ dt$

$\dfrac{1}{v}=kt+C$
• Sep 8th 2010, 05:37 AM
Quote:

Originally Posted by acevipa
Thanks. So I'm a little unsure of what to do next.

$\dfrac{dv}{dt}=-kv^2$

$-\dfrac{dv}{v^2}=k \ dt$

$\displaystyle\int -\dfrac{dv}{v^2}=\int k \ dt$

$\dfrac{1}{v}=kt+C$

You have to set up a differential equation which involves the displacement and velocity.

Hint: Notice that a=dv/dx=(dv/dx) x (dx/dt) = v (dv/dx)
• Sep 8th 2010, 05:56 AM
chisigma
Quote:

Originally Posted by acevipa
Thanks. So I'm a little unsure of what to do next.

$\dfrac{dv}{dt}=-kv^2$

$-\dfrac{dv}{v^2}=k \ dt$

$\displaystyle\int -\dfrac{dv}{v^2}=\int k \ dt$

$\dfrac{1}{v}=kt+C$

... all right!... the solution is then...

$\displaystyle v(t)= \frac{1}{k t + c}$ (1)

... where of course is $c= \frac{1}{v(0)}$. Now suppose that $v(0)$ is negative, for example $v(0) = -1$, so that the solution is...

$\displaystyle v(t)= \frac{1}{k t -1}$ (2)

A little question: what does if happend at the time $t=\frac{1}{k}$ ?(Surprised) ...

Kind regards

$\chi$ $\sigma$
• Sep 8th 2010, 06:14 AM
acevipa
Quote:

You have to set up a differential equation which involves the displacement and velocity.

Hint: Notice that a=dv/dx=(dv/dx) x (dx/dt) = v (dv/dx)

Thanks that makes a lot of sense.

$a=-kv^2 \Longrightarrow \dfrac{dv}{dt}=-kv^2$

$v\dfrac{dv}{dx}=-kv^2$

$\dfrac{dv}{dx}=-kv$

$\displaystyle\int\dfrac{dv}{v}=\int -k \ dx$

$\ln v=-kx+C$

When $x=0, v=20 \Longrightarrow C = \ln 20$

$\ln v =-kx+\ln 20$

When $x=100, v=10$

$\ln 10=-100k+\ln 20$

$100k = \ln 20 - \ln 10$

$100k = \ln 2$

$k = \dfrac{\ln 2}{100}$

Have I done everything right? If so, I'm not too sure what to do next.

The answer in the book is $20\ln 2$
• Sep 11th 2010, 09:27 PM
Quote:

Originally Posted by acevipa
Thanks that makes a lot of sense.

$a=-kv^2 \Longrightarrow \dfrac{dv}{dt}=-kv^2$

$v\dfrac{dv}{dx}=-kv^2$

$\dfrac{dv}{dx}=-kv$

$\displaystyle\int\dfrac{dv}{v}=\int -k \ dx$

$\ln v=-kx+C$

When $x=0, v=20 \Longrightarrow C = \ln 20$

$\ln v =-kx+\ln 20$

When $x=100, v=10$

$\ln 10=-100k+\ln 20$

$100k = \ln 20 - \ln 10$

$100k = \ln 2$

$k = \dfrac{\ln 2}{100}$

Have I done everything right? If so, I'm not too sure what to do next.

The answer in the book is $20\ln 2$

That's correct. Then, work from $\dfrac{dv}{dt}=-kv^2$

For velocity, integrate from 20 to 10 and for time, integrate from 0 to t.