Question 10 - Page 2

CaptainBlack · #16 $Old$ December 17th, 2006, 11:00 PM

Quote:

Originally Posted by CaptainBlack $View Post$

One day it started snowing at a heavy and steady rate. A snowplough
started out at noon, going 2 km the first hour and 1 km the second hour.
What time did it start snowing?

.

Solution:

Let $t$ be time in hours since it started to snow.
Let the rate of snowfall be $H$ metres/hr
Let the plough swept width be $W$ metres.
let the distance of the plough from its starting point be $x(t)$ km.

Then as the plough clears a fixed volume of snow per unit time we have:

$W\,(Ht)\,x'(t)=K$

where $K$ is the constant volume swept per hour in units of thousanths of a cubic metre/hr (but the units are unimportant). (this is the swept width times the snow depth times the rate of progress of the plough)

This is an ordinary differential equation of variables seperable type and has general solution:

$x(t)=\frac{K}{W\,H}\ln(t)+C$

for some constant $C$ , and times after $t=N$ which is how long Noon is after [math]t=0[math].

So from the given conditions we have:

$x(N)=\frac{K}{W\,H}\ln(N)+C=0$

$x(N+1)=\frac{K}{W\,H}\ln(N+1)+C=2$

$x(N+2)=\frac{K}{W\,H}\ln(N+2)+C=3$

The first of these equations gives:

$C=-\frac{K}{W\,H}\ln(N)$ ,

so:

$x(t)=\frac{K}{W\,H}\ln(t/N)$ .

Dividing the secon by the third equation elliminates $\frac{K}{W\,H}$ , and can be rearranged to give:

$3\ln((N+1)/N)=2\ln((N+2)/N)$ ,

or equivalently:

$\left(\frac{N+1}{N}\right)^3=\left(\frac{N+2}{N}\right)^2$ .

Simplifying this last equation gives us: $N^2+N-1=0$ , which has solutions $-\frac{1+\sqrt{5}}{2}$ and $\frac{-1+\sqrt{5}}{2}$ . The first of these is negative and so a spurious solution, so:

$N=\frac{-1+\sqrt{5}}{2}\approx 0.618$ hours $\approx 37$ minutes.

So the snow began falling $\sim 37$ minutes before noon or about $11:23$ .

RonL