Rise and shine the weathers fineĀ
Mark:
SNOW PLOW
t = the number of hours it's been snowing at 6:00
log(t+1) - log(t) = 2*[log(t+2) - log(t+1)]
log((t+1)/t) = 2*log((t+2)/(t+1))
(t+1)/t = [(t+2)/(t+1)]^2
t^2 + t - 1 = 0
t = [-1 + sqrt(5)] / 2 = 0.618034
It's been snowing for 37.082039 minutes, which is about 37 minutes and 4.9 seconds.
So, it started snowing at 5:22:55.1 a.m., or I did this wrong.
Rap:
The way I'm figgerin it the volume of snow removed is a constant (Q) equal to the product of depth and velocity(v). The snow depth, the is the product of the rate (in/hr--m) and the time since it started snowing (t), or
Q=v*m*t (1)
Now using the condition that the truck started plowing at to and that in the first hour (to+1) the truck traveled twice as far in the second hour (to+2) as the first.
Rearranging (1) to isolate the velocity
v=Q/(mt) the distance the plow traveled in a set period of time (d) is the integral with respect to time. Or
d={vdt={(Q/(mt))dt Note I'm using { as an integral sign
and since Q and m are constants then
d={(Q/(mt))dt=Q/m{dt/t=q/mln(t)+Co where Co is a constant of integration
now using the condition that the distance the first hour (between to and to+1) is twice the distance the second hour (between to+1 and to+2)
or 2Q/m[ln(to+1)-ln(to)]=Q/m[ln(to+2)-ln(to+1)]
since Q & m <>0 then
2[ln(to+1)-ln(to)]=[ln(to+2)-ln(to+1)]
rearranging
(to+1)^2/to^2=(to+2)/(to+1)
so
(to+1)^3=to^2(to+2)
to^3+3to^2+3to+1=to^3+2to^2
resulting in the quadratic
to^2+3to+1=0
solutions to=(-3+/-sqrt(5))/2
to=-2.61803 hrs~-2hrs, 37 min, 5 sec before 6:00 (3:22:55 AM)
or
to=-0.38197 hrs~22 min, 55 sec before 6:00 (5:37:05 AM)
No No No! there's only one answer. But everything in my first post is right to here
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now using the condition that the distance the first hour (between to and to+1) is twice the distance the second hour (between to+1 and to+2)
or Q/m[ln(to+1)-ln(to)]=2Q/m[ln(to+2)-ln(to+1)]
since Q & m <>0 then
[ln(to+1)-ln(to)]=2[ln(to+2)-ln(to+1)]
rearranging
(to+1)/to=(to+2)^2/(to+1)^2
so
(to+1)^3=to(to+2)^2
to^3+3to^2+3to+1=to^3+4to^2+4to
resulting in the quadratic
to^2+to-1=0
solutions are
to=(-1+/-sqrt(5))/2
crunching #
to= 0.618034 hrs
and
to= -1.61803 hrs
since the snow started before plowing the negative answer doesn't pass the sanity test and the snow started 0.61803 hrs before plowing started.
Plowing started at 6:00 AM so the snow started at 5:23:55.89 AM.
Thoh:
SNOW
i think ur BOTH wrong haha
it starts snowing at some time before 6am (time 0)
that time is y hrs before time 0
the plow will plow a distance of 3x
at 6am, the volume of snow is (3x)(y) - a rectangle
the snow keeps building up at a constant rate ahead of the plow, forming a triangle of base 3x and height 2 hrs (2) on top of the rectangle
the area of the trapezoid formed by distance 2x = area formed by last x
[y+(1+y)]*(2x)/2 = [(1+y)+(2+y)]*(x)/2....x's cancel
2y+1 = (3+2y)/2
4y+2 = 3+2y
2y=1
y= 1/2
it started snowing half hr before 6am or 5:30am
Slippy:
5 root 2 minus 1 over 2 before 6
I had a bit part in the movie
Isn't it great to see four keen minds at work, and the different ways to produce an answer. Ok, so who is right?
Let the depth of snow at time t to be t units. The speed of the plow at time t will be 1/t. Define t=0 as the time it started snowing and t=x the time the plow started.
The distance covered in the first hour is the integral from x to x+1 of 1/t dt. The antiderivative of 1/t is ln(t) so the total distance covered in the first hour is ln((x+1)/x).
By the same reasoning the distance covered in the second hour in ln((x+2)/(x+1)).
Using the fact that it the plow traveled twice as far in the first hour as the second: ln((x+1)/x) = ln((x+2)/(x+1))^2
Exp both sides and you have (x+1)/x = ((x+2)/(x+1))^2.
Solving for x you get x=(5^1/2-1)/2, which is the number of hours that elapsed between the time it started snowing and the snow plow left.
Or in English we say: It started snowing at (5^1/2-1)/2 hours before 6:00am, or 5:22:55am.
Amazing answers all:
The problem was taken from the Actuarial Review
KCLS
echeseks
