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Question:
One morning it starts to snow at a constant rate. Later, at 6:00am, a snow plow sets out to clear a straight street. The plow can remove a fixed volume of snow per
unit time, in other words its speed is inversely proportional to the depth of the snow. If the plow covered twice as much distance in the first hour as the second hour,
what time did it start snowing?
Answer:
Let the snow be falling at rate r.
The snow depth at time t is:
s =
r * ( t - ts)
where the snow has started falling at time ts (which is what we want to find).
The velocity of the plow is:
v =
c / s
=
c / ( r * ( t - ts ) ) , where c is a constant
The distance covered in the first hour is the integral from 6am to 7am of v, which is:
d1 =
( c / r )*log( ( 7 - ts ) / ( 6 - ts ) )
and the distance covered in the second hour is the integral from 7am to 8am of v, which is:
d2 =
( c / r )*log( ( 8 - ts ) / ( 7 -t s) )
Setting d1 = 2 *d2 (since the plow covered twice as much distance in the first hour as the second hour) gives the equation to solve for ts:
log( ( 7 - ts ) / ( 6 - ts ) ) =
2 * log( ( 8 - ts ) / ( 7 - ts ) )
Take Exp of boths sides to get:
( 7 - ts) / ( 6 - ts) =
( ( 8 - ts ) / ( 7 - ts ) )^2
Solving for ts, this simplifies to solving the quadratic equation:
0 =
ts^2 - 13* ts + 41
giving the solutions for when it could have started snowing:
ts =
13/2 + sqrt(5)/2 or 13/2 - sqrt(5)/2
=
(~ 7.62 or ~ 5.38)
of which only the later makes sense (since it snowed before 6am, the time at which they started plowing), so
ts =
13/2 - sqrt(5)/2
=
~5.38
=
~5:23 am
-- Tamsen Dunn
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