Snowplow problem
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5 members have voted
July 25th, 2020 at 3:21:49 PM
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One morning it starts to snow at a constant rate. Later, at 6:00am, a snowplow sets out to clear a straight street. The plow can remove a fixed volume of snow per unit time, in other words its speed it inversely proportional to the depth of the snow. If the plow covered 10 miles from 6 to 7am and 5 miles from 7 to 8am, what time did it start snowing?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 25th, 2020 at 3:37:56 PM
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Let m be the number of minutes before 6 AM when it started snowing
The amount of snow cleared in the first 10 miles = 10m + 5 x 60
The amount of snow cleared in the next 5 miles = 5 (m + 60) + 5/2 x 60
10m + 300 = 5m + 300 + 150
m = 30, so it started snowing at 5:30 AM
July 25th, 2020 at 4:41:38 PM
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Quote: Wizard...what time did it start snowing?...
that it started snowing at approximately 5:23 AM.
I see that the answer has something to do with the golden ratio--the snow started before 6:00 AM by one hour divided by the golden ratio.
I see that the answer has something to do with the golden ratio--the snow started before 6:00 AM by one hour divided by the golden ratio.
July 25th, 2020 at 4:46:19 PM
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Forgot way to much to solve it the intended way tonight...
But my excel-solution seems to converge towards (phi-1) hours before 6
So i guess it started 06:22:55
6:30 is for sure not correct
But my excel-solution seems to converge towards (phi-1) hours before 6
So i guess it started 06:22:55
6:30 is for sure not correct
July 25th, 2020 at 5:19:40 PM
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Quote: ChesterDogthat it started snowing at approximately 5:23 AM.
I see that the answer has something to do with the golden ratio--the snow started before 6:00 AM by one hour divided by the golden ratio.
I agree. Remember, for full credit you must show your work.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 25th, 2020 at 5:57:16 PM
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A try..
I assume it started to snow at time 0:
The plow have a speed at 1/X
so f(X) = 1/x
We have that the definite integral from x to (x+1) = 2 times the definite integral from (x+1) to (x+2)
The integral for 1/x = ln(x)
That gives:
ln(x+1) - ln(x) = 2*( ln(x+2) - ln(x+1) )
3*ln(x+1) -ln(x) = 2*ln(x+2)
3*ln(x+1) = 2*ln(x+2)+ln(x)
ln((x+1)^3) = ln(x) + ln((x+2)^2)
ln((x+1)^3) = ln((x+2)^2 * x)
ln(x^3+3x^2+3x+1)= ln(x^3+4x^2+4x)
x^3+3x^2+3x+1= x^3+4x^2+4x
x^2 +x-1 = 0
X = -1,618 and 0,618
As it started to snow before the plow started, the solution have to be positive and the answer is 0,618.
=it started to snow at 06:22:55
Sorry my poor english and formating.
July 25th, 2020 at 7:34:30 PM
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Quote: TorghattenForgot way to much to solve it the intended way tonight...
But my excel-solution seems to converge towards (phi-1) hours before 6
So i guess it started 06:22:55
6:30 is for sure not correct
Quote: Torghatten
A try..
I assume it started to snow at time 0:
The plow have a speed at 1/X
so f(X) = 1/x
We have that the definite integral from x to (x+1) = 2 times the definite integral from (x+1) to (x+2)
The integral for 1/x = ln(x)
That gives:
ln(x+1) - ln(x) = 2*( ln(x+2) - ln(x+1) )
3*ln(x+1) -ln(x) = 2*ln(x+2)
3*ln(x+1) = 2*ln(x+2)+ln(x)
ln((x+1)^3) = ln(x) + ln((x+2)^2)
ln((x+1)^3) = ln((x+2)^2 * x)
ln(x^3+3x^2+3x+1)= ln(x^3+4x^2+4x)
x^3+3x^2+3x+1= x^3+4x^2+4x
x^2 +x-1 = 0
X = -1,618 and 0,618
As it started to snow before the plow started, the solution have to be positive and the answer is 0,618.
=it started to snow at 06:22:55
Sorry my poor english and formating.
Really? Twice you have him plowing BEFORE it starts snowing??? I think you missed a digit or something.
This math is was beyond my abilities, but I know that’s wrong.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
July 25th, 2020 at 8:19:16 PM
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Quote: Torghatten
A try..
I assume it started to snow at time 0:
The plow have a speed at 1/X
so f(X) = 1/x
We have that the definite integral from x to (x+1) = 2 times the definite integral from (x+1) to (x+2)
The integral for 1/x = ln(x)
That gives:
ln(x+1) - ln(x) = 2*( ln(x+2) - ln(x+1) )
3*ln(x+1) -ln(x) = 2*ln(x+2)
3*ln(x+1) = 2*ln(x+2)+ln(x)
ln((x+1)^3) = ln(x) + ln((x+2)^2)
ln((x+1)^3) = ln((x+2)^2 * x)
ln(x^3+3x^2+3x+1)= ln(x^3+4x^2+4x)
x^3+3x^2+3x+1= x^3+4x^2+4x
x^2 +x-1 = 0
X = -1,618 and 0,618
As it started to snow before the plow started, the solution have to be positive and the answer is 0,618.
=it started to snow at 06:22:55
Sorry my poor english and formating.
Great answer and solution Tor! I don't recall you being a member of the Beer Club yet, in which case welcome. This means I owe you a beer upon your next visit to Vegas.
To be perfectly honest, I asked this problem because I made a video going over the solution. This is my first math video in this kind of format so please go easy on me with your reviews.
Direct: https://www.youtube.com/watch?v=33KYd7OSKjI
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
July 25th, 2020 at 10:02:24 PM
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July 26th, 2020 at 6:12:47 AM
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Quote: DJTeddyBear
Really? Twice you have him plowing BEFORE it starts snowing??? I think you missed a digit or something.
This math is was beyond my abilities, but I know that’s wrong.
Sorry, misread and used 07-09 instead of 06-08.
Quote: Wizard
Great answer and solution Tor! I don't recall you being a member of the Beer Club yet, in which case welcome. This means I owe you a beer upon your next visit to Vegas.
To be perfectly honest, I asked this problem because I made a video going over the solution. This is my first math video in this kind of format so please go easy on me with your reviews.
Direct: https://www.youtube.com/watch?v=33KYd7OSKjI
Thx :)
Idk when im able/allowed to go to Vegas again, but most likely it will be next summer :p
