Motion Puzzle [for fun and discussions]

  • #1
Simon Bridge
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“As per Inspector Lestrade, the accused house is on the bank of a river which is quarter mile wide and has current of 2 mph“. said Dr. Watson

“Don’t forget Watson, Just opposite his house, on the other bank, is the victims house. The accused has only to swim across!” Sherlock Holmes added.

“But its impossible for him to reach there in less than ten minutes”. exclaimed Dr. Watson. “The plain facts of the matter are, the accused can swim at 2.5 mph in still water and his walking speed also is 2.5 mph."

"Perhaps he can do it faster by another combination of swimming and walking, hmm Watson?" murmured Holmes, lighting his pipe.

Elementary?

Of course the accused could have skated across and into the victim's house via a certain yellow citrus fruit ...
 

Answers and Replies

  • #2
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suppose the swimming speed is v, the speed of the river r, and the width of the rivier w.

suppose the accused swims at an angle [itex] \phi [/itex] upstream of straight across.

The component of v across the river is [itex] v \cos{\phi} [/itex], and the upstream component is [itex] v \sin {\phi} [/itex].

The time to reach the other side of the river is:

[tex] \frac {w} {v \cos {\phi}} [/tex]

the downstream drift during this time is

[tex] \frac {w r} {v \cos {\phi}} [/tex]

and the upstream progress made by swimming

[tex] \frac {w v \sin {\phi}} {v \cos {\phi}} [/tex]

note that the downstream drift must be larger than the upstream swimming velocity, or you'll end up upstream of the target and then we won't get a valid solution.

The walking time will be:

[tex] \frac {1}{v} (\frac {w r} {v \cos {\phi}} - \frac {w v \sin {\phi}} {v \cos {\phi}}) [/tex]

wich is [tex] \frac {w} {v^2 \cos{\phi} }(r - v \sin {\phi}) [/tex]

The total time is:

[tex] \frac {w} {v^2 \cos {\phi} }(v + r - v \sin {\phi}) [/tex]

finding the minimum will follow, but I discovered a mistake
 
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  • #3
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differentiate the total time with respect to phi, and you'll get.

[tex] \frac {w} {v^2 \cos^2 {\phi} }(v \sin {\phi} + r \sin {\phi} - v \sin^2 {\phi} -v \cos^2 {\phi}) = \frac {w} {v^2 \cos^2 {\phi} } ( v \sin {\phi} + r \sin {\phi} - v) [/tex]

If you set this to 0 to find a minimum you'll get

[tex] ( v \sin {\phi} + r \sin {\phi} - v) = 0 [/tex] so

[tex] \sin {\phi} = \frac {v} {v+r} [/tex]

(there must be an easier way to get this)

In this case v = 2.5mph, r = 2mph, w = 0.25miles

sin(phi) = 4/9, phi = 26.39 degrees , cos(phi) = 0.8958

swimming time = 0.25/(2.5 * 0.8958) = 0.1163 hours = 6 minutes 59 seconds

walking time = 0,25/ (2,5*2,5* 0.8958) (2 - 2,5 * 4/9) = 0.0397 hours = 2 minutes 23 seconds

total time is 9 minutes and 22 seconds.
 
  • #4
cepheid
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I'm interested to find out what the OP intended for a solution (if indeed he/she had something specific in mind).
 
  • #5
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The OP posted at: Nov7-11, 12:53 AM
First reply at: Nov7-11, 11:22 AM

Is this finally concrete proof of FTL or is this more elementary?
 
  • #6
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The OP posted at: Nov7-11, 12:53 AM
First reply at: Nov7-11, 11:22 AM

Is this finally concrete proof of FTL or is this more elementary?
I suppose 12:53 AM is 53 minutes after midnight and 11:22 am is 38 minutes before noon. In my time zone the times are 6:53 AM and 5:22 PM.
 
  • #7
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If he swims angled so that his path goes straight across, his straight path progress is at 1.5mph and it takes him 10 minutes, and the heading angle comes out to 53 degrees. I guess that does not count as "less than 10 minutes".

If he swims at no angle and lets the current take him, he crosses in 6 minutes, but .2 miles downstream from his victim, so total time is 10.8 minutes, so 48 seconds to late.

If he swims at 45 degrees his progress both across and against the current is at 1.77mph and he drifts downstream at 2-1.77=.34mph. The crossing takes .25m / 1.77mph=.141hour or 8.47minutes (8m28s). Drifting at .34mph for .141 hours leaves him .048 miles from the victim. At 2.5mph walking the .048 miles takes 1.15 minutes (1m9s).

9 minutes 37 seconds
 
  • #8
Simon Bridge
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Some people were posting puzzles so I thought I'd try one.
(exactly 10mins does not allow enough time for the murder - it seems - but a nanosecond early is fine?
I didn't make this problem up - 'cause I don't write that goodly.)

The rigorous approach shows the range of angles the suspect could have attacked the crossing and still got to the victim's house in time to commit the murder.
All that is needed is one such path. This is probably the one we'd show to a jury to demonstrate opportunity.

We could ask - how far back from the river-bank could the victim's house be and still allow a travel-time less than 10mins?
 
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