Puzzle: solve this problem in your head
This problem initially had two parts but now has three.
OK, let's start with this: let's assume that we have two, trains, 100 miles apart, each going with 20 mph to the other. A fly travels between them at 50 mph, zig-zagging just before getting smashed between the trains. Question: How far does the fly fly before meeting his ultimate demise?
Got the answer? Good. Now, let's modify the original problem a little: let's also assume that the wind blows with 10 mph from east, and the trains are going on a straight railway, oriented east-west. Relative to the ground, what is the total distance traveled by the fly this time?
Now the third part. Let's assume that the fly starts from a point between the two trains (let's say "x miles from the west train"). What would be the value of "x" if the total distance is the same no matter what is the initial direction of the fly?
[update: on the second part, let's assume that the fly starts from the train from the west. Also - each train has 20 mph ground speed. Finally, another update: I added a third part.]
Comments
- Anonymous
May 19, 2005
The comment has been removed - Anonymous
May 19, 2005
The comment has been removed - Anonymous
May 19, 2005
The fly's initial position and direction should be irrelevant even with the wind. - Anonymous
May 19, 2005
- Since each travels at the speed with opposing velocity, they collide at the 50-mile marker. Each train takes 2.5 hours to travel the 50 miles to the collision point. A fly zig-zagging at 50mph would travel a total of 125 miles to reach the same collision point.
2) Assuming the easterly wind has no effect on the speed of the trains, they still take 2.5 hours to collide. With the fly starting at the western side of the course and flying into a 10mph easterly, it's groundspeed will be 40mph. So, in 2.5 hours, the fly travels 100 miles.
- Anonymous
May 19, 2005
OK, then as long as the fly really does meet his demise at the train crash, he travelled from his point of origin (the western train) to the point of train/train impact, which is a distance of 50 miles. The windspeed doesn't matter since you have fixed the two endpoints.
This also assumes that you mean "total distance traveled" in the same way that we hang glider/sailplane pilots use that phrase: the great circle distance between the launch/release (origin) and landing (destination) points. In flight, we may zig-zag quite a bit in search of the next thermal, and drift in a variety of directions as we climb in a rowdy thermal, or go back and forth many times trying to stay aloft in weak ridge lift while we wait and pray for another thermal... but we don't get credit for those "distances travelled"! - Anonymous
May 19, 2005
The comment has been removed - Anonymous
May 19, 2005
Sorry, meant 50 miles in that illustration... it's late... but you get the point. :) - Anonymous
May 19, 2005
OK - I added more precision to the text: the trains are going with 20 mph ground speed, which means that they are approachig each other with 40 mph relative speed.
Also: a hint. The fly has 50 mph constant speed relative to the air, not to the ground... - Anonymous
May 19, 2005
d = 110 miles
fly_e_to_w = 50 + 10 = 60
fly_w_to_e = 50 - 10 = 40
train_e_to_w = train_w_to_e = 20
d = fly_w_to_e * t0 + fly_e_to_w * t1 + fly_w_to_e * t2 + ...
= fly_w_to_e * (t0 + t2 + t4 + ...)
+ fly_e_to_w * (t1 + t3 + t5 + ...)
where
t0 = 100 / (fly_w_to_e + train_e_to_w)
= 5/3
t1 = (fly_w_to_e - train_w_to_e) * t0 / (fly_e_to_w + train_w_to_e)
= 20 * t0 / 80
= 1/4 * t0
t2 = (fly_e_to_w - train_e_to_w) * t1 / (fly_w_to_e + train_w_to_e)
= 40 * t1 / 60
= 2/3 * t1
t3 = 1/4 * t2
t4 = 2/3 * t3
...
And...
t0 + t2 + t4 + ...
= t0 + 2/3 * t1 + 2/3 * t3 + ...
= t0 + 2/3 * 1/4 * t0 + ((2/3 * 1/4)^2) * t0+ ....
= t0 + 1/6 t0 + (1/6)^2 t0 + ...
= t0 * ( 1 + 1/6 + (1/6)^2 + ...)
= 1.2 t0
Also...
t1 + t3 + t5 + ...
= 1/4 t0 + 1/4 * t2 + 1/4 * t4
= 1/4 * t0 * ( 1 + 1/6 + (1/6)^2 + ... )
= 0.3 t0
Therefore
d = fly_w_to_e * (t0 + t2 + t4 + ...)
+ fly_e_to_w * (t1 + t3 + t5 + ...)
= 40 * 1.2 * t0
+ 60 * .3 * t0
= 110 - Anonymous
May 19, 2005
The comment has been removed - Anonymous
May 19, 2005
>> d = 110 miles [...]
Excellent. The solution is actually correct but why did you used infinite series? There is a simpler way to find it :-) - Anonymous
May 19, 2005
The comment has been removed - Anonymous
May 20, 2005
The comment has been removed - Anonymous
May 20, 2005
The comment has been removed - Anonymous
May 20, 2005
By the way, third part is the hardest one. You can use paper and pencil if you feel so :-) - Anonymous
May 20, 2005
What do you mean "the value of x"? x is clearly not unique. - Anonymous
May 20, 2005
>> By the way, third part is the hardest one. You can use paper and pencil if you feel so :-)
Please ignore this comment - I just found out a simple solution for the third part...
>>> What do you mean "the value of x"? x is clearly not unique.
OK - that's part of the solution :-). So what are the possible values of x? - Anonymous
May 20, 2005
Ok, understood.
It should be true for every value of x.
Use the same technique as before: total travel is computed by reaching the middle and then solving the simpler ivp.
Suppose the fly is facing away from the nearest train. It will travel at M mph to the middle (A miles) then average 48 mph for the remaining trip (B miles).
Suppose the fly is facing towards the nearest train. It will travel to the train (C miles), travel back to its starting point (another C miles), travel at M mph to the middle (A miles), then average 48 mph for the remaining trip (D miles).
But one of those trips of C miles is at 40 mph and the other is at 60 mph so that's just more travel at average of 48 mph.
So in both cases you travel at M mph for A miles then average 48 mph for the remaining trip. It doesn't matter what M and A are. - Anonymous
May 20, 2005
The comment has been removed - Anonymous
May 20, 2005
The comment has been removed - Anonymous
May 20, 2005
There is something wrong with your formula. The actual distance traveled (with respect to the initial position 0 <= X <= 100) is:
D(X) = 110 + 0.2 * X
Or, if you take -50 <= Y <= 50, then:
D(Y) = 120 + 0.2 * Y
Thanks, Adi - Anonymous
May 20, 2005
I assure you the two formulas are entirely the same! :)
D(X)=|X|+48(2.5-|X|/(50+10X/|X|))
D(X)=|X|+120-48|X|/(50+10X/|X|)
D(X)=120+(|X|(50+10X/|X|)-48|X|)/(50+10X/|X|)
D(X)=120+(50|X|+10X-48|X|)/(50+10X/|X|)
D(X)=120+(2|X|+10X)/(50+10X/|X|)
D(X)=120+0.2(2|X|+10X)/(10+2X/|X|)
D(X)=120+0.2(2|X|+10X)/(10+2|X|/X)
D(X)=120+0.2(2|X|+10X)/((1/X)(10X+2|X|))
D(X)=120+0.2(1/(1/X))(2|X|+10X)/(10X+2|X|)
D(X)=120+0.21/(1/X)
D(X)=120+0.2X - Anonymous
May 20, 2005
The comment has been removed - Anonymous
May 21, 2005
OK, here is the solution:
Part 1. Already solved above. The trains have 40 mph relative speed. Since the fly flies with 50 mph, the total traveled distance will be 50/40 * 100 = 125 miles.
Part 2. This is also solved above by Eric (the otherEric) but with infinite series. The solution without infinite series is very similar, but with a twist.
The proof is actually pretty simple but here I am as verbose as possible - I don't want to present a sketchy proof.
The fly will start going east. Since the wind has opposite direction, its speed will be (50-10) = 40 mph. Now, due to the fact that this is twice the speeed of the opposite train, it will travel twice as much until to the first "collision" which means 2/3 from the total distance (i.e. 2/3 * 100 miles), while the train traveling only 1/3 * 100 miles.
So at this moment the distance between trains is 100 - 2/3 * 100 = 1/3 * 100. Now the fly goes back with 60 mph ( = 50 + 10 mph). Since it goes three times as fast as the opposing train, it will travel three times more until the second "collision". This means 3/4 * 1/3 * 100 miles, while the opposite train 1/4 * 1/3 * 100 miles. Also, both trains will travel the same distance, so now the distance between them is 1/6 * 100 miles.
At this point, let's step back a little and see the facts.
1) Now we are in the same configuration as in the beginning, just that the distance between trains is six times smaller.
3) The distance traveled by the fly is (2/3 + 3/4 * 1/3) * 100 = 11/12 * 100 miles.
So, the trains traveled until now (1 - 1/6) = 5/6 from their total distance until the collision. The fly did the same, so the total distance will be 6/5 * 11/12 * 100 miles = 110 miles. Done!
As an exercise, you can do the same calculations for the case when the fly starts from east. You will get 130 miles instead.
Part 3: already solved above in two different ways.