Last edited by richardg; 13th May 2008 at 10:31 PM.
hasta la justicia siempre
each car needs to travel 0.5 km to meet
0.5km / 10km/h = 0.05h
Since bee is twice the speed of Car A, it will meet Car B at 0.025h.
The new distance between the 2 Cars after 0.025h is : 0.025 x 20km/h = 0.5km
each car again needs to travel 0.25km
0.25km / 10km/h = 0.025h
since bee is twice the speed of Car, it will be back at Car A at 0.0125h.
BEE BACK AND FORTH ONCE
The new distance between the 2 cars after 0.0375h is 0.5-(0.0125 x 20km/h) = 0.25km
each car needs to travel 0.125 km to meet
0.125 km / 10km/h = 0.0125h
since bee is twice the speed of car, it will meet car B at 0.00625h
The new distance between the 2 cars after 0.04375h is 0.25 -(0.00625 x 20km/h) = 0.125km
each car needs to travel 0.0625km
0.0625 km / 10km/h = 0.00625h
since bee is twice the speed of car, it will be back at Car A at 0.003125h
BEE BACK AND FORTH twice
the new distance between the 2 cars after 0.046875h is 0.125 - (0.003125 x 20) = 0.0625km
each car needs to travel 0.03125km
0.03125 km / 10km/h = 0.003125h
since the bee is twice the speed of car, it will meet car B at 0.0015625h
The new distance between the 2 cars after 0.0484375h is 0.0625 - (0.0015625 x 20) = 0.03125km
each car needs to travel 0.015625km
0.015625km / 10 km/h = 0.0015625h
since the bee is twice the speed of car, it will be back at car A at 0.0078125h
The new distance between the 2 cars after 0.05625h is negative already
Since cars will meet after 0.05h it is impossible to reach 0.05625h
therefore, bee will not be able to be back at Car A
Ans: bee will be able to go back and forth 2 times
7 times....
The question is "How many times can the bee MOVE back and forth between the cars before it got crushed in between?" and the assumption is the bee can change direction with no time lack and acceleration is negligible but didn't assume a MOVE must be how long a distance.
Infinite is right if a 1/10^infinite m traveled is considered a MOVE and will be wrong if the practical MOVE distance is defined.
D300 & P5100 :bsmilie:
Ignoring the assumptions (as they make asses of you and me), the answer is none, coz the bee was trying to cross the road...
Don't Dance with the Devil in the Pale Moonlight...
Will the bee stupid enough to crash on the second car in the first place and stay in between until it got crushed? I think answer = 0 (since it can move any direction, it opts to fly away).
This is between theoretical and a practical question. In the 1st place, it is not a practical question, how can a car travel slower than a bee. However, in theory, any thing can happen be it how stupid the question is. So a stupid question will end up with a lot stupid answers.
D300 & P5100 :bsmilie:
each car needs to travel 0.5 km to meet
0.5km / 10km/h = 0.05h
bee will meet Car B at 1/30h.
The new distance between the 2 Cars after 1/30h is : 1km -(1/30 x 20km/h) = 1/3km
each car again needs to travel 1/6km
1/6km / 10km/h = 1/60h
since new distance is 1/3km, it will be back at Car A after another 1/90h.
The new distance between the 2 cars after 2/45h is 1/3 -(1/90 x 20km/h) = 1/9km
Back and forth 1 time
since new distance is 1/9km, bee will be back at Car B after another 1/270h
The new distance between the 2 cars after 13/270h is 1/9 -( 1/270 x 20) = 1/27km
since new distance is 1/27km, bee will be back at Car A after another 1/810h
The new distance between the 2 cars after 4/81h is 1/27 - (1/810 x 20 ) = 1/81km
back and forth 2 times
since new distance is 1/81km bee will be back at Car B after another 1/2430h
The new distance between the 2 cars after 121/2430h is 1/81 - (1/2430 x20) = 1/243km
since new distance is 1/243km, bee will be back at Car A after another 1/7290h
The new distance between the 2 cars after 361/7290h is 1/243 - (1/7290 x 20) = 1/729km
back and forth 3 times
since new distance is 1/729km, bee will be back at car B after another 1/21870h
The new distance between the 2 cars after 542/10935h is 1/729 - (1/21870 x 20) = 1/2187km
since new distance is 1/2187km, bee will be back at car A after another 1/65610h
The new distance between the 2 cars after 3253/65610h is 1/2187 - (1/65610 x 20) = 1/6561km
back and forth 4 times
I seriously have no patient to exhaustively test this series, and i do not wish to flip textbook to check how to prove converge or diverge....hahaha
anyway exam is over...why am i still doing maths....(take out psp)
In that case I would have 2 answers: Minimum of hit is 1 (it didn't hit at all in between until it got crush), Maximum of hit is Uncountable (I don't think it's infinite but just uncountable) It's a curve upward man (if you even bother to draw it out)....never end.
Ok... my personal answer...
Like what many have attempted to calculate, the time taken and change in distance decreased by 1/3 each time...
Hence, it follows a Geometric Series, S = a / (1 - r), where this series is infinite (S = time taken for the cars to meet, a = initial time for bee to travel from car A to car B, r = 1/3)
Mathematically, it seems like the bee can travel to and forth infinitely... But logically, it has to be a finite number isn't it?
I think the above solving method is somewhat incorrect because it is similar to "Achilles and the tortoise"...
Answer is Infiite as long as the bee is smaller than the length of the distance between the 2 cars. This is so because it is too unrealistic to have 0 reaction time required to change direction and accelerate to 20 km/hr. Even when the bee is only 0.1cm smaller than the distance between the 2 cars, it can go to and fro for millions of times/msec before the 2 car squeeze the bee together from this poor assumption.
It is essentially the problem of finding the smallest rational number. No one has ever found it and no one can ever find it.
Do you undersatnd how the Achiles and the tortise thing works?
I've read it at Wikipedia and this is what it says:
I find this difficult to understand because we know in real life that the faster runner can always overtake the slower. In the example above, if Achilles covers 100 feet in the time the tortoise takes to cover 10 feet, he will cover another 100 feet when the tortise covers another 10 feet. This makes Achilles 200 feet from his starting point, or 100 feet from the tortoise's starting point, and the tortoise is only 20 feet from the tortoise's starting point.Originally Posted by Achilles and the tortoise
“In a race, the quickest runner can never overtake the slowest, since the pursuer must first reach the point whence the pursued started, so that the slower must always hold a lead. ”
—Aristotle, Physics VI:9, 239b15
In the paradox of Achilles and the Tortoise, Achilles is in a footrace with the tortoise. Achilles allows the tortoise a head start of 100 feet. If we suppose that each racer starts running at some constant speed (one very fast and one very slow), then after some finite time, Achilles will have run 100 feet, bringing him to the tortoise's starting point. During this time, the tortoise has run a much shorter distance for example 10 feet. It will then take Achilles some further period of time to run that distance, in which period the tortoise will advance farther; and then another period of time to reach this third point, while the tortoise moves ahead. Thus, whenever Achilles reaches somewhere the tortoise has been, he still has farther to go. Therefore, because there are an infinite number of points Achilles must reach where the tortoise has already been, he can never overtake the tortoise. [5
a more open mind would help.
instead of thinking about the tortoise covering another 10 feet, think about how achilles has to cover the distance between him and the tortoise.
since the tortoise is always moving, whenever achilles catches up to the "last marked point" of the tortoise, he has a distance to go, even though he is much faster.. this catch-up distance gets smaller and smaller.
let's say eventually at the distance of 0.1 cm between them. the time taken for achilles to catch up is so minutely small that in just the next nanosecond he might have "caught up" to the tortoise 1000000000000000 (large number) times. get it?
and yes, mathematically there will be a solution as to how fast he takes to catch up, and it is not hard using relatively velocity from basic physics, or just fundamental brute force mathematics.
but if the question asks you, how many times achilles has to "catch up" to the "last marked point" of the tortoise, the answer is infinite number of points.
same for the bee thing. the crux lies in the questions you CAN answer, and the questions to which there are no answers. this is after all, a rather theoretical abstract concept which might have no practical bearing, but it isn't all that unimaginable.
Last edited by night86mare; 14th May 2008 at 06:41 PM.
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