A car is moving at 40 km/h. A fly at 100 km/h, starts from wall towards the car(20 km away)flies to car and back. How many trips can it make?'Bee flying between two trains' problemProblem: Two Trains and a FlyFly and Two Trains Riddle2 Trains and Fly Problem. Find the number of trips made by the fly back and forth.Related Rates Problem With LadderFind time when 2 cars meet?Velocity and distance problemTotal time spent travelling, given distance and speed functionsIf something is $2^N+1$, how can I get $N$ back from the end result?help with understanding the solution(time speed and distance)Determining time it takes for two approaching cars to meetUsing the digits $1$, $2$, $3$, $7$, $8$, $9$, and $0$, how many $4$-digit numbers can be created that are greater than $3718$?For how many pairs of distinct positive integers $a$ and $b$, both less than $100$, is $dfracab$ the square of an integer?Four cars A,B,C,D are moving at constant speeds on the same road
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A car is moving at 40 km/h. A fly at 100 km/h, starts from wall towards the car(20 km away)flies to car and back. How many trips can it make?
'Bee flying between two trains' problemProblem: Two Trains and a FlyFly and Two Trains Riddle2 Trains and Fly Problem. Find the number of trips made by the fly back and forth.Related Rates Problem With LadderFind time when 2 cars meet?Velocity and distance problemTotal time spent travelling, given distance and speed functionsIf something is $2^N+1$, how can I get $N$ back from the end result?help with understanding the solution(time speed and distance)Determining time it takes for two approaching cars to meetUsing the digits $1$, $2$, $3$, $7$, $8$, $9$, and $0$, how many $4$-digit numbers can be created that are greater than $3718$?For how many pairs of distinct positive integers $a$ and $b$, both less than $100$, is $dfracab$ the square of an integer?Four cars A,B,C,D are moving at constant speeds on the same road
$begingroup$
A car is moving at a constant speed 40 km/h along a straight road which heads towards a wall.A fly flying at a constant speed of 100 km/h, starts from wall the towards the car at a instant when the car is 20 km away, flies until it reaches the car and comes back to the wall at the same speed.It continues to fly between the car and the wall till the car reach the wall. How many trips has it made between the car and the wall?
I don't expect a brute force because I already did that. Some thing like arithmetic/geometric/harmonic progression will satisfy my curiosity.
algebra-precalculus
asked Mar 25 at 14:51
user654700user654700
706
$endgroup$
|
show 6 more comments
$begingroup$
A car is moving at a constant speed 40 km/h along a straight road which heads towards a wall.A fly flying at a constant speed of 100 km/h, starts from wall the towards the car at a instant when the car is 20 km away, flies until it reaches the car and comes back to the wall at the same speed.It continues to fly between the car and the wall till the car reach the wall. How many trips has it made between the car and the wall?
I don't expect a brute force because I already did that. Some thing like arithmetic/geometric/harmonic progression will satisfy my curiosity.
algebra-precalculus
asked Mar 25 at 14:51
user654700user654700
706
$endgroup$
6
$begingroup$
In theory infinitely many.
$endgroup$
– fleablood
Mar 25 at 14:53
4
$begingroup$
This 100 km/h fly is a very impressive aviator given that ordinary houseflies have a max speed of 8 km/h. Even more impressive is its ability to instantaneously go from +100 km/h to -100km/h without feeling a bit groggy for a moment.
$endgroup$
– RedGrittyBrick
Mar 25 at 17:05
2
$begingroup$
Everyone seems to forget Applied Mathematic is not as it should be when the car reaches the point where the gap is less than the size of the flies body it dies of the pressure if not exhaustion.
$endgroup$
– KJO
Mar 25 at 17:47
2
$begingroup$
At least it's not a spherical cow....
$endgroup$
– Kathy
Mar 25 at 19:36
1
$begingroup$
Possible duplicate of 'Bee flying between two trains' problem
$endgroup$
– Herohtar
Mar 25 at 20:47
|
show 6 more comments
$begingroup$
A car is moving at a constant speed 40 km/h along a straight road which heads towards a wall.A fly flying at a constant speed of 100 km/h, starts from wall the towards the car at a instant when the car is 20 km away, flies until it reaches the car and comes back to the wall at the same speed.It continues to fly between the car and the wall till the car reach the wall. How many trips has it made between the car and the wall?
I don't expect a brute force because I already did that. Some thing like arithmetic/geometric/harmonic progression will satisfy my curiosity.
algebra-precalculus
asked Mar 25 at 14:51
user654700user654700
706
$endgroup$
A car is moving at a constant speed 40 km/h along a straight road which heads towards a wall.A fly flying at a constant speed of 100 km/h, starts from wall the towards the car at a instant when the car is 20 km away, flies until it reaches the car and comes back to the wall at the same speed.It continues to fly between the car and the wall till the car reach the wall. How many trips has it made between the car and the wall?
I don't expect a brute force because I already did that. Some thing like arithmetic/geometric/harmonic progression will satisfy my curiosity.
algebra-precalculus
algebra-precalculus
asked Mar 25 at 14:51
user654700user654700
706
asked Mar 25 at 14:51
user654700user654700
706
asked Mar 25 at 14:51
user654700user654700
706
asked Mar 25 at 14:51
user654700user654700
706
asked Mar 25 at 14:51
user654700user654700
706
706
6
$begingroup$
In theory infinitely many.
$endgroup$
– fleablood
Mar 25 at 14:53
4
$begingroup$
This 100 km/h fly is a very impressive aviator given that ordinary houseflies have a max speed of 8 km/h. Even more impressive is its ability to instantaneously go from +100 km/h to -100km/h without feeling a bit groggy for a moment.
$endgroup$
– RedGrittyBrick
Mar 25 at 17:05
2
$begingroup$
Everyone seems to forget Applied Mathematic is not as it should be when the car reaches the point where the gap is less than the size of the flies body it dies of the pressure if not exhaustion.
$endgroup$
– KJO
Mar 25 at 17:47
2
$begingroup$
At least it's not a spherical cow....
$endgroup$
– Kathy
Mar 25 at 19:36
1
$begingroup$
Possible duplicate of 'Bee flying between two trains' problem
$endgroup$
– Herohtar
Mar 25 at 20:47
|
show 6 more comments
6
$begingroup$
In theory infinitely many.
$endgroup$
– fleablood
Mar 25 at 14:53
4
$begingroup$
This 100 km/h fly is a very impressive aviator given that ordinary houseflies have a max speed of 8 km/h. Even more impressive is its ability to instantaneously go from +100 km/h to -100km/h without feeling a bit groggy for a moment.
$endgroup$
– RedGrittyBrick
Mar 25 at 17:05
2
$begingroup$
Everyone seems to forget Applied Mathematic is not as it should be when the car reaches the point where the gap is less than the size of the flies body it dies of the pressure if not exhaustion.
$endgroup$
– KJO
Mar 25 at 17:47
2
$begingroup$
At least it's not a spherical cow....
$endgroup$
– Kathy
Mar 25 at 19:36
1
$begingroup$
Possible duplicate of 'Bee flying between two trains' problem
$endgroup$
– Herohtar
Mar 25 at 20:47
6
6
$begingroup$
In theory infinitely many.
$endgroup$
– fleablood
Mar 25 at 14:53
$begingroup$
In theory infinitely many.
$endgroup$
– fleablood
Mar 25 at 14:53
4
4
$begingroup$
This 100 km/h fly is a very impressive aviator given that ordinary houseflies have a max speed of 8 km/h. Even more impressive is its ability to instantaneously go from +100 km/h to -100km/h without feeling a bit groggy for a moment.
$endgroup$
– RedGrittyBrick
Mar 25 at 17:05
$begingroup$
This 100 km/h fly is a very impressive aviator given that ordinary houseflies have a max speed of 8 km/h. Even more impressive is its ability to instantaneously go from +100 km/h to -100km/h without feeling a bit groggy for a moment.
$endgroup$
– RedGrittyBrick
Mar 25 at 17:05
2
2
$begingroup$
Everyone seems to forget Applied Mathematic is not as it should be when the car reaches the point where the gap is less than the size of the flies body it dies of the pressure if not exhaustion.
$endgroup$
– KJO
Mar 25 at 17:47
$begingroup$
Everyone seems to forget Applied Mathematic is not as it should be when the car reaches the point where the gap is less than the size of the flies body it dies of the pressure if not exhaustion.
$endgroup$
– KJO
Mar 25 at 17:47
2
2
$begingroup$
At least it's not a spherical cow....
$endgroup$
– Kathy
Mar 25 at 19:36
$begingroup$
At least it's not a spherical cow....
$endgroup$
– Kathy
Mar 25 at 19:36
1
1
$begingroup$
Possible duplicate of 'Bee flying between two trains' problem
$endgroup$
– Herohtar
Mar 25 at 20:47
$begingroup$
Possible duplicate of 'Bee flying between two trains' problem
$endgroup$
– Herohtar
Mar 25 at 20:47
|
show 6 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Suppose the fly is at the wall. And suppose this will be the last trip or partial trip of the fly. Suppose the car is $h$ miles away.
The fly and and car have a combined speed of $140 frac kmhr$ so the fly reaches the car in $frac h140$ hours. In that time the car has traveled $40frac h140 = frac 27h$ and is now $h-frac 27h = frac 57h$ from the wall. So the fly heads back to the wall.
As the trip back is just as far this takes $frac h140$ hours and the car has traveled another $frac 27h$ and is now $frac 37h$ from the wall. [1]
So the fly starts another trip, contradicting that this was his last. So the fly never makes a last trip an instead there are an infinite number of trips.
Figuring out how far the fly flies is a matter of noting the car is on a straight path and travels $20km$ at $40 kmh$ so this takes $30$ minutes. The fly no matter how many times (infinitely many) it zigs will travel at $100kmh$. So in $30$ minutes it flies $50 km$.
If one wishes to set this up as an infinite sum.....
Each trip the fly flies $frac 107$ of the distance the car was away. And each trip the car is $frac 37$ of the distance it was before. So the distance the fly travels is $sum_k=0^infty frac 107*(frac 37)^k*20$ which if I did this correctly is
$frac 107*20(sum_k=0^infty (frac 37)^k)= frac 2007frac 11-frac 37 = frac 2007frac 74= 50$km.
[1](for the record, in this time, $frac 170h$ hours, the car has traveled $frac 47h$ and the fly has travelled $frac 107h$).
answered Mar 25 at 15:39
fleabloodfleablood
73.7k22891
$endgroup$
add a comment |
$begingroup$
It has made infinitely many trips. Every trip will be shorter than the last, but the fly will always reach the wall before the car, so it will always have room for one more trip. And one more. And one more.
The total length the fly flies is 50km, as the car crashes into the wall exactly 30 minutes after the whole experiment started.
answered Mar 25 at 14:53
ArthurArthur
121k7122208
$endgroup$
$begingroup$
I got that but can you please give me more rigorous proof? If you don't mind.
$endgroup$
– user654700
Mar 25 at 14:56
$begingroup$
@user654700 What exactly do you find not rigorous enough?
$endgroup$
– Arthur
Mar 25 at 15:01
8
$begingroup$
@user654700: This is rigorous. One can also write an equation for the length of each trip the fly makes and sum the series, but under the assumptions that is not needed. Don't confuse rigor with fancy mathematical symbols. There is a famous anecdote about John von Neumann and this puzzle.
$endgroup$
– Ross Millikan
Mar 25 at 15:02
$begingroup$
I suppose there are some minor details that could be added to give it some more rigor (or at least make the rigor more obvious): when the fly is at the position of the car, it has to travel the same distance to get to the wall, but its speed is greater, so it will always reach first. Once the fly is at the wall, we know that the car is not yet at the wall, due to the above reasoning. Since the fly has nonzero speed and the car does not have "infinite" speed, the fly and the car will next meet at some point between their current positions (not at the wall). Repeat.
$endgroup$
– inavda
Mar 25 at 15:08
$begingroup$
Well, one can figure that if the fly is at the wall and the car is a non-zero distance away that the fly flies to the car and that takes some non-zero time (they have a combined speed which is finite so to cover a distance takes non-zero time). The fly will then fly to the wall. As the fly is faster it reaches the wall first. The car is now a non-zero distance away. Thus is the same situation as before so by induction there will be an infinite number of trips . That is formal and rigorous.
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– fleablood
Mar 25 at 15:15
|
show 1 more comment
$begingroup$
Simple way to calculate it: It takes the car 30 minutes to travel the 20km to the wall at 40km/hr. The fly, traveling at 100km/hr will travel 50km in those same 30 minutes.
Edit: There will be an infinite number of trips. It's similar to how Zeno's paradox works where the trips get shorter and shorter and eventually take an infinitely small amount of time. But all those infinitely small trips end up as a finite amount of distance.
answered Mar 25 at 19:09
Guest3711Guest3711
312
$endgroup$
add a comment |
$begingroup$
Arthur's answer is very good, but here is another (equally valid) way of visualizing the problem:
Let us graph the positions of the wall, fly, and car over time. The wall doesn't move, so it is represented by a horizontal line. The car starts at some distance away from the wall but moves towards it at a constant speed until it hits the wall -- so we have a line that intersects the wall's line. Now for the interesting part:
The fly's path starts off as a line with greater slope than the car's line, until it hits the wall's line. The fly's speed remains the same, but it is going in the opposite direction. So now the path continues as if the wall's line was a mirror and it was reflected. When the fly hits the car, the same thing happens -- the fly's path is reflected and it continues towards the wall again (one caveat: when it bounces off the car's line, the angles of incidence and reflection are not equal so it isn't behaving exactly as light would).
So we can see that the fly's path continues bouncing up and down, always at the same or opposite slope.
The final thing to notice to grasp the intuition is that this diagram we have constructed is self-similar. If we zoom in so that the second bounce with the wall is where the first bounce used to be, we have the same exact diagram as before. I won't prove this, but if you draw it out, you can see it intuitively. Essentially, this means that no matter how close the car gets to the wall, we can zoom in and see more bounces for the fly.
answered Mar 25 at 15:04
inavdainavda
1198
$endgroup$
add a comment |
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4 Answers
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4 Answers
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$begingroup$
Suppose the fly is at the wall. And suppose this will be the last trip or partial trip of the fly. Suppose the car is $h$ miles away.
The fly and and car have a combined speed of $140 frac kmhr$ so the fly reaches the car in $frac h140$ hours. In that time the car has traveled $40frac h140 = frac 27h$ and is now $h-frac 27h = frac 57h$ from the wall. So the fly heads back to the wall.
As the trip back is just as far this takes $frac h140$ hours and the car has traveled another $frac 27h$ and is now $frac 37h$ from the wall. [1]
So the fly starts another trip, contradicting that this was his last. So the fly never makes a last trip an instead there are an infinite number of trips.
Figuring out how far the fly flies is a matter of noting the car is on a straight path and travels $20km$ at $40 kmh$ so this takes $30$ minutes. The fly no matter how many times (infinitely many) it zigs will travel at $100kmh$. So in $30$ minutes it flies $50 km$.
If one wishes to set this up as an infinite sum.....
Each trip the fly flies $frac 107$ of the distance the car was away. And each trip the car is $frac 37$ of the distance it was before. So the distance the fly travels is $sum_k=0^infty frac 107*(frac 37)^k*20$ which if I did this correctly is
$frac 107*20(sum_k=0^infty (frac 37)^k)= frac 2007frac 11-frac 37 = frac 2007frac 74= 50$km.
[1](for the record, in this time, $frac 170h$ hours, the car has traveled $frac 47h$ and the fly has travelled $frac 107h$).
answered Mar 25 at 15:39
fleabloodfleablood
73.7k22891
$endgroup$
add a comment |
$begingroup$
Suppose the fly is at the wall. And suppose this will be the last trip or partial trip of the fly. Suppose the car is $h$ miles away.
The fly and and car have a combined speed of $140 frac kmhr$ so the fly reaches the car in $frac h140$ hours. In that time the car has traveled $40frac h140 = frac 27h$ and is now $h-frac 27h = frac 57h$ from the wall. So the fly heads back to the wall.
As the trip back is just as far this takes $frac h140$ hours and the car has traveled another $frac 27h$ and is now $frac 37h$ from the wall. [1]
So the fly starts another trip, contradicting that this was his last. So the fly never makes a last trip an instead there are an infinite number of trips.
Figuring out how far the fly flies is a matter of noting the car is on a straight path and travels $20km$ at $40 kmh$ so this takes $30$ minutes. The fly no matter how many times (infinitely many) it zigs will travel at $100kmh$. So in $30$ minutes it flies $50 km$.
If one wishes to set this up as an infinite sum.....
Each trip the fly flies $frac 107$ of the distance the car was away. And each trip the car is $frac 37$ of the distance it was before. So the distance the fly travels is $sum_k=0^infty frac 107*(frac 37)^k*20$ which if I did this correctly is
$frac 107*20(sum_k=0^infty (frac 37)^k)= frac 2007frac 11-frac 37 = frac 2007frac 74= 50$km.
[1](for the record, in this time, $frac 170h$ hours, the car has traveled $frac 47h$ and the fly has travelled $frac 107h$).
answered Mar 25 at 15:39
fleabloodfleablood
73.7k22891
$endgroup$
add a comment |
$begingroup$
Suppose the fly is at the wall. And suppose this will be the last trip or partial trip of the fly. Suppose the car is $h$ miles away.
The fly and and car have a combined speed of $140 frac kmhr$ so the fly reaches the car in $frac h140$ hours. In that time the car has traveled $40frac h140 = frac 27h$ and is now $h-frac 27h = frac 57h$ from the wall. So the fly heads back to the wall.
As the trip back is just as far this takes $frac h140$ hours and the car has traveled another $frac 27h$ and is now $frac 37h$ from the wall. [1]
So the fly starts another trip, contradicting that this was his last. So the fly never makes a last trip an instead there are an infinite number of trips.
Figuring out how far the fly flies is a matter of noting the car is on a straight path and travels $20km$ at $40 kmh$ so this takes $30$ minutes. The fly no matter how many times (infinitely many) it zigs will travel at $100kmh$. So in $30$ minutes it flies $50 km$.
If one wishes to set this up as an infinite sum.....
Each trip the fly flies $frac 107$ of the distance the car was away. And each trip the car is $frac 37$ of the distance it was before. So the distance the fly travels is $sum_k=0^infty frac 107*(frac 37)^k*20$ which if I did this correctly is
$frac 107*20(sum_k=0^infty (frac 37)^k)= frac 2007frac 11-frac 37 = frac 2007frac 74= 50$km.
[1](for the record, in this time, $frac 170h$ hours, the car has traveled $frac 47h$ and the fly has travelled $frac 107h$).
answered Mar 25 at 15:39
fleabloodfleablood
73.7k22891
$endgroup$
Suppose the fly is at the wall. And suppose this will be the last trip or partial trip of the fly. Suppose the car is $h$ miles away.
The fly and and car have a combined speed of $140 frac kmhr$ so the fly reaches the car in $frac h140$ hours. In that time the car has traveled $40frac h140 = frac 27h$ and is now $h-frac 27h = frac 57h$ from the wall. So the fly heads back to the wall.
As the trip back is just as far this takes $frac h140$ hours and the car has traveled another $frac 27h$ and is now $frac 37h$ from the wall. [1]
So the fly starts another trip, contradicting that this was his last. So the fly never makes a last trip an instead there are an infinite number of trips.
Figuring out how far the fly flies is a matter of noting the car is on a straight path and travels $20km$ at $40 kmh$ so this takes $30$ minutes. The fly no matter how many times (infinitely many) it zigs will travel at $100kmh$. So in $30$ minutes it flies $50 km$.
If one wishes to set this up as an infinite sum.....
Each trip the fly flies $frac 107$ of the distance the car was away. And each trip the car is $frac 37$ of the distance it was before. So the distance the fly travels is $sum_k=0^infty frac 107*(frac 37)^k*20$ which if I did this correctly is
$frac 107*20(sum_k=0^infty (frac 37)^k)= frac 2007frac 11-frac 37 = frac 2007frac 74= 50$km.
[1](for the record, in this time, $frac 170h$ hours, the car has traveled $frac 47h$ and the fly has travelled $frac 107h$).
answered Mar 25 at 15:39
fleabloodfleablood
73.7k22891
edited Mar 25 at 15:49
answered Mar 25 at 15:39
fleabloodfleablood
73.7k22891
answered Mar 25 at 15:39
fleabloodfleablood
73.7k22891
answered Mar 25 at 15:39
fleabloodfleablood
73.7k22891
73.7k22891
add a comment |
add a comment |
$begingroup$
It has made infinitely many trips. Every trip will be shorter than the last, but the fly will always reach the wall before the car, so it will always have room for one more trip. And one more. And one more.
The total length the fly flies is 50km, as the car crashes into the wall exactly 30 minutes after the whole experiment started.
answered Mar 25 at 14:53
ArthurArthur
121k7122208
$endgroup$
$begingroup$
I got that but can you please give me more rigorous proof? If you don't mind.
$endgroup$
– user654700
Mar 25 at 14:56
$begingroup$
@user654700 What exactly do you find not rigorous enough?
$endgroup$
– Arthur
Mar 25 at 15:01
8
$begingroup$
@user654700: This is rigorous. One can also write an equation for the length of each trip the fly makes and sum the series, but under the assumptions that is not needed. Don't confuse rigor with fancy mathematical symbols. There is a famous anecdote about John von Neumann and this puzzle.
$endgroup$
– Ross Millikan
Mar 25 at 15:02
$begingroup$
I suppose there are some minor details that could be added to give it some more rigor (or at least make the rigor more obvious): when the fly is at the position of the car, it has to travel the same distance to get to the wall, but its speed is greater, so it will always reach first. Once the fly is at the wall, we know that the car is not yet at the wall, due to the above reasoning. Since the fly has nonzero speed and the car does not have "infinite" speed, the fly and the car will next meet at some point between their current positions (not at the wall). Repeat.
$endgroup$
– inavda
Mar 25 at 15:08
$begingroup$
Well, one can figure that if the fly is at the wall and the car is a non-zero distance away that the fly flies to the car and that takes some non-zero time (they have a combined speed which is finite so to cover a distance takes non-zero time). The fly will then fly to the wall. As the fly is faster it reaches the wall first. The car is now a non-zero distance away. Thus is the same situation as before so by induction there will be an infinite number of trips . That is formal and rigorous.
$endgroup$
– fleablood
Mar 25 at 15:15
|
show 1 more comment
$begingroup$
It has made infinitely many trips. Every trip will be shorter than the last, but the fly will always reach the wall before the car, so it will always have room for one more trip. And one more. And one more.
The total length the fly flies is 50km, as the car crashes into the wall exactly 30 minutes after the whole experiment started.
answered Mar 25 at 14:53
ArthurArthur
121k7122208
$endgroup$
$begingroup$
I got that but can you please give me more rigorous proof? If you don't mind.
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– user654700
Mar 25 at 14:56
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@user654700 What exactly do you find not rigorous enough?
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– Arthur
Mar 25 at 15:01
8
$begingroup$
@user654700: This is rigorous. One can also write an equation for the length of each trip the fly makes and sum the series, but under the assumptions that is not needed. Don't confuse rigor with fancy mathematical symbols. There is a famous anecdote about John von Neumann and this puzzle.
$endgroup$
– Ross Millikan
Mar 25 at 15:02
$begingroup$
I suppose there are some minor details that could be added to give it some more rigor (or at least make the rigor more obvious): when the fly is at the position of the car, it has to travel the same distance to get to the wall, but its speed is greater, so it will always reach first. Once the fly is at the wall, we know that the car is not yet at the wall, due to the above reasoning. Since the fly has nonzero speed and the car does not have "infinite" speed, the fly and the car will next meet at some point between their current positions (not at the wall). Repeat.
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– inavda
Mar 25 at 15:08
$begingroup$
Well, one can figure that if the fly is at the wall and the car is a non-zero distance away that the fly flies to the car and that takes some non-zero time (they have a combined speed which is finite so to cover a distance takes non-zero time). The fly will then fly to the wall. As the fly is faster it reaches the wall first. The car is now a non-zero distance away. Thus is the same situation as before so by induction there will be an infinite number of trips . That is formal and rigorous.
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– fleablood
Mar 25 at 15:15
|
show 1 more comment
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It has made infinitely many trips. Every trip will be shorter than the last, but the fly will always reach the wall before the car, so it will always have room for one more trip. And one more. And one more.
The total length the fly flies is 50km, as the car crashes into the wall exactly 30 minutes after the whole experiment started.
answered Mar 25 at 14:53
ArthurArthur
121k7122208
$endgroup$
It has made infinitely many trips. Every trip will be shorter than the last, but the fly will always reach the wall before the car, so it will always have room for one more trip. And one more. And one more.
The total length the fly flies is 50km, as the car crashes into the wall exactly 30 minutes after the whole experiment started.
answered Mar 25 at 14:53
ArthurArthur
121k7122208
answered Mar 25 at 14:53
ArthurArthur
121k7122208
answered Mar 25 at 14:53
ArthurArthur
121k7122208
answered Mar 25 at 14:53
ArthurArthur
121k7122208
121k7122208
$begingroup$
I got that but can you please give me more rigorous proof? If you don't mind.
$endgroup$
– user654700
Mar 25 at 14:56
$begingroup$
@user654700 What exactly do you find not rigorous enough?
$endgroup$
– Arthur
Mar 25 at 15:01
8
$begingroup$
@user654700: This is rigorous. One can also write an equation for the length of each trip the fly makes and sum the series, but under the assumptions that is not needed. Don't confuse rigor with fancy mathematical symbols. There is a famous anecdote about John von Neumann and this puzzle.
$endgroup$
– Ross Millikan
Mar 25 at 15:02
$begingroup$
I suppose there are some minor details that could be added to give it some more rigor (or at least make the rigor more obvious): when the fly is at the position of the car, it has to travel the same distance to get to the wall, but its speed is greater, so it will always reach first. Once the fly is at the wall, we know that the car is not yet at the wall, due to the above reasoning. Since the fly has nonzero speed and the car does not have "infinite" speed, the fly and the car will next meet at some point between their current positions (not at the wall). Repeat.
$endgroup$
– inavda
Mar 25 at 15:08
$begingroup$
Well, one can figure that if the fly is at the wall and the car is a non-zero distance away that the fly flies to the car and that takes some non-zero time (they have a combined speed which is finite so to cover a distance takes non-zero time). The fly will then fly to the wall. As the fly is faster it reaches the wall first. The car is now a non-zero distance away. Thus is the same situation as before so by induction there will be an infinite number of trips . That is formal and rigorous.
$endgroup$
– fleablood
Mar 25 at 15:15
|
show 1 more comment
$begingroup$
I got that but can you please give me more rigorous proof? If you don't mind.
$endgroup$
– user654700
Mar 25 at 14:56
$begingroup$
@user654700 What exactly do you find not rigorous enough?
$endgroup$
– Arthur
Mar 25 at 15:01
8
$begingroup$
@user654700: This is rigorous. One can also write an equation for the length of each trip the fly makes and sum the series, but under the assumptions that is not needed. Don't confuse rigor with fancy mathematical symbols. There is a famous anecdote about John von Neumann and this puzzle.
$endgroup$
– Ross Millikan
Mar 25 at 15:02
$begingroup$
I suppose there are some minor details that could be added to give it some more rigor (or at least make the rigor more obvious): when the fly is at the position of the car, it has to travel the same distance to get to the wall, but its speed is greater, so it will always reach first. Once the fly is at the wall, we know that the car is not yet at the wall, due to the above reasoning. Since the fly has nonzero speed and the car does not have "infinite" speed, the fly and the car will next meet at some point between their current positions (not at the wall). Repeat.
$endgroup$
– inavda
Mar 25 at 15:08
$begingroup$
Well, one can figure that if the fly is at the wall and the car is a non-zero distance away that the fly flies to the car and that takes some non-zero time (they have a combined speed which is finite so to cover a distance takes non-zero time). The fly will then fly to the wall. As the fly is faster it reaches the wall first. The car is now a non-zero distance away. Thus is the same situation as before so by induction there will be an infinite number of trips . That is formal and rigorous.
$endgroup$
– fleablood
Mar 25 at 15:15
$begingroup$
I got that but can you please give me more rigorous proof? If you don't mind.
$endgroup$
– user654700
Mar 25 at 14:56
$begingroup$
I got that but can you please give me more rigorous proof? If you don't mind.
$endgroup$
– user654700
Mar 25 at 14:56
$begingroup$
@user654700 What exactly do you find not rigorous enough?
$endgroup$
– Arthur
Mar 25 at 15:01
$begingroup$
@user654700 What exactly do you find not rigorous enough?
$endgroup$
– Arthur
Mar 25 at 15:01
8
8
$begingroup$
@user654700: This is rigorous. One can also write an equation for the length of each trip the fly makes and sum the series, but under the assumptions that is not needed. Don't confuse rigor with fancy mathematical symbols. There is a famous anecdote about John von Neumann and this puzzle.
$endgroup$
– Ross Millikan
Mar 25 at 15:02
$begingroup$
@user654700: This is rigorous. One can also write an equation for the length of each trip the fly makes and sum the series, but under the assumptions that is not needed. Don't confuse rigor with fancy mathematical symbols. There is a famous anecdote about John von Neumann and this puzzle.
$endgroup$
– Ross Millikan
Mar 25 at 15:02
$begingroup$
I suppose there are some minor details that could be added to give it some more rigor (or at least make the rigor more obvious): when the fly is at the position of the car, it has to travel the same distance to get to the wall, but its speed is greater, so it will always reach first. Once the fly is at the wall, we know that the car is not yet at the wall, due to the above reasoning. Since the fly has nonzero speed and the car does not have "infinite" speed, the fly and the car will next meet at some point between their current positions (not at the wall). Repeat.
$endgroup$
– inavda
Mar 25 at 15:08
$begingroup$
I suppose there are some minor details that could be added to give it some more rigor (or at least make the rigor more obvious): when the fly is at the position of the car, it has to travel the same distance to get to the wall, but its speed is greater, so it will always reach first. Once the fly is at the wall, we know that the car is not yet at the wall, due to the above reasoning. Since the fly has nonzero speed and the car does not have "infinite" speed, the fly and the car will next meet at some point between their current positions (not at the wall). Repeat.
$endgroup$
– inavda
Mar 25 at 15:08
$begingroup$
Well, one can figure that if the fly is at the wall and the car is a non-zero distance away that the fly flies to the car and that takes some non-zero time (they have a combined speed which is finite so to cover a distance takes non-zero time). The fly will then fly to the wall. As the fly is faster it reaches the wall first. The car is now a non-zero distance away. Thus is the same situation as before so by induction there will be an infinite number of trips . That is formal and rigorous.
$endgroup$
– fleablood
Mar 25 at 15:15
$begingroup$
Well, one can figure that if the fly is at the wall and the car is a non-zero distance away that the fly flies to the car and that takes some non-zero time (they have a combined speed which is finite so to cover a distance takes non-zero time). The fly will then fly to the wall. As the fly is faster it reaches the wall first. The car is now a non-zero distance away. Thus is the same situation as before so by induction there will be an infinite number of trips . That is formal and rigorous.
$endgroup$
– fleablood
Mar 25 at 15:15
|
show 1 more comment
$begingroup$
Simple way to calculate it: It takes the car 30 minutes to travel the 20km to the wall at 40km/hr. The fly, traveling at 100km/hr will travel 50km in those same 30 minutes.
Edit: There will be an infinite number of trips. It's similar to how Zeno's paradox works where the trips get shorter and shorter and eventually take an infinitely small amount of time. But all those infinitely small trips end up as a finite amount of distance.
answered Mar 25 at 19:09
Guest3711Guest3711
312
$endgroup$
add a comment |
$begingroup$
Simple way to calculate it: It takes the car 30 minutes to travel the 20km to the wall at 40km/hr. The fly, traveling at 100km/hr will travel 50km in those same 30 minutes.
Edit: There will be an infinite number of trips. It's similar to how Zeno's paradox works where the trips get shorter and shorter and eventually take an infinitely small amount of time. But all those infinitely small trips end up as a finite amount of distance.
answered Mar 25 at 19:09
Guest3711Guest3711
312
$endgroup$
add a comment |
$begingroup$
Simple way to calculate it: It takes the car 30 minutes to travel the 20km to the wall at 40km/hr. The fly, traveling at 100km/hr will travel 50km in those same 30 minutes.
Edit: There will be an infinite number of trips. It's similar to how Zeno's paradox works where the trips get shorter and shorter and eventually take an infinitely small amount of time. But all those infinitely small trips end up as a finite amount of distance.
answered Mar 25 at 19:09
Guest3711Guest3711
312
$endgroup$
Simple way to calculate it: It takes the car 30 minutes to travel the 20km to the wall at 40km/hr. The fly, traveling at 100km/hr will travel 50km in those same 30 minutes.
Edit: There will be an infinite number of trips. It's similar to how Zeno's paradox works where the trips get shorter and shorter and eventually take an infinitely small amount of time. But all those infinitely small trips end up as a finite amount of distance.
answered Mar 25 at 19:09
Guest3711Guest3711
312
edited Mar 25 at 19:43
answered Mar 25 at 19:09
Guest3711Guest3711
312
answered Mar 25 at 19:09
Guest3711Guest3711
312
answered Mar 25 at 19:09
Guest3711Guest3711
312
312
add a comment |
add a comment |
$begingroup$
Arthur's answer is very good, but here is another (equally valid) way of visualizing the problem:
Let us graph the positions of the wall, fly, and car over time. The wall doesn't move, so it is represented by a horizontal line. The car starts at some distance away from the wall but moves towards it at a constant speed until it hits the wall -- so we have a line that intersects the wall's line. Now for the interesting part:
The fly's path starts off as a line with greater slope than the car's line, until it hits the wall's line. The fly's speed remains the same, but it is going in the opposite direction. So now the path continues as if the wall's line was a mirror and it was reflected. When the fly hits the car, the same thing happens -- the fly's path is reflected and it continues towards the wall again (one caveat: when it bounces off the car's line, the angles of incidence and reflection are not equal so it isn't behaving exactly as light would).
So we can see that the fly's path continues bouncing up and down, always at the same or opposite slope.
The final thing to notice to grasp the intuition is that this diagram we have constructed is self-similar. If we zoom in so that the second bounce with the wall is where the first bounce used to be, we have the same exact diagram as before. I won't prove this, but if you draw it out, you can see it intuitively. Essentially, this means that no matter how close the car gets to the wall, we can zoom in and see more bounces for the fly.
answered Mar 25 at 15:04
inavdainavda
1198
$endgroup$
add a comment |
$begingroup$
Arthur's answer is very good, but here is another (equally valid) way of visualizing the problem:
Let us graph the positions of the wall, fly, and car over time. The wall doesn't move, so it is represented by a horizontal line. The car starts at some distance away from the wall but moves towards it at a constant speed until it hits the wall -- so we have a line that intersects the wall's line. Now for the interesting part:
The fly's path starts off as a line with greater slope than the car's line, until it hits the wall's line. The fly's speed remains the same, but it is going in the opposite direction. So now the path continues as if the wall's line was a mirror and it was reflected. When the fly hits the car, the same thing happens -- the fly's path is reflected and it continues towards the wall again (one caveat: when it bounces off the car's line, the angles of incidence and reflection are not equal so it isn't behaving exactly as light would).
So we can see that the fly's path continues bouncing up and down, always at the same or opposite slope.
The final thing to notice to grasp the intuition is that this diagram we have constructed is self-similar. If we zoom in so that the second bounce with the wall is where the first bounce used to be, we have the same exact diagram as before. I won't prove this, but if you draw it out, you can see it intuitively. Essentially, this means that no matter how close the car gets to the wall, we can zoom in and see more bounces for the fly.
answered Mar 25 at 15:04
inavdainavda
1198
$endgroup$
add a comment |
$begingroup$
Arthur's answer is very good, but here is another (equally valid) way of visualizing the problem:
Let us graph the positions of the wall, fly, and car over time. The wall doesn't move, so it is represented by a horizontal line. The car starts at some distance away from the wall but moves towards it at a constant speed until it hits the wall -- so we have a line that intersects the wall's line. Now for the interesting part:
The fly's path starts off as a line with greater slope than the car's line, until it hits the wall's line. The fly's speed remains the same, but it is going in the opposite direction. So now the path continues as if the wall's line was a mirror and it was reflected. When the fly hits the car, the same thing happens -- the fly's path is reflected and it continues towards the wall again (one caveat: when it bounces off the car's line, the angles of incidence and reflection are not equal so it isn't behaving exactly as light would).
So we can see that the fly's path continues bouncing up and down, always at the same or opposite slope.
The final thing to notice to grasp the intuition is that this diagram we have constructed is self-similar. If we zoom in so that the second bounce with the wall is where the first bounce used to be, we have the same exact diagram as before. I won't prove this, but if you draw it out, you can see it intuitively. Essentially, this means that no matter how close the car gets to the wall, we can zoom in and see more bounces for the fly.
answered Mar 25 at 15:04
inavdainavda
1198
$endgroup$
Arthur's answer is very good, but here is another (equally valid) way of visualizing the problem:
Let us graph the positions of the wall, fly, and car over time. The wall doesn't move, so it is represented by a horizontal line. The car starts at some distance away from the wall but moves towards it at a constant speed until it hits the wall -- so we have a line that intersects the wall's line. Now for the interesting part:
The fly's path starts off as a line with greater slope than the car's line, until it hits the wall's line. The fly's speed remains the same, but it is going in the opposite direction. So now the path continues as if the wall's line was a mirror and it was reflected. When the fly hits the car, the same thing happens -- the fly's path is reflected and it continues towards the wall again (one caveat: when it bounces off the car's line, the angles of incidence and reflection are not equal so it isn't behaving exactly as light would).
So we can see that the fly's path continues bouncing up and down, always at the same or opposite slope.
The final thing to notice to grasp the intuition is that this diagram we have constructed is self-similar. If we zoom in so that the second bounce with the wall is where the first bounce used to be, we have the same exact diagram as before. I won't prove this, but if you draw it out, you can see it intuitively. Essentially, this means that no matter how close the car gets to the wall, we can zoom in and see more bounces for the fly.
answered Mar 25 at 15:04
inavdainavda
1198
answered Mar 25 at 15:04
inavdainavda
1198
answered Mar 25 at 15:04
inavdainavda
1198
answered Mar 25 at 15:04
inavdainavda
1198
1198
add a comment |
add a comment |
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6
$begingroup$
In theory infinitely many.
$endgroup$
– fleablood
Mar 25 at 14:53
4
$begingroup$
This 100 km/h fly is a very impressive aviator given that ordinary houseflies have a max speed of 8 km/h. Even more impressive is its ability to instantaneously go from +100 km/h to -100km/h without feeling a bit groggy for a moment.
$endgroup$
– RedGrittyBrick
Mar 25 at 17:05
2
$begingroup$
Everyone seems to forget Applied Mathematic is not as it should be when the car reaches the point where the gap is less than the size of the flies body it dies of the pressure if not exhaustion.
$endgroup$
– KJO
Mar 25 at 17:47
2
$begingroup$
At least it's not a spherical cow....
$endgroup$
– Kathy
Mar 25 at 19:36
1
$begingroup$
Possible duplicate of 'Bee flying between two trains' problem
$endgroup$
– Herohtar
Mar 25 at 20:47