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













4












$begingroup$


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.










share|cite|improve this question









$endgroup$







  • 6




    $begingroup$
    In theory infinitely many.
    $endgroup$
    – fleablood
    yesterday






  • 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
    yesterday







  • 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
    yesterday






  • 2




    $begingroup$
    At least it's not a spherical cow....
    $endgroup$
    – Kathy
    yesterday






  • 1




    $begingroup$
    Possible duplicate of 'Bee flying between two trains' problem
    $endgroup$
    – Herohtar
    yesterday















4












$begingroup$


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.










share|cite|improve this question









$endgroup$







  • 6




    $begingroup$
    In theory infinitely many.
    $endgroup$
    – fleablood
    yesterday






  • 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
    yesterday







  • 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
    yesterday






  • 2




    $begingroup$
    At least it's not a spherical cow....
    $endgroup$
    – Kathy
    yesterday






  • 1




    $begingroup$
    Possible duplicate of 'Bee flying between two trains' problem
    $endgroup$
    – Herohtar
    yesterday













4












4








4


1



$begingroup$


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.










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









user654700user654700

334




334







  • 6




    $begingroup$
    In theory infinitely many.
    $endgroup$
    – fleablood
    yesterday






  • 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
    yesterday







  • 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
    yesterday






  • 2




    $begingroup$
    At least it's not a spherical cow....
    $endgroup$
    – Kathy
    yesterday






  • 1




    $begingroup$
    Possible duplicate of 'Bee flying between two trains' problem
    $endgroup$
    – Herohtar
    yesterday












  • 6




    $begingroup$
    In theory infinitely many.
    $endgroup$
    – fleablood
    yesterday






  • 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
    yesterday







  • 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
    yesterday






  • 2




    $begingroup$
    At least it's not a spherical cow....
    $endgroup$
    – Kathy
    yesterday






  • 1




    $begingroup$
    Possible duplicate of 'Bee flying between two trains' problem
    $endgroup$
    – Herohtar
    yesterday







6




6




$begingroup$
In theory infinitely many.
$endgroup$
– fleablood
yesterday




$begingroup$
In theory infinitely many.
$endgroup$
– fleablood
yesterday




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
yesterday





$begingroup$
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
yesterday





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
yesterday




$begingroup$
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
yesterday




2




2




$begingroup$
At least it's not a spherical cow....
$endgroup$
– Kathy
yesterday




$begingroup$
At least it's not a spherical cow....
$endgroup$
– Kathy
yesterday




1




1




$begingroup$
Possible duplicate of 'Bee flying between two trains' problem
$endgroup$
– Herohtar
yesterday




$begingroup$
Possible duplicate of 'Bee flying between two trains' problem
$endgroup$
– Herohtar
yesterday










4 Answers
4






active

oldest

votes


















13












$begingroup$

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 $frac 107h$).






share|cite|improve this answer











$endgroup$




















    15












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      I got that but can you please give me more rigorous proof? If you don't mind.
      $endgroup$
      – user654700
      yesterday










    • $begingroup$
      @user654700 What exactly do you find not rigorous enough?
      $endgroup$
      – Arthur
      yesterday






    • 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
      yesterday











    • $begingroup$
      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
      yesterday










    • $begingroup$
      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
      yesterday


















    3












    $begingroup$

    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.






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    New contributor




    Guest3711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    $endgroup$




















      1












      $begingroup$

      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.






      share|cite|improve this answer









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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        13












        $begingroup$

        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 $frac 107h$).






        share|cite|improve this answer











        $endgroup$

















          13












          $begingroup$

          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 $frac 107h$).






          share|cite|improve this answer











          $endgroup$















            13












            13








            13





            $begingroup$

            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 $frac 107h$).






            share|cite|improve this answer











            $endgroup$



            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 $frac 107h$).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited yesterday

























            answered yesterday









            fleabloodfleablood

            73.4k22891




            73.4k22891





















                15












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$












                • $begingroup$
                  I got that but can you please give me more rigorous proof? If you don't mind.
                  $endgroup$
                  – user654700
                  yesterday










                • $begingroup$
                  @user654700 What exactly do you find not rigorous enough?
                  $endgroup$
                  – Arthur
                  yesterday






                • 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
                  yesterday











                • $begingroup$
                  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
                  yesterday










                • $begingroup$
                  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
                  yesterday















                15












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$












                • $begingroup$
                  I got that but can you please give me more rigorous proof? If you don't mind.
                  $endgroup$
                  – user654700
                  yesterday










                • $begingroup$
                  @user654700 What exactly do you find not rigorous enough?
                  $endgroup$
                  – Arthur
                  yesterday






                • 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
                  yesterday











                • $begingroup$
                  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
                  yesterday










                • $begingroup$
                  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
                  yesterday













                15












                15








                15





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                ArthurArthur

                120k7120203




                120k7120203











                • $begingroup$
                  I got that but can you please give me more rigorous proof? If you don't mind.
                  $endgroup$
                  – user654700
                  yesterday










                • $begingroup$
                  @user654700 What exactly do you find not rigorous enough?
                  $endgroup$
                  – Arthur
                  yesterday






                • 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
                  yesterday











                • $begingroup$
                  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
                  yesterday










                • $begingroup$
                  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
                  yesterday
















                • $begingroup$
                  I got that but can you please give me more rigorous proof? If you don't mind.
                  $endgroup$
                  – user654700
                  yesterday










                • $begingroup$
                  @user654700 What exactly do you find not rigorous enough?
                  $endgroup$
                  – Arthur
                  yesterday






                • 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
                  yesterday











                • $begingroup$
                  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
                  yesterday










                • $begingroup$
                  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
                  yesterday















                $begingroup$
                I got that but can you please give me more rigorous proof? If you don't mind.
                $endgroup$
                – user654700
                yesterday




                $begingroup$
                I got that but can you please give me more rigorous proof? If you don't mind.
                $endgroup$
                – user654700
                yesterday












                $begingroup$
                @user654700 What exactly do you find not rigorous enough?
                $endgroup$
                – Arthur
                yesterday




                $begingroup$
                @user654700 What exactly do you find not rigorous enough?
                $endgroup$
                – Arthur
                yesterday




                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
                yesterday





                $begingroup$
                @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
                yesterday













                $begingroup$
                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
                yesterday




                $begingroup$
                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
                yesterday












                $begingroup$
                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
                yesterday




                $begingroup$
                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
                yesterday











                3












                $begingroup$

                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.






                share|cite|improve this answer










                New contributor




                Guest3711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$

















                  3












                  $begingroup$

                  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.






                  share|cite|improve this answer










                  New contributor




                  Guest3711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$















                    3












                    3








                    3





                    $begingroup$

                    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.






                    share|cite|improve this answer










                    New contributor




                    Guest3711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    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.







                    share|cite|improve this answer










                    New contributor




                    Guest3711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited yesterday





















                    New contributor




                    Guest3711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered yesterday









                    Guest3711Guest3711

                    312




                    312




                    New contributor




                    Guest3711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





                    New contributor





                    Guest3711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    Guest3711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





















                        1












                        $begingroup$

                        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.






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          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.






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            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.






                            share|cite|improve this answer









                            $endgroup$



                            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.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            inavdainavda

                            1198




                            1198



























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