Pigeon Problem Solved: Calculating Car Velocities and Meeting Time

[rudransh verma]

Homework Statement

A pigeon flies at 36 km/h to and fro between two cars moving towards each other on the straight road, starting from first car when the car separation is 40km. The first car has the speed of 16km/h and the second has 25km/h. By the time the cars meet head on what are the @ total distance (b) net displacement flown by the pigeon?

Relevant Equations

Avg velocity = displacement/time
Avg speed= total distance/time

The solution was done adding the two velocities of the car while finding the total time it took for the two cars to meet. I don’t really get the solution.

 

[Frabjous]

The pigeon is flying at v_p so it flies a total distance of v_pt where t is determined from when the cars meet: d=v₁t+v₂t=(v₁+v₂)t.
Notice that t=t₁+t₂+… where t_i is time it takes to fly from one car to another. Convince yourself that this is true. Drawing a picture will help.

For b, draw a good picture. The actual calculation is easy once you see the positions.

 

[sysprog]

This is a version of a famous anecdote about the 'two trains' problem, and Prof. Von Neumann:

​

The following problem can be solved either the easy way or the hard way.​

​

Two trains 200 miles apart are moving toward each other; each one is going at a speed of 50 miles per hour.​

​

A fly starting on the front of one of them flies back and forth between them at a rate of 75 miles per hour.​

​

It does this until the trains collide and crush the fly to death.​

​

What is the total distance the fly has flown?​

​

​

The fly actually hits each train an infinite number of times before it gets crushed, and one could solve the problem the hard way with pencil and paper by summing an infinite series of distances.​

​

The easy way is as follows: Since the trains are 200 miles apart and each train is going 50 miles an hour, it takes 2 hours for the trains to collide.​

​

Therefore the fly was flying for two hours.​

​

Since the fly was flying at a rate of 75 miles per hour, the fly must have flown 150 miles.​

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That's all there is to it.​

​

​

When this problem was posed to John von Neumann, he immediately replied, "150 miles."​

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"It is very strange," said the poser, "but nearly everyone tries to sum the infinite series."​

​

"What do you mean, strange?" asked Von Neumann. "That's how I did it!"​

 

[kuruman]

This reminds me of another well-known train problem.

Cities A and B are 240 km apart. A train leaves from A traveling non-stop towards B at 60 km/h. Another train leaves B on the same track 20 minutes later and travels towards A at 80 km/h. It is scheduled to make 4-minute stops at town C which is 60 km from B and town D which is 160 km from B. When the trains collide, which train is closer to city B?

Spoiler

When the trains collide, they will be at the same distance from city A. Oops, the problem asked for the distance from city B .

 

[rudransh verma]

For b, draw a good picture. The actual calculation is easy once you see the position

To calculate the displacement we take the velocity of first car 16km/h and multiply by 40/41 h. But the speed of the pigeon is actually 36. So why do we take the speed 16. It should be that the displacement and distance are same.
In about 1 hour two cars travel 16km and 24 km and the pigeon travel 36km. So there is no back and forth motion for the pigeon.

 

[kuruman]

In about 1 hour two cars travel 16km and 24 km and the pigeon travel 36km. So there is no back and forth motion for the pigeon.

Before 1 hour has gone by, the cars will collide because they start with a separation of 40 km and one car moves towards the other with a relative speed of 41 km/h. This means that the pigeon will be crushed between them. That's plenty back and forth motion. As has already been suggested, draw a picture.

 

[Father_Ing]

It should be that the displacement and distance are same.

This is not true in every situation. Let's use the same event as what the question implied, but instead of having both cars moving, here, the cars will have 0 velocity, and the bird will have speed 20 km/h

To travel from one car to the other, the bird needs to fly for 2 hour.

Now, after 2 hours, are the distance and displacement the same? Yes, both are 40 km.

Now, how about if the bird has flown for 3 hour?

Obviously, this means that the bird had covered the distance 60 km. How about the displacement? It will be 20 km from the point where the bird initially stand

 

[rudransh verma]

This is a version of a famous anecdote about the 'two trains' problem, and Prof. Von Neumann:

​

The following problem can be solved either the easy way or the hard way.​

​

Two trains 200 miles apart are moving toward each other; each one is going at a speed of 50 miles per hour.​

​

A fly starting on the front of one of them flies back and forth between them at a rate of 75 miles per hour.​

​

It does this until the trains collide and crush the fly to death.​

​

What is the total distance the fly has flown?​

​

​

The fly actually hits each train an infinite number of times before it gets crushed, and one could solve the problem the hard way with pencil and paper by summing an infinite series of distances.​

​

The easy way is as follows: Since the trains are 200 miles apart and each train is going 50 miles an hour, it takes 2 hours for the trains to collide.​

​

Therefore the fly was flying for two hours.​

​

Since the fly was flying at a rate of 75 miles per hour, the fly must have flown 150 miles.​

​

That's all there is to it.​

​

​

When this problem was posed to John von Neumann, he immediately replied, "150 miles."​

​

"It is very strange," said the poser, "but nearly everyone tries to sum the infinite series."​

​

"What do you mean, strange?" asked Von Neumann. "That's how I did it!"​

I did it like that and I got close to 35km. My answer was coming 1.66Km+9.06km+23.6km= 34.32km
We need to repeat the same few steps over and over again and we will get smaller and smaller values for distance. Add all of them and there is your answer.

 

[jbriggs444]

I did it like that and I got close to 35km. My answer was coming 1.66Km+9.06km+23.6km= 34.32km
We need to repeat the same few steps over and over again and we will get smaller and smaller values for distance. Add all of them and there is your answer.

You'll have to explain your calculation for that figure. I could not reverse engineer a reasonable calculation that gives that result.

 

[rudransh verma]

You'll have to explain your calculation for that figure. I could not reverse engineer a reasonable calculation that gives that result.

First find at what distance the bird collides with the train 2. We get a value of distance for bird(23.6 km) and train 2(16.4 km) whose total will be 40 km. now calculate time taken(0.6 hrs) in this collision. Now find the distance traveled by train 1 in this time(10.5 km). The new distance between the two trains will be 40-10.5-16.4=13.1 km.
Now repeat the same process with new distance 13.1 km and find the distance where the bird collides(9.06 km) with train 1(which traveled 4.03 km). total 13.1 km. The new distance between two trains will be 13.1-4.03-6.25km=2.82 km
Go on like that and keep adding the distances traveled by the bird before collisions and the value will converge to 35 km.

 

[jbriggs444]

First find at what distance the bird collides with the train 2. We get a value of distance for bird(23.6 km) and train 2(16.4 km) whose total will be 40 km. now calculate time taken(0.6 hrs) in this collision. Now find the distance traveled by train 1 in this time(10.5 km). The new distance between the two trains will be 40-10.5-16.4=13.1 km.
Now repeat the same process with new distance 13.1 km and find the distance where the bird collides(9.06 km) with train 1(which traveled 4.03 km). total 13.1 km. The new distance between two trains will be 13.1-4.03-6.25km=2.82 km
Go on like that and keep adding the distances traveled by the bird before collisions and the value will converge to 35 km.

So you were summing an infinite series and rounding off the first few terms to one decimal place. And apparently doing so numerically rather than using a formula. No wonder the result was not accurate.

 

[rudransh verma]

So you were summing an infinite series and rounding off the first few terms to one decimal place. And apparently doing so numerically rather than using a formula. No wonder the result was not accurate.

Not really an infinite series. Few values will get us very close to 35. Who cares it to be exact. Its just a distance traveled by a dumb bird flying between two cars. It’s a question from 2 chapter(motion in 1D)of resnik after all. And I like the way I did it.
Well we use the formula of distance, time, etc.

 

[jbriggs444]

Not really an infinite series. Few values will get us very close to 35. Who cares it to be exact. Its just a distance traveled by a dumb bird flying between two cars. It’s a question from 2 chapter(motion in 1D)of resnik after all. And I like the way I did it.
Well we use the formula of distance, time, etc.

It is really an infinite series. You can truncate it if you like. Indeed you have to if you insist on simply adding terms together.

Recognizing that the situation after one round trip for the bird is exactly the same as the original problem, just scaled down, leads quickly to a formula for the sum of the infinite series.

 

[rudransh verma]

leads quickly to a formula for the sum of the infinite series.

Which one?

 

[kuruman]

Recognizing that the situation after one round trip for the bird is exactly the same as the original problem, just scaled down, leads quickly to a formula for the sum of the infinite series.

Also, writing the equations in the inertial frame of the car where the bird starts, simplifies the problem and makes it easier to see what's going on.

 

[jbriggs444]

Which one?

Which infinite series? Let me go into painful detail.

You can split the bird's trip into a sequence of round trips from car A to car B and back to car A.

You start with the bird sitting on car A a certain distance from car B. The bird flies to car B. You calculate the distance flown and the time elapsed. Now the bird flies back to car A. You calculate the distance flown and the time elapsed.

At this point you are back in the original situation. You calculate the new distance between the two cars. You figure out what fraction ($r$) the new distance is compared to the old distance.

You write down an equation that exploits the way the original problem recursively appears in the scenario.

Total_Distance_Flown_By_Bird = Distance_Flown_During_First_Round_Trip + r $\times$ Total_Distance_Flown_By_Bird

Solve for Total_Distance_Flown_By_Bird.

This will be equal to the sum of the infinite series of round trip distances.

 

[bob012345]

I tried the geometric approach. Seems to work but my precision wasn't enough to get the same answer beyond three decimals. I get XX.XXX69 vs. XX.XXX95. Edit: I fixed it so they now match to high precision. Time is the vertical axis in hours and distance is the horizontal axis in $km$.