another math question

This is a tough one.

A train is travelling on a straight line with speed 60 kmph. At a particular time instance, a bird which is 30 km away from the train, exactly on a line perpendicular to the trains trajectory, drawn from the point where the train is at that moment, starts travelling towards the train at some constant speed. The bird always heads towards the train, changing its direction continuously, and finally meets it when the train has traveleld 60 km from the point the bird started its flight.

What is the birds speed?

Ha A new instance of the dog-chase problem, aka 'Dog and Cat', 'Hunter and Deer', 'Pig and Farmer', 'Lion and Christian'... So now we have 'Bird and Train'.

I posted the general solution and how I found it in sci.physics on 24-Jun-2001 20:43, title 'Re: Still another'.

At the bottom of the post you find the following equation: C = A/2 * [ S/(S-1) - S/(S+1) ] where: A = Initial distance Hunter to Prey C = Distance ran by a A new instance of the dog-chase problem, aka 'Dog and Cat', 'Hunter and Deer', 'Pig and Farmer', 'Lion and Christian'... So now we have 'Bird and Train'.

I posted the general solution and how I found it in sci.physics on 24-Jun-2001 20:43, title 'Re: Still another'.

At the bottom of the post you find the following equation: C = A/2 * [ S/(S-1) - S/(S+1) ] where: A = Initial distance Hunter to Prey C = Distance ran by Prey before being caught S = <Speed of Hunter> / <Speed of Prey>

When A and C are known, we can solve for S: S = 1/2 * ( A/C +/- sqrta A new instance of the dog-chase problem, aka 'Dog and Cat', 'Hunter and Deer', 'Pig and Farmer', 'Lion and Christian'... So now we have 'Bird and Train'.

I posted the general solution and how I found it in sci.physics on 24-Jun-2001 20:43, title 'Re: Still another'.

At the bottom of the post you find the following equation: C = A/2 * [ S/(S-1) - S/(S+1) ] where: A = Initial distance Hunter to Prey C = Distance ran by a A new instance of the dog-chase problem, aka 'Dog and Cat', 'Hunter and Deer', 'Pig and Farmer', 'Lion and Christian'... So now we have 'Bird and Train'.

I posted the general solution and how I found it in sci.physics on 24-Jun-2001 20:43, title 'Re: Still another'.

At the bottom of the post you find the following equation: C = A/2 * [ S/(S-1) - S/(S+1) ] where: A = Initial distance Hunter to Prey C = Distance ran by Prey before being caught S = <Speed of Hunter> / <Speed of Prey>

When A and C are known, we can solve for S: S = 1/2 * ( A/C +/- sqrt[ (A/C)^2 + 4 ] )

So plugging in A = 30, C = 60, we get (taking the positive solution): S = 1/2 * ( 1/2 + sqrta A new instance of the dog-chase problem, aka 'Dog and Cat', 'Hunter and Deer', 'Pig and Farmer', 'Lion and Christian'... So now we have 'Bird and Train'.

I posted the general solution and how I found it in sci.physics on 24-Jun-2001 20:43, title 'Re: Still another'.

At the bottom of the post you find the following equation: C = A/2 * [ S/(S-1) - S/(S+1) ] where: A = Initial distance Hunter to Prey C = Distance ran by a A new instance of the dog-chase problem, aka 'Dog and Cat', 'Hunter and Deer', 'Pig and Farmer', 'Lion and Christian'... So now we have 'Bird and Train'.

I posted the general solution and how I found it in sci.physics on 24-Jun-2001 20:43, title 'Re: Still another'.

At the bottom of the post you find the following equation: C = A/2 * [ S/(S-1) - S/(S+1) ] where: A = Initial distance Hunter to Prey C = Distance ran by Prey before being caught S = <Speed of Hunter> / <Speed of Prey>

When A and C are known, we can solve for S: S = 1/2 * ( A/C +/- sqrta A new instance of the dog-chase problem, aka 'Dog and Cat', 'Hunter and Deer', 'Pig and Farmer', 'Lion and Christian'... So now we have 'Bird and Train'.

I posted the general solution and how I found it in sci.physics on 24-Jun-2001 20:43, title 'Re: Still another'.

At the bottom of the post you find the following equation: C = A/2 * [ S/(S-1) - S/(S+1) ] where: A = Initial distance Hunter to Prey C = Distance ran by a A new instance of the dog-chase problem, aka 'Dog and Cat', 'Hunter and Deer', 'Pig and Farmer', 'Lion and Christian'... So now we have 'Bird and Train'.

I posted the general solution and how I found it in sci.physics on 24-Jun-2001 20:43, title 'Re: Still another'.

At the bottom of the post you find the following equation: C = A/2 * [ S/(S-1) - S/(S+1) ] where: A = Initial distance Hunter to Prey C = Distance ran by Prey before being caught S = <Speed of Hunter> / <Speed of Prey>

When A and C are known, we can solve for S: S = 1/2 * ( A/C +/- sqrt[ (A/C)^2 + 4 ] )

So plugging in A = 30, C = 60, we get (taking the positive solution): S = 1/2 * ( 1/2 + sqrt[ 1/4+4 ] ) = 1.2808 (rounded) So the bird flies at 1.2808 * 60 kmph = 76.85 kmph Right?

Dirk Vdm

Dirk Van de moortel wrote:

Is there a more 'elegant' way to solve this ?

ObPuzzle : 4 , 5 , 35 , 56 , 83 , ?

In Kmph, approximately: (2 * Pi * SquareRoot(0.5 * (30^2 + 60^2))) / 4 = 0.5 * 3.14159265358979 * SquareRoot(2250) = 74.5094119934707

****** I worked this out with the help of geometry,, using small increments of flaps of the bird’s wings, which can be arbitrarily small.

Let s = the speed of the bird relative to the train. The train is moving west to east. The bird begins 30 km, north of the train.

Take a sheet of paper and draw a horizontal line near the bottom to indicate the eastward path of the train. Mark any point A somewhere north of this line to indicate a where the bird is at some time during its flight. Draw line AB perpendicular to the track.

Mark point C on the train’s path somewhat east of point B to indicate where the train is when the bird is at point A. Draw line AC, which is the direction that the bird flies in its next wing flap. The bird cannot change direction during a flap of its wings.

Let the train travel a distance of one unit during a flap of the bird’s wings. Then the bird will travel s units per flap. Mark off a distance of s on line AC and the draw a horizontal line westward to AB from that point. This will be the horizontal distance that the bird flies while the train is moving one unit eastward.

Now mark off point D one unit east of point C. The train will be at D when the bird finishes its flap toward C. Draw AD and draw CE perpendicular to AD. Given very small flaps, he triangle CDE will be similar to the little triangle you drew near A for the bird. Distance ED will be h/s. Thus is the incremental distance that the bird must fly owing to the movement of the train while the bird was making its flap of s units toward C.

In order to catch the train, the bird must fly a total of 60 km. horizontally = the sum of the individual small h’s. Thus. the movement of the train will add 60/s to the 30 km. that the bird was from the train originally. Thus, the bird flies a total of 30 + 60/s km to catch the train.

Also, while the train travels 60 km., the bird flies 60*s km. Equating:

30 + 60/s = 60s

Solving: s = 1.2808

Multiplying by the train’s speed: 60*1.2808 = 76.85 km/hr = the speed of the bird.

Bill Ryan

Where exactly do you put the point E? I'd like to go on but I can't find this point

Dirk Vdm

' Let the train travel a distance of one unit during a flap of the bird’s wings. Then the bird will travel s units per flap. Mark off a distance of s on line AC and the draw a horizontal line westward to AB from that point. This will be the horizontal distance that the bird flies while the train is moving one unit eastward.

'Now mark off point D one unit east of point C. The train will be at D when the bird finishes its flap toward C. Draw AD and draw CE perpendicular to AD. Given very small flaps, he triangle CDE will be similar to the little triangle you drew near A for the bird.' ****** Dirk Vdm asked: *******

Point E is the point at which a line drawn from point C forms a right angle CED. It is easier to visualize this if you rotate your drawing clockwise so that line AD is horizontal and C is above this line. You would drop the perpendicular line from point C to line AD.

I should also note that for arbitrarily small incremental distances angle ACB is arbitrarily close to angle ADB.

I hope this helps. Frankly, I found the problem interesting but rather complicated. It took me a while to realize that the fact that line ED = h/s would lead to a solution.

Bill Ryan

Okay, got it. Really brilliant!

I solved this thing somewhere in January. Had a jolly good fight with a differential equation... and I won ) But it was a tough one.

If I feel like it, I'll have a look at Matthew Russotto' challenge tomorrow, the same problem but with a uniformly accelerating chaser. Won't be easy I guess.

Again, you method is super. Chapeau!

Dirk Vdm

Great solution!

I had not seen this twist of the 1-d pursuit problem. I was surprised at the simplicity of the formulae for this version.

Let d is the initial separation, V the speed of the quarry (here, the train) and v that of the pursuer (here, the bird). If the quarry goes a distance L before capture

L = r*d/(1 - r*r)

where r = V/v, the speed ratio.

Thus, r = sqrt(1 + b*b) - b

where b = d/(2*L).

With d/L = 60/30 = 1/2, b = 1/4. Hence, r = 0.78 and so v = 60/0.78 kmph = 1.28*60 kmph.

No, it is not! I do not think that this has been solved.

Not by me anyway: had to give it up Besides, being too busy in the physics newsgroups

Dirk Vdm