Distance, velocity...?

Question

I am absolutely stumped regarding this questions. Could someone please assist me? I would be extremely grateful.

A runner is jogging at a steady v = 8.4 km/hr. When the runner is L = 4.9 km from the finish line, a bird begins flying from the runner to the finish line at V = 16.8 km/hr (2 times as fast as the runner). When the bird reaches the finish line, it turns around and flies back to the runner. We will assume that the bird occupies only one point in space, ie., a zero length bird.

How far does the bird travel? Answer in units of km.

ALSO:

After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. The bird repeats the back and forth trips until the runner reaches the finish line.

How far does the bird travel from the beginning? (ie include the distance traveled to the first encounter) Answer in units of km.

Answer 1

Easy: Just assume the bird never stops flying.
The runner is 4.9km from the finish line, which means he needs another 4.9/8.4 hours to finish.
In that time the bird will have flown 4.9/8.4*16.8 km, or 9.8km

You are tempted to run some impossible calculations to determine how many km the bird travels on each round trip between the runner and the finish line, but all you need to know is the bird's speed and for how long it flies.
How long is a round trip for the bird is irrelevant.

Edit: Good one about Gauss, GI!
There is a similar story about Poincare scrambling to finish a math exam for admission at the Ecole Polytechnique. The guy tells him it's ok if he's not done, he can try again next year.
Poincare replied: "But I'm almost done with the third method to solve this problem!"

Edit2: Integration method (groan...)

the first Distance the bird travels is D0 = 4.9Km, at his speed 2V = 16.8km/h (the runner runs at a speed V = 8.4 Km/h)
On his trip back, he travels D1.
D1 is the distance between the finish line and the runner, who traveled V*T1 Km, T1 is the time the bird took to do the round trip, so:
D1 = D0 - VT1 and
T1 = (D0+D1)/2V
so D1 = D0 -(D0+D1)/2
D1 = D0/3
Let's call R0 the round trip distance covered by the bird:
R0 = D0+D1 = 4D0/3
On his subsequent trips:
Rn = 4Dn/3. Since D1=D0/3, Dn=D0/3^n and:
Rn=4D0/3*(1/3)^n
The sum of those (infinite) round trips is:
Sn = ΣRn = 4D0/3* (1- (1/3)^infinite)/(1-1/3) = 4D0/3*(1/(1-1/3)) = 2D0 = 9.8km.
Hurray.

Answer 2

at t0, bird and runner are both 4.9 km from the finish line. at 16.8 km/h, bird reaches it in 4.9/16.8 = 0.2916666... h, not that we care about the time. since the runner is half as fast, he'll cover half the distance, 2.95 km. now it's t1. on the next leg of the trip, moving toward each other, the runner will cover 1/3 of the 2.95 km, the bird 2/3 of it, since he's twice as fast [8.4t + 16.8t = 2.95, solve for t, compute 16.8t, very much the hard way]. so when they meet again, t2, runner has gone 1/2 + (1/3)(1/2) = 2/3 of the 4.9 km, the bird will have gone 1 + (2/3)(1/2) of it, 4/3 of 4.9 = 6.533333333.... km, 6 and 8/15 km exactly.

Now all this is to sucker you into trying to construct and sum up an infinite series. But it will take the runner 4.9/8.4 = 7/12 hr to reach the finish line. Flying at a constant rate of 16.8 km/h, in that time the bird will go (7/12)(16.8) = [guess what!??] 9.8 km. This is still too much math. Same time, twice the speed, twice the distance. 4.9 x 2 = 9.8 km.

According to legend, a student in the audience asked the great John von Neumann (inventor of game theory, father of computer programming, etc) this question. He hesitated a fraction of a second and gave the right answer. The student said that was amazing, that most people with Dr. von Neumann's education would have tried to sum up the infinite series. Von Neumann blinked and said, "but I did."

Answer 3

We know that the bird initially travels the finish line at 16 km/h (from the runner), we know that the finish line is 4.9km from the runner as well. The bird then flies back to the runner. First of all, you must understand that in the time it takes the bird to travel to the finish line, the runner is still moving at that steady 8.4 km/h.

Bird -> Finish line:

v=d/t
t=d/v
=4.9/16.8
=0.2916 h

So how far did the runner go in that time?

Runner:

v=d/t
d=vt
=8.4*0.2916
=2.45 km

The birds travel (in km) is his distance from the runner to the finish line, and from the finish line to the runners 'new distance' from the finish line.

4.9 km + 2.45 km = 7.35 km ... ans

Answer 4

A variant of this question was supposedly asked to Gauss (or Von Neuman, or ... ) to see if he was a mathematician or a physicist.

If it takes you a long time to answer, it means you set the problem up as an integral with the counting measure, i.e. an infinite series (which is correct).

If you answer quickly, you just figured out the time until the runner gets to the finish line, then multiplied it by the bird's velocity (also correct).

Gauss answered very quickly, and the asker said "Aha! So you are a physicist!" Gauss replied "But I integrated...."

It turns out that Gauss was a lightning-fast integrator, as well.