[ hopeless ] Wile E. Coyote patiently waits behind a cactus in a desert, as Road Runner runs along?

Question

a straight line with constant velocity u = 10 m/s.

The shortest distance between the path of RR and the cactus is 50 ft. When he runs past the cactus WEC ambushes him, and begins chasing him with speed v = u = 10 m/s.

The velocity of WEC is always directed towards RR.

What is the distance between them after sufficiently long chase?


Position when the chase begins:
WEC
--|
--|
--v
50ft
--|
--|
--|
--RR---u> ------ ----- ------ ----- ------ -----

Despite this problem is very difficult,
about 1% of 16yo students are able
to solve it in the course of elementary
kinematics, even before they learn
Newton's laws.

Hint:

The trajectory of WEC wrt RR is a
simple geometric curve.

Answer 1

Choose units of time so that the speed (of both) is 1. Let r be the distance between the two with x the component of this distance along the road and y the perpendicular component.

WEC's speed along and perpendicular to the road are respectively x/r and y/r. RR's speed along the road is 1 -->

dx/dt = 1 - x/r
dy/dt = -y/r

Substituting these expressions into:

dr/dt = (x/r)dx/dt + (y/r)dy/dt

gives:

dr/dt = (x/r) - 1 = -dx/dt
-->
x + r = Constant = 50 ft

After an infinite length of time r = x = 50/2 = 25ft.

A difficult problem (I used a much more arduous process to calculate the answer originally) but I imagine that for a teenager looking at it, the most common guess (after 50 and 0) would be 25.

Answer 2

If Wile E Coyote is really the wily coyote he is supposed to be, he can actually catch the Road Runner. All he has to do is to anticipate the arrival of the Road Runner, and begin running almost parallel to the Road Runner, but just slightly ahead of it. But let's assume that Dumb E Coyote, his brother, instead runs straight for the Road Runner at all times. Then the pursuit curve has the equation, x = 0 being the path of the Road Runner:

y = (1/100)x² - 50Log(x/(50√2)) - 25(1+Log(1/√2))

I'll have to get back to you later on the arclength.

Addendum: Hmm, I can't seem to get an exact answer. The best I can do is to numerically come to 18.4594 feet as the final minimum distance between Dumb E Coyote and the Road Runner. Am I close? But I gotta leave now. I'll have another look at this and see where I could have gone wrong.

Addendum: I must be dumber than 1% of 16 year olds, because I still can't find the "simple" function describing this pursuit curve. I'll give this one more thought.

Answer 3

lol, i presumed there would be greater exciting solutions to this (different than for bradley's). Anyhoo, heehee it would be like this....... First i'd flow discover Taz the tazmanian devil and tell him that the line Runner is the main delicious ingredient on the earth (which he then couldnt help yet desire) and get the hell out of there till now he ate me! Then i'd flow tell Daffy Duck and Foghorn Leghorn that the line Runner became into boasting approximately how he became into the main superiour chicken around (which might piss them off). Then i'd tell Samitty Sam and the Bald hunter guy that there became right into a great fee on highway Runner's head. ultimately i'd flow tell Mr bugs Bunny that highway Runner is making plans to overthrow him and alter into the lead Looney Tunes character. Then i'd watch the mayhem as all of them tried to seize highway Runner. of direction they wouldnt, inspite of the incontrovertible fact that it would in basic terms make me experience greater effective understanding I wasnt the only failure. case closed. thats all human beings.

Answer 4

Here!!! Get a ruler and measure it!!!!

http://www.bb-sbl.de/_pics/Dichte_Exponential.jpg

WHAT DOES THE SPEED MATTER ANYWAY IF THEY'RE THE SAME?!?!?!