Monday Madness

One fine day - out on the great plains - a pair of bull buffalo are approaching each other at great speed. They are exactly 1/4 mile apart and Bull A is traveling at a steady eight miles an hour. Bull B is traveling at a steady twelve miles an hour. They are heading directly for each other. There is a fly that can not decide which bull it wants to land on. It is continually traveling back and forth between them at a stead 30 miles per hour. The fly has just left bull A heading for Bull B. This continues until the two bulls crash head on into each other and the fly dies a horrible death as it is squished between them. How far does the fly travel before the spectacular crash?

20mph total speed of bulls so that's 60sec to impact over a 1/4 mile. Fly traveling at 30mph, so it has traveled 1/3rd more than 1/4 mile (?) Something like that anyway. Cheers, Lee

i don't think "Something like that anyway." will get you a cookie lee.. i'll wait a few days before i start working it out, probably better to convert the 1/4 mile into feet and inches. it only takes 45 seconds before the bulls crash head on, according to my calcs.

I did it without a calculator though !

so did i, after i googled how many feet where in a mile.. anyway all you need to do now is calculate how far it can fly in 45 secs(if that's correct) at 30mph to get the answer tomk wants, i do believe.

I agree terry1966 The fly has to fly for 45 secs or 1/80 hr So distance traveled by the fly is 30/80 mile = 3/8 mile = 1980 feet Lets explain s[Bull A] = 0 + 8t where s is displacement and t is time elapsed. s[Bull B] = 1/4 - 12t At collision s[Bull A] = s[Bull B] So 8t = 1/4 - 12t and we get 20t = 1/4 or t = 1/80 hours Fly is flying at constant 30mph So distance the fly travels is 30 * 1/80 or 3/8 mile

1,980 feet sounds right to me.

I agree terry1966

The fly has to fly for 45 secs or 1/80 hr
So distance traveled by the fly is 30/80 mile = 3/8 mile = 1980 feet

Lets explain

s[Bull A] = 0 + 8t where s is displacement and t is time elapsed.
s[Bull B] = 1/4 - 12t
At collision
s[Bull A] = s[Bull B]
So
8t = 1/4 - 12t and we get 20t = 1/4 or t = 1/80 hours

Fly is flying at constant 30mph So distance the fly travels is 30 * 1/80 or 3/8 mile

I tips my hat to you sir.
A well deserved cookie!

Cheers,
Lee

Well done again otisx!

The bulls are indeed converging from 1/4 mile apart at a rate of 20 mph. The fly is traveling at 30 mph or one and one-half times the rate of the bulls. Therefore the fly will travel one and one-half times the distance or 3/8 of a mile.

1.5 x 0.25 x 5280 = 1980 feet. never was any good trying to work things out in fractions. lee was thinking along the right lines in his first post just got the details slightly wrong. well done again otisx.

Thanks guys. When TomK gives me the cookie ill make sure and give half to Lee and terry1966 And these two glasses one for each of you since you guys were both going the right way first. I really have to start watching my figure too. Too many cookies will cause cavities.