SOLUTION: A horse and a chicken are are standing at a point on a 1/4 mile track. The horse heads in one direction at 26mph while the chicken heads in the opposite direction at 4mph. How many

Algebra ->  Equations -> SOLUTION: A horse and a chicken are are standing at a point on a 1/4 mile track. The horse heads in one direction at 26mph while the chicken heads in the opposite direction at 4mph. How many      Log On


   

Question 923465: A horse and a chicken are are standing at a point on a 1/4 mile track. The horse heads in one direction at 26mph while the chicken heads in the opposite direction at 4mph. How many seconds will pass before they meet again
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A horse and a chicken are are standing at a point on a 1/4 mile track.
The horse heads in one direction at 26mph while the chicken heads in the opposite direction at 4mph.
How many seconds will pass before they meet again
:
Convert their speeds to feet/sec
%2826%2A5280%29%2F3600 = 38.13 ft/sec for the horse
%284%2A5280%29%2F3600 = 5.87 ft/sec
Convert 1/4 mi to 5280%2F4 = 1320 ft
When the horse meets the chicken they will have traveled a total 1320 ft
let t = no. of seconds for this to happen
38.13t + 5.87t = 1320
44t = 1320
t = 1320/44
t = 30 seconds
:
much easier way to do it, since they are going towards each other the relative speed is 26 + 4 = 30 mph
How long to go 1/4 mi at 30 mph: Use .25 for 1/4, convert to seconds
.25%2F30 * 3600 = 30 sec