SOLUTION: Joe leaves his house to go jogging when his wrist watch reads exactly 1:00pm. On the way to his destination, he stops for a break. He arrives at Pete's house, knocks on the door, t

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Joe leaves his house to go jogging when his wrist watch reads exactly 1:00pm. On the way to his destination, he stops for a break. He arrives at Pete's house, knocks on the door, t      Log On


   



Question 816907: Joe leaves his house to go jogging when his wrist watch reads exactly 1:00pm. On the way to his destination, he stops for a break. He arrives at Pete's house, knocks on the door, then immediately turns around and jogs back home at the same pace, taking another break along the way. At the time of his arrival back home, his watch reads exactly 1:40pm.
Upon returning home, he calls Pete and asks what time he knocked on the door. Pete answers,"1:15pm according to my clock".
What did Pete's clock read when Joe left his house?
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This is a real-world problem abstracted from trying to get the exact time of a server in relation to a client, with an unknown amount of latency on both the request and the response.
Joe needs to get his watch exactly in sync with Pete's clock. Joe knows that because of his breaks, the rate of travel to and from Pete's house varies. Since he doesn't know what his watch read when Pete's clock read 1:15, the only way to get his watch in sync is to determine the length of his breaks.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Joe can ask Pete what his clock reads now and compare them and adjust the watches accordingly.
t+b+c+r=40/60
we have four unknowns which add up to 40/60
two jogging times and two breaks.
we have one consistent jogging pace but unknown distances and times