A typical black lab
A person is walking their black lab on a leash towards home at a steady 3mph. When they are 7 miles from home, the person lets the lab off the leash. The lab immediately runs towards home at 8mph. When the lab reaches the house, he turns around and runs back to the person at the same speed. When the lab reaches the person, he turns back for the house. This is repeated until the person gets home and lets the lab in. How many miles does the lab cover from being let off the lead to being let in the house?
Note: No one is home to let the lab in, plus he'd much rather run back to his owner.
Note: No one is home to let the lab in, plus he'd much rather run back to his owner.
Keenan
Labels: SharedPuzzle
posted by Mr Fahrenheit at
2:30 AM
8 Comments:
The person will take 2hrs 20mins to get home, so the dog was left to run back and forth for 2hrs 20mins with the speed of 8mph. At that speed it would have covered 18.66667 miles...at least I think so!
-gilumar-
I got the same answer as web user. 18.66666666... miles
Only problem is that the person might have been walking faster as the dog was straining on the lead, so when the dog gets let off the lead, the 3mph walk becomes 2.5mph.
How did a dog turn around on a dime? It has 4 legs. It is running in one direction, but yearning to be in another. It hesitates to allow for the turn, so the steady state speed within the question is a lie. Or the dog is robot. Or cut out of a dog on a stick that flips at the house and points in back in the direction of the master once the mechanism it is mounted on reaches an endpoint at the house (and does the same upon reaching the master).
I was working to solve it, but I gave up as it took to long.
Basically you have to do the math each time the dog runs home:
7mi/8mph x 60 = minutes it took the dog the first time. Which equals 52.5min.
In that time the owner would have covered 2.652mis(52.5/60 x 3).
That leaves 4.375mi for the person to walk home.
Now as the dog runs back to the owner, they are walking towards each other so 3mph+8mph=11mph.
so 4.375mi/11mph x 60 = 23.8636min
The person would have covered 1.19318mi(23.8636/60 x 3)
The dog would have covered 3.18181mi.
So now the person has 2.4432mi to go.
The dog then takes off again towards the house and makes it in 18.324mins.
In that time the person traveled .9162mi.
that leaves 1.527mi to get home.
Once again the dog runs back to the owner and the owner walks towards the dog. this time it takes 8.32909mins to meet up.
The person traveled .416455miles
The dog traveled 1.11055miles
Total miles for the dog traveled so far = 13.7356 (7HB + 3.18181RE + 2.4432HB + 1.11055).
HB=homebound, RE=return.
That leaves 1.11055mi for the owner to get home.
The dog runs home in 8.36625min
The owner traveled .41831mi
That leaves .69224mi
This is when I realized I messed up at the bold(it's 3.18181 instead of 2.4432. so I quit... but u get the idea, if someone has an easier way to figure it out other than doing tedious math then please show me >.<
duc, the easy way is the person is walking 7 miles. calculate how long the person walks the 7 miles at 3mph (2hours 20minutes) and multiply that by how fast the dog is going (8mph). Obviously our canine challenged optimist has not seen a lab move, especially after a tennis ball. haha, give it up, Timothy.
16.34 miles or somewhat
79 elephants!
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