Old Dog

Finally found a job! I'll be a clinical research coordinator at a teaching hospital/medical center, whoohoo. (The major downside is that I have to commute for 2 hours and get up at 5:45...)

Also I bought one of these, so I'm pretty excited this week ^o^ (Farewell, Beetle ._,)


This was on Car Talk the other day:

A man and his dog live at the foot of a tall hill, and for exercise they walk up to the top and then down the hill every morning. But as the years go by, the dog has gotten old, and now can only walk at half his original speed. So the man and the dog still walk the hill every day, but the man will get to the hilltop and then turn around. The dog meets him on the way down, and due to the power of gravity, is able to make it down the hill at his original young dog pace.

So the question is, if the hilltop is 3 miles away from their house, how many miles does the old dog walk every day?




Answer:
You can always make up values and plug it in, but here's the algebra answer that I worked out :D

Since distance = rate x time, the time that the dog and owner walked are the same, so t=d/r. And let's set D for the distance to the hilltop, and d for the dog's distance when he meets his owner. Rate will be r for the man and r/2 for the dog, since he walks half as fast.

The trick is the set up for the equation. Man makes it to the hilltop and turns around, so the distance he walked is the total plus where he meets his dog.
(D + (D-d)) / r = d / (r/2)

then,
(2D - d) / r = 2d / r
(2D - d) / d = 2
2 D/d - 1 = 2
D/d = 3/2
so the distance the dog walked will be d/D, or 2/3 the trip, and since the hilltop is 3 miles away, the dog walks 4 miles.