Math puzzle - Hunter and his dog

Drakar

A hunter is in the woods with his dog, and wants to return to his cabin. The hunter walks back to the cabin at a constant speed. His excitable dog runs to the cabin, then back to the hunter, then back to the cabin and back to the hunter again repeatedly until the hunter arrives back at the cabin. What total distance does the dog travel when the hunter reaches the cabin if the dog travels twice as fast as the hunter?

Maybe the question is too easily phrased, the key is to think in times not distance. If you imagine it takes the hunter 10 minutes, it also took the dog 10 minutes (since the dog will keep going back and forth but they arrive together). Since the both took the same time, and the dogs speed is twice the hunters we can tell that the dogs travels twice the distance the hunter travels, ie twice the distance to the cabin.

I'd like to see an alternative solution involving distances, but I keep coming up with icky limit equations...

Rubecula

The dog covers twice the distance of the man for any given position. ie to the cabin and back again.

If the dog runs twice as fast too, then it must travel twice as far in the same time, For each position it will run twice the distance.

One trip = twice as far.
two trips (the hunter has moved forward) = three times the original distance.

and so on.

The distance the dog covers depends on the number of times it runs back to the hunter.

If both travel just the once then the dog covers twice the distance at twice the speed.