A Man & His Dog
One day, a man's dog waited by the door to see which way his master was going. As soon as the master appeared and started along a familiar road, the dog raced along to the end of the road, immediately returning to the man and then again racing to the end of the road, back and forth.
The dog did this four times in all—four round trips between the man and the end of the road—at a uniform speed, and then walked by the man's side the remaining distance of 81 feet at the man's speed, 4 miles per hour.
If the distance between the door and the end of the road is 625 feet, what was the uniform speed of the dog when racing back and forth?
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1 solution
Assume Vm = Speed of man and Vd=Speed of dog. Let Vd/Vm = k (>1) & d be the distance between the door and road-end. After the dog runs once to the road-end and returns to the man in 't' units of time, we can say that: Vm t+Vd t=2d or t=2d/(Vd+Vm) and the distance walked by the man in time 't' = 2 d Vm/(Vd+Vm)=2 d/(k+1). Now the new distance between the man and road-end, d1=d - 2d/(k+1)=d(k-1)/(k+1). And then d2 =d[(k-1)/(k+1)]², d3 =d[(k-1)/(k+1)]³ and lastly after the 4th meeting, d4 = d [(k-1)/(k+1)]^4. Now we're given that d4=81 ft. while d=625 ft. and Vm=4 mph. Thus, 81=625[(k-1)/(k+1)]^4, which yields k=1/4 or 4 of which the first root is inadmissible since we’ve assumed that k > 1. So k=4 or Vd=4*4 =16 mph.