The geometric series method is straightforward and you need to practice setting up such methods yourself....
To solve it the easy way, suppose there was another dog Trot2 who started out running with Trot, however Trot2 forgets to make the turns and just continues to run in a straight line. How far will Trot2 have run by the time the two runners meet. And at any given time what is the relation between the total distance run by Trot and Trot2?
Tom and Joel are good runners, both able to run at a constant speed of 10 mph. Their amazing dog Trot can do even better, he runs at 20 mph. Starting from towns 60 miles apart, tom and joel run toward each other while Trot runs back and forth between them. How far does Trot run by the time the boys meet? Assume that Trot started with Tom running toward Joel and that he is able to make instants turnarounds.
Solve it using a geometric series then solve it an easier way....
I'm never good at these kinds of questions.
How do i do this?
2 answers
Solve the problem two ways.
(a) Use a geometric series.
(b) Find a shorter way to do the problem.
Based on dog cycles, the time between dog turn arounds.
If the distance between the runners is d,
(2/3) d (1/3) d
*-----------------------------|--------------*
(a) The dog will run 2/3 of the distance
(b) Each runner will run 1/3 of the distance.
(c) The new distance will be d minus the distance the runners ran.
(d) The dog will run (2/3) d.
(e) The new distance will be (1/3) d.
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Distance d d/3 d/3^2 d/3^3 ....
The dog will run
(2/3)d + (2/3)(d/3) + (2/3) (d/3^2) + (2/3) (d/3^3) + ......
(2/3) d
(2/3) d ( 1 + 1/3 + 1/3^2 + 1/3^3 + ,.....) = -----------
1-1/3
2d
= ----------- = d = 60 miles.
3-1
(b) The runners have to run for 3 hours. The dog runs for
3 hours so he runs 60 miles.
or The dog runs twice as fast as the runner. Each runner runs 30
miles, so the dog runs sixty miles.
(a) Use a geometric series.
(b) Find a shorter way to do the problem.
Based on dog cycles, the time between dog turn arounds.
If the distance between the runners is d,
(2/3) d (1/3) d
*-----------------------------|--------------*
(a) The dog will run 2/3 of the distance
(b) Each runner will run 1/3 of the distance.
(c) The new distance will be d minus the distance the runners ran.
(d) The dog will run (2/3) d.
(e) The new distance will be (1/3) d.
�
Distance d d/3 d/3^2 d/3^3 ....
The dog will run
(2/3)d + (2/3)(d/3) + (2/3) (d/3^2) + (2/3) (d/3^3) + ......
(2/3) d
(2/3) d ( 1 + 1/3 + 1/3^2 + 1/3^3 + ,.....) = -----------
1-1/3
2d
= ----------- = d = 60 miles.
3-1
(b) The runners have to run for 3 hours. The dog runs for
3 hours so he runs 60 miles.
or The dog runs twice as fast as the runner. Each runner runs 30
miles, so the dog runs sixty miles.