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24 Oct 2007, 17:43
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Could I solve this by setting johns time to D-2 and Jacob to D. Also set John's time to T and Jacobs time to t+1 with their rates? It seems to give me the right answer but it is not the way the book solves it so I am not sure if it is just a fluke or it can be answered either way. Also is there a way to solve this using only the difference in thier rates instead of their actual rates?
John and Jacob set out together on bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, John stops to fix a flat tire. If it takes John one hour to fix the flat tire and Jacob continues to ride during this time, how many hours will it take John to catch up to Jacob assuming he resumes his ride at 15 miles per hour? (consider John's deceleration/acceleration before/after the flat to be negligible)
Thanks!!
John and Jacob set out together on bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, John stops to fix a flat tire. If it takes John one hour to fix the flat tire and Jacob continues to ride during this time, how many hours will it take John to catch up to Jacob assuming he resumes his ride at 15 miles per hour? (consider John's deceleration/acceleration before/after the flat to be negligible)
Thanks!!
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24 Oct 2007, 18:21
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I looked at the problem the following way:
John stopped for 1 hour, and then started again. During that time, Jacob continuously rides, at 12 mph, covering a distance of 12 miles. If X is the time to catch up, then John needs to go X PLUS 1 hour @ Jacob's speed:
15(x)=12(x+1)
15x=12x+12
3x=12, x=4 hours
John stopped for 1 hour, and then started again. During that time, Jacob continuously rides, at 12 mph, covering a distance of 12 miles. If X is the time to catch up, then John needs to go X PLUS 1 hour @ Jacob's speed:
15(x)=12(x+1)
15x=12x+12
3x=12, x=4 hours
24 Oct 2007, 22:36
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First, you must find the element they have in common (distance), so therefore D = R*T.
Second, you must set both sides of the following equation to equal each other, accounting for (1) the initial 40 mins; and (2) John's 1 hour break:
15x = 12(x+1) - (2/3)(15-12)
15x = 12x + 12 - 2
15x = 12x + 10
3x = 10
x = 10/3
x = 3hrs 20 mins.
Any suggestions?
Second, you must set both sides of the following equation to equal each other, accounting for (1) the initial 40 mins; and (2) John's 1 hour break:
15x = 12(x+1) - (2/3)(15-12)
15x = 12x + 12 - 2
15x = 12x + 10
3x = 10
x = 10/3
x = 3hrs 20 mins.
Any suggestions?
