SOLUTION: At 6:00 A.M. a Pony Express rider leaves San Antonio for Las Cruces. His horse travels at 10 miles per hour. A second Pony Express Rider leaves at 9:00 A.M. with the sole purpose

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Question 111939: At 6:00 A.M. a Pony Express rider leaves San Antonio for Las Cruces. His horse travels at 10 miles per hour. A second Pony Express Rider leaves at 9:00 A.M. with the sole purpose of giving an additional package for the first Pony Express rider to deliver. The second rider’s horse travels at 18 miles per hour. At what time will the second Pony Express rider catch up to the first Pony Express rider.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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t 6:00 A.M. a Pony Express rider leaves San Antonio for Las Cruces. His horse travels at 10 miles per hour. A second Pony Express Rider leaves at 9:00 A.M. with the sole purpose of giving an additional package for the first Pony Express rider to deliver. The second rider’s horse travels at 18 miles per hour. At what time will the second Pony Express rider catch up to the first Pony Express rider.
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One thing to remember on all these "catch-up" problems is when this occurs they will have traveled the same distance. You can make a simple distance equation from this fact.
:
Let t = time (in hrs) traveled by the 1st rider when the 2nd rider catches up.
Then
(t-3) = time traveled by the 2nd rider
:
Distance = speed * time
:
2nd rider dist = 1st rider dist
18(t-3) = 10t
18t - 54 = 10t
18t - 10t = +54
8t = 54
t = 54/8
t = 6.75 hrs
Change the decimal portion to minutes: .75*60 = 45 min
:
6 hr 45 min from 6 am = 12:45 when the 2nd rider catches up
:
:
We can check our solution by confirming that both riders traveled the same distance:
2nd rider travel time 3.75 hr
3.75 * 18 = 67.5 mi
6.75 * 10 = 67.5 mi
:
Did this make sense, any questions?