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

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) (Show Source): You can put this solution on YOUR website!
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?
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