SOLUTION: I don't know why I just can't "get" distance/rate/time problems! I am really trying but am stuck again! Thanks for any help. A mountain climber is injured. His partner radios fo

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Question 178884: I don't know why I just can't "get" distance/rate/time problems! I am really trying but am stuck again! Thanks for any help.
A mountain climber is injured. His partner radios for help, then the two begin to descend from the 12,500 foot mountain at a speed of 2000 feet per hour. The rescue team starts at an altitude of 6,500ft and climbs 1000 feet per hour. How long does it take the resuers to reach the climbers? At what height will they meet?
I came up with 171.43 hours - can't be right!!! Please set me right! Thanks.
Found 2 solutions by Alan3354, Mathtut:
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
A mountain climber is injured. His partner radios for help, then the two begin to descend from the 12,500 foot mountain at a speed of 2000 feet per hour. The rescue team starts at an altitude of 6,500ft and climbs 1000 feet per hour. How long does it take the resuers to reach the climbers? At what height will they meet?
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They're 6000 feet apart (12500 - 6500)
They're closing at 3000 feet/hr (2000 + 1000)
6000/3000 = 2 hours.
12500 - 2000*2 = 8500 feet.
6500 + 1000*2 = 8500 feet also, the meeting point.

Answer by Mathtut(3670) (Show Source):
You can put this solution on YOUR website!
distance = rate * time
:
can be frustrating at times, cant it!!!!
:
we know that the total distance will be what seperates them which is 6000 ft. At this point we dont know how far each will travel. lets call the rescuers distance d and the climbers 6000-d. We know time will be equal and we will call that t. rate for climber is 2000ft/hr and rescuers 1000ft/hr
:
climber:6000-d=2000t....eq 1
rescuers: d=1000t.......eq 2
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take d's value from eq 2 and plug it into eq 1
:
6000-1000t=2000t
:
3000t=6000
:
hours
:
how far will the rescuers ascend or how far will the climber descend in 2 hours
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climber will descend:2000(2)=4000 ft
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rescuers will ascend:1000(2)=2000 ft
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so they will meet at 12500-4000=8500ft or 6500=2000=8500ft
:
feet