SOLUTION: at 2:00 p.m. a runner heads north on a highway jogging at 10mph. at 2:30 p.m. a driver heads north on the same highway to pick up the runner. if the car travels at 55 mph, how long
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Question 303501: at 2:00 p.m. a runner heads north on a highway jogging at 10mph. at 2:30 p.m. a driver heads north on the same highway to pick up the runner. if the car travels at 55 mph, how long will it take the driver to catch the runner? Found 3 solutions by Alan3354, stanbon, mananth:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! at 2:00 p.m. a runner heads north on a highway jogging at 10mph. at 2:30 p.m. a driver heads north on the same highway to pick up the runner. if the car travels at 55 mph, how long will it take the driver to catch the runner?
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In 30 min, the runner goes 5 miles.
The driver gains on the runner at 45 mph (55-10)
5/45 = 1/9 hour = 20/3 minutes
= 6 min 40 seconds
You can put this solution on YOUR website! at 2:00 p.m. a runner heads north on a highway jogging at 10mph.
at 2:30 p.m. a driver heads north on the same highway to pick up the runner.
if the car travels at 55 mph, how long will it take the driver to catch the runner?
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Runner DATA:
rate = 10 mph ; time = x hrs ; distance = rt = 10x miles
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Driver DATA:
55 mph ; time = x-(1/2) hrs ; distance = rt = 55(x-(1/2))
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Equation:
distance = distance
10x = 55(x-(1/2))
10x = 55x - 27.5
45x = 27.5
x = 0.6111 hrs
x = 0.6111(60)= 36 2/3 minutes
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Cheers,
Stan H.
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You can put this solution on YOUR website! jogger runs at 10 mph
so in half hour he goes 5 miles
Let the distance from this 5 mile point to the point they meet be x
time taken by jogger = x/10
car travels at 55 mph
it has to travel x+5 miles to catch up the jogger
time car takes = x+5 / 55
both these times are same
x+5/55 = x/10
10(x+5) = 55x
10x+50 = 55x
45x=50
x= 50/45 =10/9
time = distance / speed
distance car goes = 10/9 +5 miles
time = 55/9 / 55
1/9
= 1 /9
-= 1/9 *60 minutes
the truck cathches up in 6.66 minutes
ananth
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