SOLUTION: at 7:00 am joe starts jogging at 6mi/h . At 7:10 Ken starts off after him. How fast must ken run in order to overtake him at 7:30
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Question 562282: at 7:00 am joe starts jogging at 6mi/h . At 7:10 Ken starts off after him. How fast must ken run in order to overtake him at 7:30
Found 2 solutions by lwsshak3, mananth:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
at 7:00 am joe starts jogging at 6mi/h . At 7:10 Ken starts off after him. How fast must ken run in order to overtake him at 7:30
**
let x=ken's speed (mph)
ken's travel time=7:30-7:10=20 min=1/3 hr
joe's speed=6 mph
joe's travel time=7:30-7:00=30 min=1/2 hr
Both runners covered the same distance
distance=travel time*speed
x*(1/3)=6*(1/2)=3
x=9 mph
ans:
How fast must ken run in order to overtake him at 7:30=9 mph
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Joe speed = 6 mph
Ken speed = x mph
Joe starts 10 minutes early
so he has run 6*10/60 = 1 mile before ken starts
distance Ken has to run to catch up = 6-1=5 miles
catchup time = 20 minutes ( from 7:10 to 7:30)= 2/3 hours
speed = d/t
speed = 5/(2/3)
speed = 5 *3/2
speed = 15/2 = 7.5 mph
CHECK
catchup time = 2/3
Joe started 1/6 hour early
so he ran for 2/3 +1/6 = 5/6 hours
his speed = 6 mph
5/6 hour *6 = 5 miles
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