SOLUTION: Vehicle A and B, moving at a constant velocities, are 8km apart at the starting time of 3:40pm. If after 40 mins, Vehicle B overtakes A, at what time will they be 10 km apart?
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Question 1033112: Vehicle A and B, moving at a constant velocities, are 8km apart at the starting time of 3:40pm. If after 40 mins, Vehicle B overtakes A, at what time will they be 10 km apart?
Is there a strategy to do these problems? Plus, it seems like they are only a few numerical values. How do I deal with these problems as well? Thank you very much.
Found 2 solutions by ankor@dixie-net.com, mananth:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Vehicle A and B, moving at a constant velocities, are 8km apart at the starting time of 3:40pm.
If after 40 mins, Vehicle B overtakes A, at what time will they be 10 km apart?
:
We are dealing with their relative speed here.
40 min = 2/3 hr
speed = dist/time
We know from the given information that the difference in their velocities is:
= 12 km/hr the relative speed to each other
:
How long will it take to travel 10 km at 12 km/hr?
= hr which is * 60 = 50 min
:
Travel time from 3:40. 50 min + 40 min = 90 min from 3:40: 5:10 pm
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let speed of B be x km/h
and that of A = y
after 2/3 hour B&A are together
catchup speed *catchup time =8
catchup speed = x-y
catchup time = 2/3
d=8
2/3(x-y) =8
x-y =12
d=10
let t be the time they are 10 km apart
t(x-y) = 10
12t = 10
t = 10/12 = 5/6
5/6 +2/3 = 9/6 = 1.5 hours
3:40 +1:30 = 5:10 pm
At 5:10 they are 10 km apart
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