Question 152760: At 5:00 PM a plane leaves an airport and flies due north at 588 km/h. At 6:00 PM a second plane leaves the airport, also flying north but at 732 km/h. When does the second plane overtake the first?
Found 2 solutions by Alan3354, nerdybill:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! At 5:00 PM a plane leaves an airport and flies due north at 588 km/h. At 6:00 PM a second plane leaves the airport, also flying north but at 732 km/h. When does the second plane overtake the first?
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The 1st plane has is 588 km away when the 2nd plane leaves.
The 2nd plane is going 732 - 588 kph faster, = 144 kph faster.
The time to overtake is 588/144, = 4.08333 = 4:05
So it overtakes it at 2205, or 10:05 PM.
Answer by nerdybill(7384)  (Show Source):
You can put this solution on YOUR website! At 5:00 PM a plane leaves an airport and flies due north at 588 km/h. At 6:00 PM a second plane leaves the airport, also flying north but at 732 km/h. When does the second plane overtake the first?
.
You need to apply the "distance formula":
d = rt
where
d is distance
r is rate or speed
t is time
.
Let x = amount of time it takes for second plane to catch up to the first plane
.
"distance traveled by plane one" = "distance traveled by plane two"
588(x+1) = 732(x)
588x + 588 = 732x
588 = 144x
588/144 = x
4.083 hours = x
or in terms of hours and minutes:
4 hrs and .083(60) mins
4 hrs and 4.98 mins
or, in terms of hours, minutes and seconds:
4 hrs, 4 mins and .98(60) seconds
4 hrs, 4 mins and 59 seconds
.
Conclusion:
6 PM plus 4 hrs, 4 mins and 59 seconds
is
10:04:59 PM
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