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Question 493883: Hi my name is Diala. I'm 13 in Algebra 1 (8th grade). My problem is kind of long, but here it is (I'm stuck on it!):
11. A jet leaves the Charlotte, North Carolina, airport traveling at an average rate of 564 km/hr. Another jet leaves the airport on half hour later traveling at
744 km/hr in the same direction. How long will it take the second jet to overtake the first?
I can probably solve it, but in my head with a scratch sheet of paper just to do any addition, division, etc. Can you please show/tell the work? Cause my problem is to show the work the way my teacher wants it. Thanks!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! rate * time = distance.
the first plane has traveled at 564 km/hr for 1/2 hours before the second plane starts.
this puts the first plane at 1/2 * 564 = 282 kilometers ahead of the second plane when it starts.
the second plane needs to close the gap.
the speed of the second plane is equal to 744 km/hr.
subtract 564 km/hr from 744 km/hr and you find that the second plane is traveling 180 km/hr faster than the first plane.
the second plane needs to cover the gap of 282 kilometers.
rate * time = distance.
this translates to 180 km/hr * time = 282 km.
divide both sides of this equation by 180 km/hr to get:
time = 282 / 180 = 1.566667 hours.
the second plane will catch up to the first plane in 1.566667 hours.
another way of looking at this problem is:
first plane is traveling at 564 km / hr.
second plane is traveling at 744 km / hr.
first plane is in the air 1/2 hour longer than the first plane.
when the second plane catches up with the first plane, they will both have traveled the same distance.
we call that distance "d".
the first plane is in the air a half hour longer than the first plane, so:
we let x = hours the second plane is in the air.
this make x + .5 the number of hours the first plane is in the air.
note that 30 minutes is equal to 1/2 hour = .5 hours.
rate * time = distance.
for the first plane, this equation becomes:
564 * (x + .5) = d
for the second plane, this equation becomes:
744 * x = d
both equations are equal to d, so we can make them equal to each other to get:
564 * (x + .5) = 744 * x
simplify this equation to get:
564 * x + 282 = 744 * x
subtract 564 * x from both sides of this equation to get:
282 = 744 * x - 564 * x
this becomes:
282 = 180 * x
divide both sides of this equation by 180 to get:
282 / 180 = x
simplify to get x = 1.566666667 hours
both ways get the same answer.
in 1.56666667 hours, the first plane will have traveled 564 * 1.566666657 = 883.6 kilometers.
in 1.56666667 hours, the second plane will have traveled 744 * 1.566666667 = 1165.6 kilometers.
the second plane has traveled 1165.6 - 883.6 = 282 kilometers more than the first plane in 1.5666667 hours.
at that point in time, it has caught up with the first plane and is getting set to pass it.
this looks something like this:
hour 0 .5 2.0666667
first plan xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx (564 km/hr)
second plane xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx (744 km/hr)
first plane starts at hour 0
second plane starts at hour .5
second plane catches up to first plane at hour 2.1666667
both planes travel 1165.6 kilometers.
first plane takes 2.0666667 hours to travel that distance at 564 km/hr.
second plane takes 1.5666667 hours to travel that distance at 744 km/hr.
in the first 1/2 hour, the first plane has traveled .5 * 564 = 282 kilometers.
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