SOLUTION: A plane leaves an airport at 3:00 PM. At 4:00 PM another plane leaves the same airport traveling in the same direction at a speed 150 mph faster than that of the first plane. Fou
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Question 193092This question is from textbook Intermediate Algebra
: A plane leaves an airport at 3:00 PM. At 4:00 PM another plane leaves the same airport traveling in the same direction at a speed 150 mph faster than that of the first plane. Four hours after the first planes takes off, the second plane is 250 mi ahead of the first plane. How far does the second plane travel?
my try:
4x + 4(x+150) =250
4x + 4x + 600 = 250
8x + 600 = 250
8x = 350
x = 350/8
This question is from textbook Intermediate Algebra
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A plane leaves an airport at 3:00 PM. At 4:00 PM another plane leaves the same airport traveling in the same direction at a speed 150 mph faster than that of the first plane. Four hours after the first planes takes off, the second plane is 250 mi ahead of the first plane. How far does the second plane travel?
my try:
4x + 4(x+150) =250
4x + 4x + 600 = 250
8x + 600 = 250
8x = 350
x = 350/8
---------------------
Going in the same direction, the speed relative to each other is the difference, not the sum.
x is the speed of the slower plane. It flies 4 hrs at x = 4x miles.
x+150 is the speed of the faster plane. It flies 3 hrs at x+150.
3(x+150) = 4x + 250
3x+450 = 4x + 250
x = 200 mph
The faster plane goes 350 mph for 3 hours, so the distance is
1050 miles.
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