You can
put this solution on YOUR website!I think your actual first step was to assign x to represent the speed of the eastbound aircraft and y to represent the speed of the one going to California. Ok, as far as that goes, but from there on, you went completely off in the wrong direction.
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The basic formula you need for distance-rate-time questions like this is
. let's use these three variables instead of x and y because it is easier to keep things straight in our mind. Let's call the plane going to New York number 1 and the other one number 2.
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So, to start, we can write two equations:
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, and
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But we know that both planes have been flying for 1 hour, so we can now write:
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, and
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We also know that aircraft 2 was traveling 25 mph faster than aircraft 1, so we can write:
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, and
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Finally, we know that the SUM of the distances the two aircraft traveled was 575 miles, or:
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, but we can substitute the right side expressions to obtain:
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Now all we have to do is solve:
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:
So the New York flight is traveling 275 mph, and the California flight is traveling 25 mph faster than that or 300 mph. After one hour, the New York flight will have gone 275 miles, and the California flight, 300 miles: 275 plus 300 is 575, so the answer checks.
You can
put this solution on YOUR website!QUESTION -- ARE THESE PLANES FLYING DUE WEST & DUE EAST?
IF NOT -- WHAT IS THE ANGLE BETWEEN THEIR FLIGHT PATHS?
THIS INFORMATION WOULD HELP DETERMINE HOE TO SOLVE THIS PROBLEM.
IF THEY ARE ON OPOSITE (180 DEGREE) FLIGHT PATHS THEN:
X+(X+25)=575
2X+25=575
2X=575-25
2X=550
X=550/2
X=275 MPH FOR THE SLOWER PLANE.
275+25=300MPH FOR THE FASTER PLANE.
PROOF
275+300=575
575=575
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