SOLUTION: I must say word problems or problems of this nature are impossible to calculate this particular one doesnt make a whole lot of sense any help would be appreciated:
Two airplanes
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Two airplanes
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Question 287927: I must say word problems or problems of this nature are impossible to calculate this particular one doesnt make a whole lot of sense any help would be appreciated:
Two airplanes are leaving O'Hare at the same time, one bound for New York and the other for California. The one heading to California has a tail wind and is thus going 25 MPH faster than the one headed for New York. If after one hour they are 575 miles apart, what are their speeds?
Eric
You can put this solution on YOUR website! Whatz up baby boy!
Let's take riiiide!
OK RATE X TIME =DISTANCE
Both planes have travelled (thus far) one hour.
So our time will be a KONSTANT= One hour
Now we have different rates. Buckle-up.
let x+25 represent the speed of the plane going to Calif.
Let then let x represent the speed of the plane going to New York.
THE MAJOR CONCEPT TO THIS PROBLEM IS THAT WE WILL BE ADDING DISTANCES.
LET'S LOOK BACK AT THE FORMULA--BIG MAMMA SAYS RATE X TIME =DISTANCE
WE ARE ADDING TWO DIFFERENT DISTANCES, SO WE HAVE A) RATE X TIME PLUS ANOTHER B) RATE X TIME.
STOP HERE. GO BACK OVER WHAT I HAVE WRITTEN AND MAKE SURE YOU UNDERSTAND IT BEFORE GOING FURTHER.
OK NOW A) (X+ 25) TIMES ONE(OUR KONSTANT) PLUS B) X TIMES ONE(OUR KONSTANT) =575 MILES
(X+25)+X=575 (WE ARE APPROACHING, PLEASE FASTEN SEAT BELTS.)
2x + 25 = 575
2x=550
x=275mph.
not finished yet...x+25 =300mph is speed of plane going to Calif. It goes 300 miles(remember one hour!) the other going to the Big Apple goes 275 miles(remember one hour). They made the problem simple by giving the travel time one hour--remember this. Don't lose your lunch over a little turbulence. Thanks for flying with Grinnell.
You can put this solution on YOUR website! The difficulty is that we don't know the angles of departure from Chicago.
But do we need them?
Yes, because we are told they are 575 miles apart.
If the problem had just stated that the planes had traveled 575 miles then we could solve the problem.
I will assume that that was really meant.
(x+25)+x=575
2x+25=575
2x=550
x=275 mph
x+25=300 mph
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