Question 331440: If two planes leave the same airport at 1:00 PM, how many miles apart will they be at 3:00 PM if one travels directly north at 150 mph and the other travels directly west at 200 mph?
I try the following but did not get the correct solution:
Plane A travels North 150 mph and Plane B travels West at 200 mph.
From 1:00 PM to 3:00 PM--> Plane A has traveled for two hours (150 + 150) = 300
Plane B has traveled for two hours (200 + 200) = 400
So, 400 - 300 = 100 miles apart?
Or is it a pythagoream problem...using the a^2 + b^2 = c^2 which would be
150^2 + b^2 = 300^2
= 22500 + b^2 = 90000
= b^2 = 9000-22500
= b^2 = 67500
= b = square root of 67500
which would be 259.8 miles apart
or could it be the following...
Plane A is 50 miles behind which means Plane B is 50 miles ahead...
so Plane B will be 100 miles ahead and that gives me 150 + 200 + 150 = 500 miles apart?
If so, is there a quicker way of finding it using an equation? Or can you answer it in a more simpler way?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! If two planes leave the same airport at 1:00 PM, how many miles apart will they be at 3:00 PM if one travels directly north at 150 mph and the other travels directly west at 200 mph?
I try the following but did not get the correct solution:
Plane A travels North 150 mph and Plane B travels West at 200 mph.
From 1:00 PM to 3:00 PM--> Plane A has traveled for two hours (150 + 150) = 300
Plane B has traveled for two hours (200 + 200) = 400
So, 400 - 300 = 100 miles apart?
Or is it a pythagoream problem...using the a^2 + b^2 = c^2 which would be
150^2 + b^2 = 300^2
= 22500 + b^2 = 90000
= b^2 = 9000-22500
= b^2 = 67500
= b = square root of 67500
---------------------------------
It can be solved using Pythagoras, but the distance is the hypotenuse.
You solved for one of the legs.
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c^2 = 300^2 + 400^2
c = 500 miles
--------------------------------
which would be 259.8 miles apart
or could it be the following...
Plane A is 50 miles behind which means Plane B is 50 miles ahead...
so Plane B will be 100 miles ahead and that gives me 150 + 200 + 150 = 500 miles apart?
If so, is there a quicker way of finding it using an equation? Or can you answer it in a more simpler way?
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