SOLUTION: Please help me solve my assignment. Thank you. Two planes take off from an airport at 9:15 AM. One flies on a course of 86 deg. at a constant rate of 190 mi/h.The second one flies

Algebra ->  Trigonometry-basics -> SOLUTION: Please help me solve my assignment. Thank you. Two planes take off from an airport at 9:15 AM. One flies on a course of 86 deg. at a constant rate of 190 mi/h.The second one flies      Log On


   



Question 90501: Please help me solve my assignment. Thank you.
Two planes take off from an airport at 9:15 AM. One flies on a course of 86 deg. at a constant rate of 190 mi/h.The second one flies on a course of 176 deg. at 160 mi/h. How far apart are they at 10:45 AM?

Found 2 solutions by kev82, Earlsdon:
Answer by kev82(151) About Me  (Show Source):
You can put this solution on YOUR website!
Hi,

The planes' flightpath can be modelled by a straight line. The position of plane i after time t is given by:



Where is the position of plane , is the starting position of plane , and is the direction(vector) in which plane flys in a time unit.

The disatnce between the planes is obviously .

Since both planes start at the airport . Substituting and tidying up gives the distance as .

Now in an hour the first plane moves 190miles, so and similarly . That's about as simple as we can get, so now we have to actually deal with some vectors to calculate .

By definition and similarly . So their dot product is:



At this point you realise all your hard work is for nothing because the angles differ by 90 degrees so you could have just used Pythagorous theorem, but if they didn't differ by 90 degrees, this is how you would do it. Anyway, because they differ by 90 degress .

The planes have been in the air for 1.5 hours so . That gives the answer as:



I'll leaave you to do the math.

Kev

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The two planes leave from the same airport at 9:15 am.
One flies a course of 86 degrees (from north) at 190 mph.
The other flies a course of 176 degrees at 160 mph.
Now if you draw a diagram of this situation, place a point representing the airport and draw two lines representing the paths of the two airplanes.
The first line will be drawn at an angle of 86 degrees (clockwise) from north while the second line will be drawn at an angle of 179 degrees (clockwise) from north.
You will notice one thing right away, the two line make a 90-degree angle between them. 176 degrees - 86 degrees = 90 degrees.
Now, if you knew the length of each line you could use the Pythagorean Theorem to solve the problem.
The length of each line (path) represents the distance traveled by each plane during the time between 9:15 am and 10:45 am. Well, this is just 1.5 hours.
To find the distance, use d = rt, or distance = rate(speed) times time (1.5 hours)
So, for the first plane, the distance traveled is: d = (190 mph)(1.5 hrs) = 285 miles.
For the second plane, d = (160 mph)(1.5 hrs) = 240 miles.
To find their distance from each other (D) at 10:45 am, you want to find the length of the hypotenuse of the right triangle, so...
D+=+sqrt%28285%5E2%2B240%5E2%29
D+=+sqrt%2881225%2B57600%29
D+=+sqrt%28138825%29
D+=+372.59miles.
The distance between the two planes at 10:45 am is 372.6 miles.