# In what direction should the pilot travel in this direct flight?

I'm new to physics and am having a problem with a homework questions

An airplane starting from airport A flies 300 km east, then 340 km at 34.0° west of north, and then 150 km north to arrive finally at airport B.

A.)The next day, another plane flies directly from A to B in a straight line. In what direction should the pilot travel in this direct flight?

Thanks

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please tell us what effort you yourself have put into the problem before we can help you.

ALso try drawing a diagram... it makes it significantly easier to figure out from there

I spent my entire brake in between classes today on it which is 3 hours and I have worked on it about an hour after class. I have drawed a couple of different pics and I just can't figure it out.

HallsofIvy
Homework Helper
Draw a picture! Draw a line 340 km long due East from A (scale it if you prefer!), i.e. horizontal, then draw a line at "34 degrees W of N" which would be 90- 34= 56 degrees above the horizontal line (back above the horizontal line) and mark that off at 340 km (and I'm going to call that point C). Finally draw a third line due N (vertical) from that point and mark off distance 150 km. The final point is B. Finally draw in the line from A to B. That's the straight flight back.
You could do this problem purely trigonmetrically. Draw the line from A to B and you should see two triangles- one ABC and the other AC and an unlabled point. The angle at that unlabled point is 56 degrees and the two sides adjacent (SAS) have lengths 300 and 340. You can use the cosine law to find the length of AC, then use the sine law to find the other two angles in that triangle and so angle C in the other triangle. Use the cosine law (since you know the length of CB and have just found the length of CA) to find the length of side AB and then use the sine law to find the angle ABC- the direction to fly.
Of course, most people might prefer to use "components". The components of the first leg are 300i+ 0j and the components of the third are 0i+ 150j. The components of the second leg are -340cos(56)i+ 340 sin(56)j. Add those vectors to find the vector from A to B, reverse it to find the vector from B to A and get the direction from that.

let us know when you have found the answer and what it is to make sure that youre on the right track. the methods shown here are superb for your cause. the key here is really to draw a diagram. its tricky the first time, but youll see that it all makes sense.

good luck!

Bobo