Linear Algebra - Vectors

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Homework Statement



A plane flies at 400 km/h [south 30 degrees east]. The wind is blowing from the east at 100 km/h. Find the resulting speed of the plane.

Homework Equations



Not sure

The Attempt at a Solution



I've drawn a diagram similar to this:

|\
| \
| \
|30\
| \
| \ 400 km/h
| \
|_____\
100km/h

I thought this was just a simple question of finding the left side of the triangle, but I don't seem to get the answer. Any hints is greatly appreciated.

Thank you in advance.
 
Last edited:

Answers and Replies

  • #2
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You have drawn a 30-60-90 right triangle. If the hypotenuse is 400 km/hr, the base would be 200 km/hr, not 100 km/hr.

Decompose the plane's velocity into a south component and and east component.
 
  • #3
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Hmm so would this work?

use 30 degrees and 400 km/h to find the speed going straight south

cos (30) * (south_speed/400)
south_speed = cos(30) * 400 = 346

Then use speed going straight south and incorporate the 100km/h wind blowing west, use Pythagorean theorem and find the resultant speed(hypotenuse)?

346^2 + 100^2 = 130000^2
= 360.5

Does that seem right?
 
  • #4
LCKurtz
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Set up a vector for the plane velocity: P = < ?, ?>
Write a vector for the wind velocity: W = <?, ?>

The resultant velocity is V = P + W and its length is the groundspeed.
 
  • #5
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Can you please show me how? I was only only taught how to do it when the points were given.
 
  • #6
LCKurtz
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Can you please show me how? I was only only taught how to do it when the points were given.

Think of the plane as at the origin with the x direction East and y direction North. The direction described as "South 30 degrees East" is the polar coordinate angle -60 degrees. So the plane velocity is given to you in polar coordinates, r = 400 and [itex]\theta = -\pi/3[/itex]. Use the polar coordinate equations to get P = <x, y> for the airplane. And you should easily be able to write the wind vector since you are given its length 100 and direction "from the East".

By the way, I didn't say your answer was incorrect. I'm just trying to show you how you should set up such problems.
 
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  • #7
35,009
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Hmm so would this work?

use 30 degrees and 400 km/h to find the speed going straight south

cos (30) * (south_speed/400)
south_speed = cos(30) * 400 = 346

Then use speed going straight south and incorporate the 100km/h wind blowing west, use Pythagorean theorem and find the resultant speed(hypotenuse)?

346^2 + 100^2 = 130000^2
= 360.5

Does that seem right?
Yes, this is pretty close. I get a slightly larger figure since I didn't round off what you're calling south_speed, which I get as 346.41 km/hr (approx.).
 

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