anonymous
  • anonymous
An airplane flies at airspeed (speed relative to the air) 187 km/h. The pilot wishes to fly due West but there is a 57.7 km/h wind blowing in from North to South. In what direction should the pilot head the plane? Answer in units of  (north of West)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
What would be the ground speed of the plane (its speed relative to the ground)? Answer in units of km/h
anonymous
  • anonymous
can you help on the question prior to this one... when your finished with this one.
anonymous
  • anonymous
This is a little tricky. We can't fly due west because the wind will blow us off target to the south. Therefore, we must vectorize our flight such that our velocity to the north equals the velocity of the wind. The angle we are interested in is north of west. Let's draw a diagram. |dw:1327623668551:dw|\(v_a\) is the velocity actually traveling west, \(v_t\) is the velocity of the airplane, and \(v_w\) is the velocity of the wind. From the diagram, we can see that \[v_a = v_t \cos(\theta)\]and\[v_w = v_t \sin(\theta)\]We know \(v_w\) and \(v_t\). Solve the second expression for \(\theta\). The velocity of the airplane relative to the ground can be had from the first expression.

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More answers

anonymous
  • anonymous
oh my word.
anonymous
  • anonymous
\[\theta = 17.972\] then we are looking for \[v_{a}\]?? \[v_{a}=187\sin(17.972)\]
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
57.5
anonymous
  • anonymous
how do you answer in degrees N W
anonymous
  • anonymous
That answer is north of west. Notice how I drew my diagram.
anonymous
  • anonymous
ahaaa oh
anonymous
  • anonymous
but im guessing i made an error in my calculations.....
anonymous
  • anonymous
ok can you please actually work this out.... i got v_a=177.88
anonymous
  • anonymous
All answers look good.
anonymous
  • anonymous
ok back to this problem.
anonymous
  • anonymous
i know it specifically asks for degrees
anonymous
  • anonymous
Hop on chat
anonymous
  • anonymous
I am 98% positive I made a calculation error.
anonymous
  • anonymous
So we established that\[v_x = v \sin(\theta) \rightarrow 57.7 = 187 \sin(\theta)\]We can solve this for \(\theta\).
anonymous
  • anonymous
What do we get for \(\theta\)? I get 17.9722 degrees.
anonymous
  • anonymous
yes. that is what I got
anonymous
  • anonymous
v sub a = 57.5
anonymous
  • anonymous
That should be v_y above. Sorry. Now, let's find out the speed to the west. \[v_x = 187 \cos(17.9722^\circ)\]
anonymous
  • anonymous
ok i have v in respect to the wind right???
anonymous
  • anonymous
which is on the y axis... thats what the y stands for in this case so v sub y and v sub w are one in the same?
anonymous
  • anonymous
Sorry. I'm changing notations. Let me revert. \[v_w = 57.7 = 187 \sin(\theta)\]\[v_a = 187 \cos(\theta)\]
anonymous
  • anonymous
ok yes.
anonymous
  • anonymous
What do you get for \(v_a\)? I get 177.876 km/hr
anonymous
  • anonymous
i rounded up to 177.88 but yes
anonymous
  • anonymous
that could be the error though
anonymous
  • anonymous
Doubtful. 5 sig figs is plenty.
anonymous
  • anonymous
This HAS TO be the velocity on the airplane relative to the ground.
anonymous
  • anonymous
It is the only vector in which displacement is occurring.
anonymous
  • anonymous
ok.
anonymous
  • anonymous
theta = 17.972 and the final answer is 177.88
anonymous
  • anonymous
Yes. I'm thinking. If your teacher defined east as positive, it would be a negative velocity.
anonymous
  • anonymous
ok let me try this....
anonymous
  • anonymous
and yep...its right... FINALLY
anonymous
  • anonymous
that was ridiculous
anonymous
  • anonymous
What made it right?
anonymous
  • anonymous
the rounding.....
anonymous
  • anonymous
actually

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