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anonymous

  • 4 years ago

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)

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  1. anonymous
    • 4 years ago
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    What would be the ground speed of the plane (its speed relative to the ground)? Answer in units of km/h

  2. anonymous
    • 4 years ago
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    can you help on the question prior to this one... when your finished with this one.

  3. anonymous
    • 4 years ago
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    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.

  4. anonymous
    • 4 years ago
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    oh my word.

  5. anonymous
    • 4 years ago
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    \[\theta = 17.972\] then we are looking for \[v_{a}\]?? \[v_{a}=187\sin(17.972)\]

  6. anonymous
    • 4 years ago
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    Yes.

  7. anonymous
    • 4 years ago
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    57.5

  8. anonymous
    • 4 years ago
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    how do you answer in degrees N W

  9. anonymous
    • 4 years ago
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    That answer is north of west. Notice how I drew my diagram.

  10. anonymous
    • 4 years ago
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    ahaaa oh

  11. anonymous
    • 4 years ago
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    but im guessing i made an error in my calculations.....

  12. anonymous
    • 4 years ago
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    ok can you please actually work this out.... i got v_a=177.88

  13. anonymous
    • 4 years ago
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    All answers look good.

  14. anonymous
    • 4 years ago
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    ok back to this problem.

  15. anonymous
    • 4 years ago
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    i know it specifically asks for degrees

  16. anonymous
    • 4 years ago
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    Hop on chat

  17. anonymous
    • 4 years ago
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    I am 98% positive I made a calculation error.

  18. anonymous
    • 4 years ago
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    So we established that\[v_x = v \sin(\theta) \rightarrow 57.7 = 187 \sin(\theta)\]We can solve this for \(\theta\).

  19. anonymous
    • 4 years ago
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    What do we get for \(\theta\)? I get 17.9722 degrees.

  20. anonymous
    • 4 years ago
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    yes. that is what I got

  21. anonymous
    • 4 years ago
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    v sub a = 57.5

  22. anonymous
    • 4 years ago
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    That should be v_y above. Sorry. Now, let's find out the speed to the west. \[v_x = 187 \cos(17.9722^\circ)\]

  23. anonymous
    • 4 years ago
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    ok i have v in respect to the wind right???

  24. anonymous
    • 4 years ago
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    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?

  25. anonymous
    • 4 years ago
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    Sorry. I'm changing notations. Let me revert. \[v_w = 57.7 = 187 \sin(\theta)\]\[v_a = 187 \cos(\theta)\]

  26. anonymous
    • 4 years ago
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    ok yes.

  27. anonymous
    • 4 years ago
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    What do you get for \(v_a\)? I get 177.876 km/hr

  28. anonymous
    • 4 years ago
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    i rounded up to 177.88 but yes

  29. anonymous
    • 4 years ago
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    that could be the error though

  30. anonymous
    • 4 years ago
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    Doubtful. 5 sig figs is plenty.

  31. anonymous
    • 4 years ago
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    This HAS TO be the velocity on the airplane relative to the ground.

  32. anonymous
    • 4 years ago
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    It is the only vector in which displacement is occurring.

  33. anonymous
    • 4 years ago
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    ok.

  34. anonymous
    • 4 years ago
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    theta = 17.972 and the final answer is 177.88 <is that in km/hr?

  35. anonymous
    • 4 years ago
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    Yes. I'm thinking. If your teacher defined east as positive, it would be a negative velocity.

  36. anonymous
    • 4 years ago
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    ok let me try this....

  37. anonymous
    • 4 years ago
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    and yep...its right... FINALLY

  38. anonymous
    • 4 years ago
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    that was ridiculous

  39. anonymous
    • 4 years ago
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    What made it right?

  40. anonymous
    • 4 years ago
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    the rounding.....

  41. anonymous
    • 4 years ago
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    actually

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