- 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)

- jamiebookeater

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- anonymous

What would be the ground speed of the plane
(its speed relative to the ground)?
Answer in units of km/h

- anonymous

can you help on the question prior to this one... when your finished with this one.

- 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

oh my word.

- anonymous

\[\theta = 17.972\] then we are looking for \[v_{a}\]??
\[v_{a}=187\sin(17.972)\]

- anonymous

Yes.

- anonymous

57.5

- anonymous

how do you answer in degrees N W

- anonymous

That answer is north of west. Notice how I drew my diagram.

- anonymous

ahaaa oh

- anonymous

but im guessing i made an error in my calculations.....

- anonymous

ok can you please actually work this out.... i got v_a=177.88

- anonymous

All answers look good.

- anonymous

ok back to this problem.

- anonymous

i know it specifically asks for degrees

- anonymous

Hop on chat

- anonymous

I am 98% positive I made a calculation error.

- anonymous

So we established that\[v_x = v \sin(\theta) \rightarrow 57.7 = 187 \sin(\theta)\]We can solve this for \(\theta\).

- anonymous

What do we get for \(\theta\)? I get 17.9722 degrees.

- anonymous

yes. that is what I got

- anonymous

v sub a = 57.5

- 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

ok i have v in respect to the wind right???

- 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

Sorry. I'm changing notations. Let me revert.
\[v_w = 57.7 = 187 \sin(\theta)\]\[v_a = 187 \cos(\theta)\]

- anonymous

ok yes.

- anonymous

What do you get for \(v_a\)? I get 177.876 km/hr

- anonymous

i rounded up to 177.88
but yes

- anonymous

that could be the error though

- anonymous

Doubtful. 5 sig figs is plenty.

- anonymous

This HAS TO be the velocity on the airplane relative to the ground.

- anonymous

It is the only vector in which displacement is occurring.

- anonymous

ok.

- anonymous

theta = 17.972 and the final answer is 177.88

- anonymous

Yes.
I'm thinking. If your teacher defined east as positive, it would be a negative velocity.

- anonymous

ok let me try this....

- anonymous

and yep...its right... FINALLY

- anonymous

that was ridiculous

- anonymous

What made it right?

- anonymous

the rounding.....

- anonymous

actually

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