A community for students.
Here's the question you clicked on:
 0 viewing
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)
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)

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What would be the ground speed of the plane (its speed relative to the ground)? Answer in units of km/h

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you help on the question prior to this one... when your finished with this one.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\theta = 17.972\] then we are looking for \[v_{a}\]?? \[v_{a}=187\sin(17.972)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how do you answer in degrees N W

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That answer is north of west. Notice how I drew my diagram.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but im guessing i made an error in my calculations.....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok can you please actually work this out.... i got v_a=177.88

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0All answers look good.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok back to this problem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i know it specifically asks for degrees

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am 98% positive I made a calculation error.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So we established that\[v_x = v \sin(\theta) \rightarrow 57.7 = 187 \sin(\theta)\]We can solve this for \(\theta\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What do we get for \(\theta\)? I get 17.9722 degrees.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes. that is what I got

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That should be v_y above. Sorry. Now, let's find out the speed to the west. \[v_x = 187 \cos(17.9722^\circ)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok i have v in respect to the wind right???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which 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
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry. I'm changing notations. Let me revert. \[v_w = 57.7 = 187 \sin(\theta)\]\[v_a = 187 \cos(\theta)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What do you get for \(v_a\)? I get 177.876 km/hr

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i rounded up to 177.88 but yes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that could be the error though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Doubtful. 5 sig figs is plenty.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This HAS TO be the velocity on the airplane relative to the ground.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It is the only vector in which displacement is occurring.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0theta = 17.972 and the final answer is 177.88 <is that in km/hr?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. I'm thinking. If your teacher defined east as positive, it would be a negative velocity.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok let me try this....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and yep...its right... FINALLY
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.