2D motion: frames of refrence: calculating velocity,heading and time

  • Thread starter rahrahrah1
  • Start date
  • #1
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Homework Statement



A pilot wishes to fly to city A, 800km north of the present location. The plane is capable of an air velocity of 300km/h. There is a wind blowing 120 km/h [S40°W]
a) if she flies directly to city A, what will be her observed ground velocity?
b) what should her orignal heading have been?
c) what will her flight time to city A be?

Homework Equations


cosine law
basic trig (SOH CAH TOA, Pythagorean)
v = d/t

The Attempt at a Solution



a) I realize that the question can either be answered using cosine law or broken up into components. I decided to use components.
I therefore made a vector table and broke up the wind vector. So that I got 77.1km/h south and 92km/h west. I knew the pilot wanted to travel north and could do so at a velocity of 300 km/h. I added this vector to the table. I came up with an right angle triangle:
- 223 km/h north
- 92km/h west
I then solved for the Hypotenuse, which came out to be 241 km/h. I assumed this was the velocity I was looking for.

b) I'm guessing they are asking for the heading that corresponds to the velocity that was calculated in a)
using the same right traingle that I had built from components, I used basic trig and calculated the heading to be: [N 22.4° E]

c) I know this is where my 800 km comes into play, however, I'm stumped as where to go from here and concerned whether my results calculated for a) and b) are actually right.
 

Answers and Replies

  • #2
I was reading over my post and I can't remember, do I include the 300km/h in my orignal component triangle?
 

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