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

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SUMMARY

The discussion focuses on calculating the ground velocity, heading, and flight time for a pilot flying to city A, located 800 km north, with an airspeed of 300 km/h and a wind blowing at 120 km/h from the S40°W direction. The pilot's observed ground velocity was determined to be 241 km/h using vector components, with a calculated heading of N 22.4° E. The flight time to city A, based on the effective ground speed, remains to be calculated using the formula v = d/t.

PREREQUISITES
  • Understanding of vector components and their application in physics
  • Proficiency in basic trigonometry, including SOH CAH TOA
  • Familiarity with the cosine law for triangle calculations
  • Knowledge of the formula v = d/t for calculating velocity and time
NEXT STEPS
  • Calculate the flight time to city A using the effective ground speed of 241 km/h
  • Explore vector addition techniques for resolving wind effects on flight paths
  • Study advanced trigonometric applications in navigation and flight planning
  • Learn about the impact of varying wind conditions on flight dynamics
USEFUL FOR

Aerospace engineers, pilots, physics students, and anyone involved in navigation and flight planning will benefit from this discussion.

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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.
 
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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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