Answer:
θ = 14° ........ North of West
V_A/G = 300 km/h ........ West
Explanation:
Given:
- The speed of wind V_A = 75 km/h ( South )
- The speed of plane relative to air V_P/A = 310 km/h
- The plane wants to go in westwards.
Find:
In which direction should the pilot head?
What is the speed of the air relative to a person standing on the ground, v_A/G?
Solution:
- The plane wants to go westwards; however, the wind would push the plane down south. To combat the effect of wind the plane needs to travel somewhat North West just enough such that wind pushes it down to a point westwards. The angle the plane travels North of west is θ.
- Construct a velocity vector triangle on a coordinate system where unit vectors +i = East , -i = West , +j = North and -j = South. With plane's initial position as origin.
- We know that the plane travels relative to air at angle θ North of west we have the following velocity vector for V_P/A:
vector (V_P/A) = -V_P/A*cos(θ) i + V_P/A*sin(θ) j
- Similarly for velocity of wind V_w and plane relative to ground or stationary observer on ground V_A/G is:
vector (V_w) = -V_w j = -75 j km/h
vector (V_A/G) = -V_A/G i = -V_A/G i km/h
- Use the relative velocity formulation:
vector (V_A/G) = vector (V_P/A) + vector (V_w)
- Plug the respective expressions developed above:
-V_A/G i km/h = -310*cos(θ) i + 310*sin(θ) j -75 j km/h
- Now compare coefficients of i and j unit vectors we have:
For j unit vector: 310*sin(θ) -75 km/h = 0
sin(θ) = 75 / 310 = 0.2419354839
θ = arcsin(0.2419354839)
θ = 14° ........ North of West
For i unit vector: -V_A/G = -310*cos(θ)
V_A/G = 310*cos(14)
V_A/G = 300 km/h ........ West
Answer: The plane must travel at 14 degrees north of west if it wants to end up at any point west of its direction. The stationary observer at ground will see the plane moving west at a speed of 300 km/h.
