consider east-west direction along x-axis with east pointing towards positive x-axis.
consider north-south direction along y-axis with north pointing towards positive y-axis
\underset{v_{p}}{\rightarrow} = velocity of the plane = 0 i + 100 j
\underset{v_{w}}{\rightarrow} = velocity of the wind = (30 Cos315) i + (30 Sin315) j = 21.2 i - 21.2 j
net velocity of the plane is given as
\underset{v_{net}}{\rightarrow} = \underset{v_{p}}{\rightarrow} + underset{v_{w}}{\rightarrow}
\underset{v_{net}}{\rightarrow} = (0 i + 100 j) + (21.2 i - 21.2 j ) = 21.2 i + 78.8 j
t = time of travel = 3 hours
position of the plane is given as
\underset{X}}{\rightarrow} = \underset{v_{net}}{\rightarrow} t
\underset{X}}{\rightarrow} = (21.2 i + 78.8 j ) (3)
\underset{X}}{\rightarrow} = 63.6 i + 236.4 j
magnitude of distance from the initial starting point is given as
d = sqrt((63.6)² + (236.4)²) = 244.81 m
direction is given as
θ = tan⁻¹(236.4/63.6) = 75 deg north of east.
Answer:
244.78 km, 74.9⁰ North East
Explanation:
Step 1: identify the given parameters
Velocity of plane = 100 km/h
Velocity of wind = 30 km/h
Time of flight = 3 hours
Distance traveled by plane = (100 km/h)*(3 hours) = 300 km
Distance blew by wind = (30 km/h)*(3 hours) = 90 km
Step 2: construct a triangle with the distance of plane and wind as show in the imaged uploaded.
Step 3: calculate the final position (R) of the plane
Using cosine rule, calculate R
R² = 300² + 90² -2(300X90)cos315
R² = 98,100 - 54,000cos315
R² = 98,100 - 38,183.766
R² = 59,916.234
R = √59,916.234
R = 244.78 km
Step 4: calculate the direction of the plane(Ф)
Using sine rule, calculate the direction of the plane
[tex]\frac{90}{sine \theta} =\frac{244.78}{sine 315}[/tex]
[tex]sine\theta =\frac{90*sine315}{244.78}[/tex]
[tex]sine\theta = -0.25998[/tex]
[tex]\theta = -15.07[/tex]
[tex]\theta = 90-15.07[/tex] = 74.9⁰ North East
