A plane flies due north (90° from east) with a velocity of 100 km/h for 3 hours. During this time, a steady wind blows southeast at 30 km/h at an angle of 315° from due east. After 3 hours, where will the plane’s position be relative to its starting point? Show your work.
v=sqrt{v(p)²+v(w)²-2v(p)v(w)cos α}=
=sqrt{100²+30²-2•100•30cos45} =81.6 km/h
s= vt=81.6•3=244.8 km
β= 85° (northeast)
To determine the final position of the plane, we need to consider both the velocity of the plane and the velocity of the wind. We can break down the velocities into their east and north components.
The plane is flying due north with a velocity of 100 km/h, so its north component of velocity is 100 km/h.
The wind is blowing southeast at an angle of 315° from due east. To find the velocity components of the wind, we can consider a right-angled triangle formed by the east and north directions. Since the angle is given as 315°, it is equivalent to -45° from due east (since clockwise rotations are considered negative). The magnitude of the wind velocity is given as 30 km/h.
To find the east component of the wind velocity, we can use trigonometry. The angle (-45°) is in the fourth quadrant, so we can use the cosine function:
cos(-45°) = adjacent/hypotenuse
east component/30 km/h = cos(-45°)
cos(-45°) is equal to cos(45°), so we have:
east component/30 km/h = cos(45°) ≈ 0.707
Solving for the east component, we find:
east component ≈ 0.707 * 30 km/h ≈ 21.21 km/h
Similarly, to find the north component of the wind velocity, we can use the sine function:
sin(-45°) = opposite/hypotenuse
north component/30 km/h = sin(-45°)
sin(-45°) is equal to -sin(45°), so we have:
north component/30 km/h = -sin(45°) ≈ -0.707
Solving for the north component, we find:
north component ≈ -0.707 * 30 km/h ≈ -21.21 km/h
Now, let's consider the time and the velocities to find the displacement of the plane.
The plane flies due north for 3 hours with a velocity of 100 km/h. The displacement is equal to the velocity multiplied by the time:
displacement of plane = velocity of plane * time
displacement of plane = 100 km/h * 3 hours
displacement of plane = 300 km north
The wind blows southeast at an angle of 315° from due east with an east component of 21.21 km/h and a north component of -21.21 km/h. Considering the wind's velocity for 3 hours, the displacement is:
displacement of wind = east component * time + north component * time
displacement of wind = 21.21 km/h * 3 hours - 21.21 km/h * 3 hours
displacement of wind = 0 km
To find the position relative to the starting point, we add the displacements of the plane and the wind:
final position = initial position + displacement of plane + displacement of wind
final position = 0 km + 300 km north + 0 km
final position = 300 km north
Therefore, after 3 hours, the plane's position relative to its starting point will be located 300 km north.