An airplane is heading due south at a speed of 590km/h. If a wind begins blowing from the southwest at a speed of 65.0km/h (average).
Calculate magnitude of the plane's velocity, relative to the ground.
Calculate direction of the plane's velocity, relative to the ground.
Calculate how far from its intended position it will be after 15.0min if the pilot takes no corrective action.
Express the velocity of the air plane into x (east) and y (north) directions.
v = (0, -590) km/h
Do the same for 65 km/hour wind from south-west, therefore at 45° towards NE:
w = (65√2/2,65√2/2)
Now add the components of v and w to get the resultant velocity.
The direction of the velocity is the angle atan2(y,x) taking into account of the quadrants. It should be in the fourth quadrant.
The distance from the "intended" position is the magnitude of the vector
(v -w)*15/60
since the duration is 15 minutes, and the speeds are in km/hour.
I don't understand what you mean by the commas v=(0,-590)?
To solve this problem, we can break it down into three steps:
Step 1: Calculate the resultant velocity of the airplane:
The resultant velocity is the vector sum of the airplane's velocity and the wind's velocity. We can find the resultant velocity using vector addition.
First, let's convert the speeds to meters per second (m/s):
Airplane's speed = 590 km/h * (1000 m/km) / (60 min/h) / (60 s/min) = 163.89 m/s
Wind's speed = 65.0 km/h * (1000 m/km) / (60 min/h) / (60 s/min) = 18.06 m/s
To add these two vectors, we can break them down into their horizontal and vertical components.
Airplane's velocity in the south direction = 163.89 m/s * (-1) = -163.89 m/s (since south is opposite to the positive y-axis)
Wind's velocity in the southwest direction = 18.06 m/s * cos(45°) ≈ 12.75 m/s (cosine component of the southwest direction)
Wind's velocity in the east direction = 18.06 m/s * sin(45°) ≈ 12.75 m/s (sine component of the southwest direction)
Now, add the horizontal and vertical components separately:
Resultant velocity in the east direction = 0 + 12.75 m/s = 12.75 m/s
Resultant velocity in the south direction = -163.89 m/s + (-12.75 m/s) = -176.64 m/s
Step 2: Calculate the magnitude (speed) of the plane's velocity, relative to the ground:
The magnitude of the plane's velocity relative to the ground is simply the square root of the sum of the squares of its horizontal and vertical components:
Magnitude = sqrt((12.75 m/s)^2 + (-176.64 m/s)^2) ≈ 177.04 m/s
Therefore, the magnitude of the plane's velocity, relative to the ground, is approximately 177.04 m/s.
Step 3: Calculate the direction of the plane's velocity, relative to the ground:
The direction of the plane's velocity relative to the ground can be found using the inverse tangent function:
Direction = atan2((horizontal component), (vertical component))
Direction = atan2(12.75 m/s, -176.64 m/s) ≈ -3.54°
Therefore, the direction of the plane's velocity, relative to the ground, is approximately 3.54° west of south.
Step 4: Calculate how far from its intended position it will be after 15.0 min if the pilot takes no corrective action:
To find out how far from its intended position the plane will be, we need to calculate the displacement.
Displacement = (magnitude of velocity) * (time)
Displacement = 177.04 m/s * 15.0 min * (60 s/min) ≈ 159528 m
Therefore, if the pilot takes no corrective action, the plane will be approximately 159528 meters (159.528 km) from its intended position after 15.0 minutes.
Note: Please double-check the calculations, rounding, and unit conversions as necessary.