Well, this is kind of a "plane" in the neck question. Let's break it down!
To find the magnitude of the plane's velocity relative to the ground, we need to use some fancy vector math. We'll have to calculate the resultant velocity by adding the airplane's velocity to the velocity of the wind.
To do this, we can use the Pythagorean theorem. The magnitude of the plane's velocity can be found using the formula:
Magnitude = β(plane's velocity^2 + wind's velocity^2)
So, plugging in the given values, we get:
Magnitude = β(590^2 + 65^2) km/h
Calculating that, we find the magnitude of the plane's velocity relative to the ground is approximately 597.23 km/h. Phew, that's fast!
Now, to determine the direction of the plane's velocity relative to the ground, we need to find the angle. We can use trigonometry for this. The angle can be found using the formula:
Angle = tan^(-1)(wind's velocity / plane's velocity)
Plugging in the given values, we get:
Angle = tan^(-1)(65 / 590) degrees
Calculating that, we find the angle to be approximately 6.18 degrees. So, the plane is headed southward with a slight southwestern deviation.
Finally, to calculate how far off course the plane will be after 15.0 minutes, we need to find the distance traveled by the plane in that time. Remember, distance = speed Γ time.
So, distance off course = (plane's velocity - wind's velocity) Γ time
Plugging in the given values, we get:
Distance off course = (590 - 65) km/h Γ (15 / 60) h
Calculating that, we find the plane will be approximately 144.58 km off course. Looks like the pilot might want to make a u-turn!
I hope this helps you navigate through this windy situation. Safe travels!