An airplane is heading due south at a speed of 590 km/h. If a wind begins blowing from the southwest at a speed of 65.0 km/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.
I typed in the answer for the direction (85 and 175) and they are both wrong according to the problem. I want to try 4.8 degrees since it says east of south, but I don't know though.
wind blows plane Northeast
V = + 65 cos 45 i -590 j + 65 sin 45 j
= +46 i - 590 j + 46 j
= 46 i - 544 j
|V| = sqrt (46^2+544^2) = 546
in quadrant 4
tan theta = 544/46
theta = 85 degrees south of East
= 85+90 = 175 clockwise from North or about South by 1/2 point East
In 1/4 hour should have gone 147.5 South
in fact went
(1/4)546 sin 85 South = 136 South
and (1/4)546 cos 85 East = 11.9 East
error vector = 11.9 i + (147.5-136) j
= 11.9 i + 11.5 j
error = sqrt(11.9^2 + 11.5^2)
=16.5 km
What the last comment said is wrong although the reasoning that the southwest wind will get a 45 degree angle is true, the next step is addition between the vector of the wind and the vector of the airplane. Then pitagoras you get the magnitude
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 = √(plane's velocity^2 + wind's velocity^2)
So, plugging in the given values, we get:
Magnitude = √(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 × time.
So, distance off course = (plane's velocity - wind's velocity) × time
Plugging in the given values, we get:
Distance off course = (590 - 65) km/h × (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!
To determine the magnitude of the plane's velocity, relative to the ground, we need to consider the components of the plane's velocity and the wind velocity.
1. Find the horizontal component of the plane's velocity:
The plane is heading due south, so its horizontal component of velocity is zero.
2. Find the vertical component of the plane's velocity:
The plane's speed is given as 590 km/h, and it is heading due south. Since the vertical direction is opposite to the direction of the wind, we subtract the wind speed from the plane's speed: 590 km/h - 65 km/h = 525 km/h.
3. Calculate the magnitude of the plane's velocity:
The magnitude of the plane's velocity is the vector sum of its horizontal and vertical components. In this case, it is equal to the vertical component alone since the horizontal component is zero. So, the magnitude of the plane's velocity, relative to the ground, is 525 km/h.
To determine the direction of the plane's velocity, relative to the ground:
The direction will be the direction of the vector formed by the vertical component of the plane's velocity, which is due south.
To calculate how far from its intended position it will be after 15.0 minutes (0.25 hours) if the pilot takes no corrective action:
To find the displacement, we multiply the magnitude of the plane's velocity by the time. Displacement = velocity x time.
Displacement = 525 km/h x 0.25 h = 131.25 km
Therefore, the plane will be 131.25 km from its intended position after 15.0 minutes if the pilot takes no corrective action.