An airplane flies at an airspeed of 445km/hr. The pilot wants to fly 785km to the north. She knows that she must head 18.0° east of north to fly directly there. If the plane arrives in 1.97hr, find the magnitude of the wind velocity vector.
the plane flies 445*1.97 = 876.65 km
So, in still air, its position would be
(270.9,833.7)
So, you want
Vx * 1.97 = 0-270.9
Vy * 1.97 = 785-833.7
Now you know Vx and Vy, so you can find V and its direction
To find the magnitude of the wind velocity vector, we can break down the airplane's velocity into its components.
Let's consider the airplane's velocity vector as V, and the wind velocity vector as W.
First, we need to find the northward component of the airplane's velocity. We can use trigonometry to solve this:
Northward component of airplane's velocity = airspeed x sin(18°)
= 445 km/hr x sin(18°) = 445 km/hr x 0.3090
= 137.9 km/hr (rounded to one decimal place)
Next, we need to find the eastward component of the airplane's velocity:
Eastward component of airplane's velocity = airspeed x cos(18°)
= 445 km/hr x cos(18°) = 445 km/hr x 0.9511
= 423.6 km/hr (rounded to one decimal place)
Now, let's assume the wind has a northward component of Wn and an eastward component of We.
The total northward component of the airplane's velocity is the sum of its own northward component and the wind's northward component:
Total northward velocity = Airplane's northward velocity + Wind's northward velocity
785 km = 137.9 km/hr + Wn
Wn = 785 km - 137.9 km/hr
Wn = 647.1 km/hr
Similarly, the total eastward component of the airplane's velocity is the sum of its own eastward component and the wind's eastward component:
Total eastward velocity = Airplane's eastward velocity + Wind's eastward velocity
0 km/hr = 423.6 km/hr + We
We = -423.6 km/hr
Since the wind is blowing from the west (negative eastward component), the magnitude of the wind velocity vector is:
Magnitude of the wind velocity vector = sqrt(Wn^2 + We^2)
= sqrt(647.1 km/hr)^2 + (-423.6 km/hr)^2
= sqrt(418454.41 km^2/hr^2 + 179516.96 km^2/hr^2)
= sqrt(598971.37 km^2/hr^2)
= 773.9 km/hr (rounded to one decimal place)
Therefore, the magnitude of the wind velocity vector is approximately 773.9 km/hr.
To find the magnitude of the wind velocity vector, we need to use vector addition. The airplane has an airspeed of 445 km/hr, and it is heading 18.0° east of north. This means that the airplane has a component of its velocity in the north direction and a component in the east direction.
Let's break down the components of the airplane's velocity:
The north component of the airplane's velocity is given by V_north = 445 km/hr * cos(18.0°).
The east component of the airplane's velocity is given by V_east = 445 km/hr * sin(18.0°).
Now, we know that the airplane needs to fly 785 km to the north, and it takes 1.97 hours to get there. So, the magnitude of the airplane's north velocity component is given by V_north = 785 km / 1.97 hr.
Now, since the wind is affecting the airplane's trajectory, we can assume that the wind is blowing in the west direction. Let's represent the wind velocity as V_wind.
To find the wind velocity, we need to subtract the airplane's velocity components from the actual velocity (which is the velocity measured in the north direction).
The magnitude of the wind velocity vector can be found using the Pythagorean theorem:
Magnitude of the wind velocity vector = √((V_north - V_wind)^2 + V_east^2).
By substituting the given values, we can solve for the magnitude of the wind velocity vector.