First, let's break down each part of the question and provide a detailed, step-by-step solution.
### Part (a)
Express the velocity [tex]\( v_{a} \)[/tex] of the jet relative to the air and the velocity [tex]\( v_{w} \)[/tex] of the jet stream in terms of [tex]\( \mathbf{i} \)[/tex] and [tex]\( \mathbf{j} \)[/tex]
1. Velocity of the jet relative to the air, [tex]\( v_{a} \)[/tex]:
The jet is moving due north at a constant airspeed of 550 miles per hour. In vector notation:
[tex]\[
v_{a} = 0\mathbf{i} + 550\mathbf{j}
\][/tex]
So, [tex]\( v_{a} \)[/tex] is:
[tex]\[
v_{a} = 0\mathbf{i} + 550\mathbf{j}
\][/tex]
2. Velocity of the jet stream, [tex]\( v_{w} \)[/tex]:
The jet stream is moving in a northeasterly direction at 110 miles per hour. Since northeast is at a 45-degree angle, we can break this vector down into its components.
Both the [tex]\( \mathbf{i} \)[/tex] (east) and [tex]\( \mathbf{j} \)[/tex] (north) components can be found using the cosine and sine of 45 degrees, respectively, multiplied by the speed.
[tex]\[
v_{w,i} = 110 \cos(45^\circ) = 110 \left(\frac{\sqrt{2}}{2}\right) = 77.78
\][/tex]
[tex]\[
v_{w,j} = 110 \sin(45^\circ) = 110 \left(\frac{\sqrt{2}}{2}\right) = 77.78
\][/tex]
Therefore, the velocity of the jet stream [tex]\( v_{w} \)[/tex] is:
[tex]\[
v_{w} = v_{w,i}\mathbf{i} + v_{w,j}\mathbf{j} = 77.78\mathbf{i} + 77.78\mathbf{j}
\][/tex]
### Part (b)
Find the velocity of the jet relative to the ground
To find the velocity of the jet relative to the ground [tex]\( v_{g} \)[/tex], we add the vectors [tex]\( v_{a} \)[/tex] and [tex]\( v_{w} \)[/tex]:
[tex]\[
v_{g} = v_{a} + v_{w}
\][/tex]
Given:
[tex]\[
v_{a} = 0\mathbf{i} + 550\mathbf{j}
\][/tex]
[tex]\[
v_{w} = 77.78\mathbf{i} + 77.78\mathbf{j}
\][/tex]
[tex]\[
v_{g} = (0 + 77.78)\mathbf{i} + (550 + 77.78)\mathbf{j}
\][/tex]
[tex]\[
v_{g} = 77.78\mathbf{i} + 627.78\mathbf{j}
\][/tex]
### Part (c)
Find the actual speed and direction of the jet relative to the ground
1. Actual Speed:
The actual speed of the jet relative to the ground can be found using the magnitude of [tex]\( v_{g} \)[/tex]:
[tex]\[
\text{Speed} = \sqrt{v_{g,i}^2 + v_{g,j}^2}
\][/tex]
[tex]\[
\text{Speed} = \sqrt{(77.78)^2 + (627.78)^2}
\][/tex]
[tex]\[
= 632.58 \text{ miles per hour}
\][/tex]
2. Direction:
The direction (angle) of the velocity vector relative to the ground can be found using the arctangent function:
[tex]\[
\theta = \tan^{-2}\left(\frac{v_{g,j}}{v_{g,i}}\right)
\][/tex]
[tex]\[
\theta = \tan^{-1}\left(\frac{627.78}{77.78}\right)
\][/tex]
[tex]\[
\theta \approx 82.94^\circ \text{ (measured from the positive x-axis, east, towards the positive y-axis, north)}
\][/tex]
### Summary
(a) The velocity [tex]\( v_{a} \)[/tex] of the jet relative to the air:
[tex]\[
v_{a} = 0\mathbf{i} + 550\mathbf{j}
\][/tex]
The velocity [tex]\( v_{w} \)[/tex] of the jet stream:
[tex]\[
v_{w} = 77.78\mathbf{i} + 77.78\mathbf{j}
\][/tex]
(b) The velocity of the jet relative to the ground:
[tex]\[
v_{g} = 77.78\mathbf{i} + 627.78\mathbf{j}
\][/tex]
(c) The actual speed and direction of the jet relative to the ground:
[tex]\[
\text{Speed} = 632.58 \text{ miles per hour}
\][/tex]
[tex]\[
\text{Direction} \approx 82.94^\circ
\][/tex]
