Respuesta :
Answer: [tex]24.02\ mph[/tex]
Step-by-step explanation:
Given
Boat is traveling due to north at [tex]v_{bf}=27\ mph[/tex]
The current is flowing [tex]60^{\circ}[/tex] west of south at [tex]v_f=8\ mph[/tex]
Suppose the actual velocity of boat is [tex]\vec{v_b}[/tex]
In vector notation we can write
[tex]\Rightarrow \vec{v_{bf}}=\vec{v_b}-\vec{v_f}\\\\\Rightarrow \vec{v_b}=\vec{v_{bf}}+\vec{v_f}[/tex]
Flow of current can be written as
[tex]\Rightarrow \vec{v_f}=-8\sin 60^{\circ}\hat{i}-8\cos 60^{\circ}\hat{j}[/tex]
Insert the values of vectors
[tex]\Rightarrow \vec{v_b}=27\hat{j}-8\sin 60^{\circ}\hat{i}-8\cos 60^{\circ}\hat{j}\\\\\Rightarrow \vec{v_b}=-4\sqrt{3}\hat{i}+27\hat{j}-4\hat{j}\\\\\Rightarrow \vec{v_b}=-4\sqrt{3}\hat{i}+23\hat{j}[/tex]
Magnitude of the velocity is
[tex]=\sqrt{(4\sqrt{3})^2+(23)^2}\\=\sqrt{48+529}\\=24.02\ mph[/tex]
The actual direction and magnitude of the velocity of the boat is the sum of
the velocity of the boat and the velocity of the current both in vector form.
- The actual velocity and direction of the boat are approximately 24.02 m/s at a bearing of 16.76° west of north.
Reasons:
Direction of the boat = North
Speed of the both = 27 mph
Direction of the current = 60° west of south
Speed of the current = 8 mph
Required:
The actual speed and direction of the boat written in magnitude and
direction.
Solution:
The velocities given in vector form are;
Velocity of the boat, [tex]\vec {v}_{boat}[/tex] = 27·j
60° west of south = 30° south of west
Therefore;
Velocity of the current, [tex]\vec{v}_{current}[/tex] = -8 × cos(30°)·i - 8×sin(30°)·j
The actual velocity of the boat, [tex]\vec{v}_{actual}[/tex] = [tex]\mathbf{\vec {v}_{boat}}[/tex] + [tex]\mathbf{\vec{v}_{current}}[/tex]
Which gives;
[tex]\vec{v}_{actual}[/tex] = 27·j - 8 × cos(30°)·i - 8 × sin(30°)·j
[tex]\vec{v}_{actual}[/tex] = 23·j - (4·√3)·i
Therefore;
[tex]\vec{v} _{actual}[/tex] = -(4·√3)·i + 23·j
Which gives;
[tex]|v_{actual}|[/tex] = √((-4·√3)² + 23²) = √(577) ≈ 24.02
- The magnitude of the velocity of the boat, [tex]\mathbf{|v_{actual}|}[/tex] ≈ 24.02 m/s
The actual direction of the boat, θ, is given by the arctangent of the ratio
of the vertical to the horizontal component of the velocity as follows;
[tex]\displaystyle \theta = arctan \left(\frac{23}{-4 \cdot \sqrt{3} } \right) \approx \mathbf{-73.24^{\circ}}[/tex]
Based on the horizontal and vertical component, the above angle θ is
approximately 73.24° relative to the negative x-axis, which is 90° - 73.24° =
16.76° west of north.
Therefore;
- The actual direction of the both is on a bearing of approximately 16.76° west of north.
Learn more about vector quantities here:
https://brainly.com/question/1614684
