Question

A bird is flying south at a rate of 45 mph
while being pushed east by wind with a
speed of 12 mph. What is the direction
of the bird's resultant vector?
Hint: Draw a vector diagram.
R = 46.57 mph
0 = [ ? ] °
Round your answer to the nearoct

Answer 1

Answer:

284.93°

Step-by-step explanation:

To find the direction of the bird's resultant vector, we can use vector addition.

The bird's velocity due to its southward motion can be represented as a vector pointing directly south with a magnitude of 45 mph. The wind's velocity, pushing the bird eastward, can be represented as a vector pointing directly east with a magnitude of 12 mph. (See attachment)

To find the resultant vector, we connect the tail of the bird's velocity vector to the head of the wind's velocity vector. This line represents the resultant vector, which is the bird's actual velocity considering both its southward motion and the eastward push from the wind.

To find the magnitude of the resultant vector, we can use the Pythagorean theorem because the two vectors form a right triangle:

$|\mathbf{R}|=\sqrt{x^2+y^2}\\\\|\mathbf{R}|=\sqrt{12^2+(-45)^2}\\\\|\mathbf{R}|=\sqrt{144+2025}\\\\|\mathbf{R}|=\sqrt{2169}\\\\|\mathbf{R}|=46.5725240887...\\\\|\mathbf{R}|=46.57\; \rm mph$

The direction of a vector R is the angle, θ, between a line parallel to the x-axis and R, where 0° ≤ θ < 360°. This angle is typically measured counterclockwise from the x-axis, indicating the degree of rotation of the vector about its tail from due east.

$\theta=\tan^{-1}\left(\dfrac{y}{x}\right)\\\\\\\theta=\tan^{-1}\left(\dfrac{-45}{12}\right)\\\\\\\theta=-75.0685828218...^{\circ}$

Since the angle represents the direction, we express it as positive within the range where 0° ≤ θ < 360°. So, in this case, we need to add 360°:

$\theta=-75.0685828218...^{\circ}+360^{\circ}\\\\\\\theta=284.93^{\circ}\; \rm (nearest\;hundredth)$

Therefore, the direction of the bird's resultant vector is:

$\Large\boxed{\boxed{284.93^{\circ}}}$