Questions: An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when a strong wind from the south imparts a constant acceleration of 0.500 m / s^2. If the wind's acceleration lasts for 2.10 s, find the magnitude r and direction θ (measured counterclockwise from the easterly direction) of the bird's displacement over this time interval. Assume that the bird is originally traveling in the +x direction and that there are 1609 m in 1 mi. r= m θ=

An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when a strong wind from the south imparts a constant acceleration of 0.500 m / s^2. If the wind's acceleration lasts for 2.10 s, find the magnitude r and direction θ (measured counterclockwise from the easterly direction) of the bird's displacement over this time interval. Assume that the bird is originally traveling in the +x direction and that there are 1609 m in 1 mi. r= m θ=

Transcript text: An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when a strong wind from the south imparts a constant acceleration of $0.500 \mathrm{~m} / \mathrm{s}^{2}$. If the wind's acceleration lasts for 2.10 s , find the magnitude $r$ and direction $\theta$ (measured counterclockwise from the easterly direction) of the bird's displacement over this time interval. Assume that the bird is originally traveling in the $+x$ direction and that there are 1609 m in 1 mi . $r=$ $\square$ m \[ \theta= \]

Solution

Solution Steps

Step 1: Identify the given information

Initial velocity of the bird, $ v_0 = 4.00 $ mph (eastward)
Acceleration due to wind, $ a = 0.500 $ m/s² (northward)
Time interval, $ t = 2.00 $ s

Step 2: Convert initial velocity to consistent units

Convert the initial velocity from mph to m/s: \[ 1 \text{ mph} = 0.44704 \text{ m/s} \] \[ v_0 = 4.00 \text{ mph} \times 0.44704 \text{ m/s per mph} = 1.78816 \text{ m/s} \]

Step 3: Calculate displacement in the x-direction (eastward)

Since there is no acceleration in the x-direction: \[ x = v_0 \times t \] \[ x = 1.78816 \text{ m/s} \times 2.00 \text{ s} = 3.57632 \text{ m} \]

Step 4: Calculate displacement in the y-direction (northward)

Using the formula for displacement under constant acceleration: \[ y = \frac{1}{2} a t^2 \] \[ y = \frac{1}{2} \times 0.500 \text{ m/s}^2 \times (2.00 \text{ s})^2 \] \[ y = \frac{1}{2} \times 0.500 \times 4.00 \] \[ y = 1.00 \text{ m} \]

Step 5: Calculate the resultant displacement

Using the Pythagorean theorem: \[ R = \sqrt{x^2 + y^2} \] \[ R = \sqrt{(3.57632 \text{ m})^2 + (1.00 \text{ m})^2} \] \[ R = \sqrt{12.79 + 1.00} \] \[ R = \sqrt{13.79} \] \[ R \approx 3.71 \text{ m} \]

Step 6: Calculate the direction of the displacement

Using the tangent function to find the angle: \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \] \[ \theta = \tan^{-1}\left(\frac{1.00 \text{ m}}{3.57632 \text{ m}}\right) \] \[ \theta = \tan^{-1}(0.2796) \] \[ \theta \approx 15.6^\circ \]