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To solve this problem, let's break it down step by step.

### Given:
1. Initial velocity of the bird: \( v_x = 4.00 \, \text{mph} \)
2. Wind acceleration in the y-direction (north): \( a_y = 0.400 \, \text{m/s}^2 \)
3. Time duration of the wind: \( t = 4.00 \, \text{s} \)
4. Conversion factor: \( 1 \, \text{mile} = 1609 \, \text{m} \)

We are tasked with finding:
- The magnitude \( r \) of the bird's displacement after 4 seconds.
- The direction \( \theta \) (counterclockwise from the easterly direction).

### Step 1: Convert initial velocity from mph to m/s
We first need to convert the bird's initial velocity from miles per hour to meters per second.

\[
v_x = 4.00 \, \text{mph} \times \frac{1609 \, \text{m}}{1 \, \text{mile}} \times \frac{1 \, \text{hour}}{3600 \, \text{s}} = 1.788 \, \text{m/s}
\]

### Step 2: Calculate the bird's displacement in the x-direction
Since there is no acceleration in the x-direction (the bird was coasting), the displacement in the x-direction can be calculated as:

\[
x = v_x \cdot t = 1.788 \, \text{m/s} \times 4.00 \, \text{s} = 7.152 \, \text{m}
\]

### Step 3: Calculate the bird's displacement in the y-direction
The bird accelerates northward (y-direction) due to the wind. The displacement in the y-direction is given by the equation:

\[
y = \frac{1}{2} a_y t^2
\]

Substituting the given values:

\[
y = \frac{1}{2} \times 0.400 \, \text{m/s}^2 \times (4.00 \, \text{s})^2 = 3.200 \, \text{m}
\]

### Step 4: Calculate the magnitude of the displacement
The total displacement \( r \) is the resultant vector from the x and y components:

\[
r = \sqrt{x^2 + y^2} = \sqrt{(7.152)^2 + (3.200)^2} = \sqrt{51.146 + 10.240} = \sqrt{61.386} \approx 7.836 \, \text{m}
\]

### Step 5: Calculate the direction of the displacement
The direction \( \theta \) is the angle counterclockwise from the easterly direction. It can be found using the tangent function:

\[
\theta = \tan^{-1} \left( \frac{y}{x} \right) = \tan^{-1} \left( \frac{3.200}{7.152} \right)
\]

\[
\theta \approx \tan^{-1}(0.447) \approx 24.05^\circ
\]

### Final Answer:
- The magnitude of the bird's displacement \( r \) is approximately **7.836 m**.
- The direction \( \theta \) is approximately **24.05° counterclockwise** from the easterly direction.

Would you like more details or have any questions?

Here are some related questions to explore:
1. How would the result change if the wind had lasted for 8 seconds instead of 4?
2. What if the wind had a higher acceleration, say \( 0.800 \, \text{m/s}^2 \)?
3. How would the displacement change if the bird started traveling at a higher initial velocity?
4. What is the bird's velocity in the y-direction after 4 seconds?
5. How can we calculate the bird's total velocity at the end of the 4-second interval?

**Tip:** Always ensure that your velocity and acceleration units are consistent (e.g., meters per second) before calculating displacements and directions.

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 0.400 m/s². If the wind's acceleration lasts for 4.00 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.

Learn how to calculate the displacement of a bird traveling easterly under wind acceleration. This problem involves kinematics, vectors, and trigonometry, with a step-by-step solution for finding both magnitude and direction. Suitable for students in Grades 10-12.

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