Now, assume the same bird is moving along again at 2.00 mph in an easterly direction, but this time, the acceleration given by the wind is at a 𝜙=33.0° angle to the original direction of motion, measured counterclockwise from the easterly direction. If the magnitude of the acceleration is 0.300 m/s2 , find the displacement vector 𝑟⃗ and the angle of the displacement 𝜃1 .

Question

Now, assume the same bird is moving along again at 2.00 mph in an easterly direction, but this time, the acceleration given by the wind is at a 𝜙=33.0° angle to the original direction of motion, measured counterclockwise from the easterly direction. If the magnitude of the acceleration is 0.300 m/s2 , find the displacement vector 𝑟⃗ and the angle of the displacement 𝜃1 .

Enter the components of the vector, 𝑟𝑥 and 𝑟𝑦 , and the angle. Assume the time interval is still 3.50 s .

Answer 1

To find the displacement vector 𝑟⃗ and the angle of displacement 𝜃₁, we need to break down the motion into its x and y components.

Given:
- Bird's velocity = 2.00 mph
- Wind's acceleration magnitude = 0.300 m/s²
- Wind's acceleration angle 𝜙 = 33.0° (counterclockwise from the easterly direction)
- Time interval = 3.50 s

First, let's convert the bird's velocity from mph to m/s:
Bird's velocity = 2.00 mph = 0.894 m/s (approx.)

Next, let's find the x and y components of the displacement vector 𝑟⃗:

𝑟_𝑥 component: Since the bird's motion is along the easterly direction, there is no acceleration in the x-direction. Therefore, the 𝑟_𝑥 component of the displacement vector remains constant throughout the motion. The 𝑟_𝑥 component is given by:

𝑟_𝑥 = 𝑣ₓ × 𝑡 = (Bird's velocity) × (Time interval)
= 0.894 m/s × 3.50 s
≈ 3.129 m

𝑟_𝑦 component: The 𝑟_𝑦 component is influenced by the wind's acceleration. We need to consider the angle 𝜙 and the magnitude of acceleration. The 𝑟_𝑦 component can be calculated using the equation:

𝑟_𝑦 = v_y₀ × 𝑡 + (1/2) × 𝑎 × 𝑡²

To find the initial vertical velocity 𝑣_y₀, we can use the trigonometric relationship between the bird's velocity and the angle 𝜙:
sin(𝜙) = 𝑣_y₀ / 𝑣

Solving for 𝑣_y₀:
𝑣_y₀ = sin(𝜙) × 𝑣
= sin(33.0°) × 0.894 m/s
≈ 0.47 m/s (approx.)

Now we can calculate the 𝑟_𝑦 component:
𝑟_𝑦 = (0.47 m/s × 3.50 s) + (0.5 × 0.300 m/s² × (3.50 s)²)
≈ 5.31 m

Finally, we can find the magnitude of the displacement vector 𝑟:

|𝑟⃗| = √(𝑟_𝑥² + 𝑟_𝑦²)
= √((3.129 m)² + (5.31 m)²)
≈ 6.16 m

And the angle of the displacement 𝜃₁ can be found using the inverse tangent function:

𝜃₁ = atan(𝑟_𝑦 / 𝑟_𝑥)
= atan(5.31 m / 3.129 m)
≈ 59.2°

Therefore, the components of the displacement vector 𝑟⃗ and the angle of displacement 𝜃₁ are:
𝑟_𝑥 ≈ 3.129 m
𝑟_𝑦 ≈ 5.31 m
𝜃₁ ≈ 59.2°