Question 1183038
Here's how to solve this rotational motion problem:

**(a) Position after 1.15 seconds:**

1. **Angular displacement (θ):**  The particle travels a distance (s) of 15 m along an arc of radius (r) 2 m. The angular displacement is given by:
   θ = s / r = 15 m / 2 m = 7.5 radians

2. **Initial position:** The initial position is 45 degrees from the negative x-axis.  We need to convert this to radians measured counterclockwise from the positive x-axis:

   * 45 degrees from the negative x-axis is equivalent to 180° + 45° = 225°
   * 225° * (π/180°) = 5π/4 radians

3. **Final position:** The final position is the initial position plus the angular displacement:

   * Final position = 5π/4 + 7.5 radians
   * Final position ≈ 3.927 + 7.5 ≈ 11.427 radians

4. **Convert to degrees (optional):**  If you want the final position in degrees:

   * 11.427 radians * (180°/π) ≈ 654.7°

   * To find the angle from the +x-axis, you may need to reduce this value to its principal value, by subtracting multiples of 360 until the value is between 0 and 360 degrees.

**(b) Initial and final angular velocity:**

1. **Initial angular velocity (ω₀):** The particle starts from rest, so ω₀ = 0 rad/s.

2. **Final angular velocity (ω):** We can use the following equation of rotational motion:
   ω = ω₀ + αt
   where α is the angular acceleration and t is the time.

   ω = 0 + (0.15π rad/s²)(1.15 s)
   ω ≈ 0.542 rad/s

**(c) Tangential velocity and acceleration at t = 2 s:**

1. **Angular velocity at t = 2 s:**
   ω(t=2) = ω₀ + αt = 0 + (0.15π rad/s²)(2 s) ≈ 0.942 rad/s

2. **Tangential velocity (v):**
   v = ωr = (0.942 rad/s)(2 m) ≈ 1.884 m/s

3. **Tangential acceleration (a_t):**
   a_t = αr = (0.15π rad/s²)(2 m) ≈ 0.942 m/s²

4. **Radial (centripetal) acceleration (a_r):**
   a_r = ω²r = (0.942 rad/s)²(2 m) ≈ 1.776 m/s²

5. **Total acceleration (a):** The total acceleration is the vector sum of the tangential and radial accelerations.
   a = sqrt(a_t² + a_r²) = sqrt((0.942 m/s²)² + (1.776 m/s²)²) ≈ 2.00 m/s²

   The direction of the total acceleration can be found using trigonometry:
   angle = arctan(a_t / a_r) = arctan(0.942/1.776) ≈ 28 degrees from the radial direction.

**Summary of Results:**

* (a) Position after 1.15 seconds: ≈ 11.427 radians (or ≈ 654.7 degrees) from the +x-axis
* (b) Initial angular velocity: 0 rad/s; Final angular velocity (at 1.15 s): ≈ 0.542 rad/s
* (c) Tangential velocity at 2 s: ≈ 1.884 m/s
* (c) Tangential acceleration at 2 s: ≈ 0.942 m/s²
* (c) Total acceleration at 2 s: ≈ 2.00 m/s² at an angle of approximately 28 degrees from the radial direction.