5 The initial and final velocities of a particle are shown in Fig. 3-25. Find the particle's average acceleration if the change in velocity takes place in a 10.0 s interval. 30.0 m/s 45.0° X 20.0 m/s 30.0° Fig. 3-25 13

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#5

### Understanding Components of Velocity and Acceleration

---

**Problem Statement:**

The initial and final velocities of a particle are shown in Fig. 3-25. Find the particle’s average acceleration if the change in velocity takes place in a 10.0 s interval.

**Diagram Description (Fig. 3-25):**

The diagram provided is a vector diagram showing the initial and final velocities of a particle. It is divided into four quadrants with the \(x\)- and \(y\)-axes.

- The initial velocity vector \(\mathbf{v_i}\) is shown as a green arrow pointing upwards and to the right from the origin. It has a magnitude of 30.0 m/s at an angle of 30.0° above the positive \(x\)-axis.
- The final velocity vector \(\mathbf{v_f}\) is shown as a green arrow pointing downwards and to the left from the origin. It has a magnitude of 20.0 m/s at an angle of 45.0° below the negative \(x\)-axis.

---

**Task Explanation:**

To find the particle’s average acceleration, we need to perform the following steps:

1. **Determine the components of the initial and final velocities:**
   - For the initial velocity \(\mathbf{v_i}\):
     \[
     v_{ix} = 30.0 \cos(30.0^\circ), \quad v_{iy} = 30.0 \sin(30.0^\circ)
     \]

   - For the final velocity \(\mathbf{v_f}\):
     \[
     v_{fx} = 20.0 \cos(45.0^\circ), \quad v_{fy} = -20.0 \sin(45.0^\circ)
     \]
     Note: The negative sign for \(v_{fy}\) is because the vector points downward.

2. **Calculate the change in velocity components:**
   - Change in \(x\)-component (\(\Delta v_x\)):
     \[
     \Delta v_x = v_{fx} - v_{ix}
     \]

   - Change in \(y\)-component (\(\Delta v_y\)):
     \[
     \Delta v_y = v_{fy} - v_{iy}
     \]

3. **Calculate the average acceleration components:**
   - For the average acceleration in \(x\
Transcribed Image Text:### Understanding Components of Velocity and Acceleration --- **Problem Statement:** The initial and final velocities of a particle are shown in Fig. 3-25. Find the particle’s average acceleration if the change in velocity takes place in a 10.0 s interval. **Diagram Description (Fig. 3-25):** The diagram provided is a vector diagram showing the initial and final velocities of a particle. It is divided into four quadrants with the \(x\)- and \(y\)-axes. - The initial velocity vector \(\mathbf{v_i}\) is shown as a green arrow pointing upwards and to the right from the origin. It has a magnitude of 30.0 m/s at an angle of 30.0° above the positive \(x\)-axis. - The final velocity vector \(\mathbf{v_f}\) is shown as a green arrow pointing downwards and to the left from the origin. It has a magnitude of 20.0 m/s at an angle of 45.0° below the negative \(x\)-axis. --- **Task Explanation:** To find the particle’s average acceleration, we need to perform the following steps: 1. **Determine the components of the initial and final velocities:** - For the initial velocity \(\mathbf{v_i}\): \[ v_{ix} = 30.0 \cos(30.0^\circ), \quad v_{iy} = 30.0 \sin(30.0^\circ) \] - For the final velocity \(\mathbf{v_f}\): \[ v_{fx} = 20.0 \cos(45.0^\circ), \quad v_{fy} = -20.0 \sin(45.0^\circ) \] Note: The negative sign for \(v_{fy}\) is because the vector points downward. 2. **Calculate the change in velocity components:** - Change in \(x\)-component (\(\Delta v_x\)): \[ \Delta v_x = v_{fx} - v_{ix} \] - Change in \(y\)-component (\(\Delta v_y\)): \[ \Delta v_y = v_{fy} - v_{iy} \] 3. **Calculate the average acceleration components:** - For the average acceleration in \(x\
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