The diagram shows the all of the forces acting on a body of mass 2.13 kg. The three forces have magnitudes F1 = 63 N, F2 = 22.1 N, and F3 = 77.8 N, with directions as indicted in the diagram, where θ = 46.1 degrees and φ = 32.4 degrees. The dashed lines are parallel to the x and y axes. At t = 0, the body is moving at a speed of 8.33 m/s in the positive x direction.
      Part (a) 
What is the x component of the acceleration in m/s2? 
      Part (b) 
What is the y component of the acceleration in m/s2? 
      Part (c)   
What is the speed of the body, in m/s, at t = 8.3 s? 
      Part (d)   
What is the magnitude of the displacement of the body, in meters, between t = 0 and 8.3 s? 

Let's solve this step by step for each part.

We are given:
- Mass of the body: \( m = 2.13 \, \text{kg} \)
- Forces: \( F_1 = 63 \, \text{N}, F_2 = 22.1 \, \text{N}, F_3 = 77.8 \, \text{N} \)
- Angles: \( \theta = 46.1^\circ \) and \( \phi = 32.4^\circ \)
- Initial speed: \( v_0 = 8.33 \, \text{m/s} \) (in the positive x direction)

We will start by calculating the components of the forces along the x and y directions.

### Step 1: Resolving forces into components

1. **\( F_1 \)** is along the positive x-axis, so its components are:
   \[
   F_{1x} = F_1 = 63 \, \text{N}, \quad F_{1y} = 0
   \]

2. **\( F_2 \)** is at an angle \( \theta = 46.1^\circ \) above the negative x-axis:
   \[
   F_{2x} = -F_2 \cos(\theta) = -22.1 \cos(46.1^\circ)
   \]
   \[
   F_{2y} = F_2 \sin(\theta) = 22.1 \sin(46.1^\circ)
   \]

3. **\( F_3 \)** is at an angle \( \phi = 32.4^\circ \) below the positive x-axis:
   \[
   F_{3x} = F_3 \cos(\phi) = 77.8 \cos(32.4^\circ)
   \]
   \[
   F_{3y} = -F_3 \sin(\phi) = -77.8 \sin(32.4^\circ)
   \]

### Step 2: Net force in the x and y directions

1. **Net force in the x direction**:
   \[
   F_{\text{net},x} = F_{1x} + F_{2x} + F_{3x}
   \]
   
2. **Net force in the y direction**:
   \[
   F_{\text{net},y} = F_{1y} + F_{2y} + F_{3y}
   \]

### Step 3: Acceleration components

Using Newton's second law, \( F = ma \), we can find the acceleration components.

1. **x component of the acceleration**:
   \[
   a_x = \frac{F_{\text{net},x}}{m}
   \]

2. **y component of the acceleration**:
   \[
   a_y = \frac{F_{\text{net},y}}{m}
   \]

### Step 4: Speed at \( t = 8.3 \, \text{s} \)

The total velocity components after time \( t = 8.3 \, \text{s} \) can be found using:
1. \( v_x(t) = v_{0x} + a_x \cdot t \), where \( v_{0x} = 8.33 \, \text{m/s} \)
2. \( v_y(t) = v_{0y} + a_y \cdot t \), where \( v_{0y} = 0 \, \text{m/s} \)

The speed at \( t = 8.3 \, \text{s} \) is:
\[
v(t) = \sqrt{v_x(t)^2 + v_y(t)^2}
\]

### Step 5: Displacement magnitude

The displacement components are found using:
1. \( x(t) = v_{0x} t + \frac{1}{2} a_x t^2 \)
2. \( y(t) = v_{0y} t + \frac{1}{2} a_y t^2 \)

The magnitude of the displacement is:
\[
d = \sqrt{x(t)^2 + y(t)^2}
\]

---

Let's now compute each part with the numerical values!

### Part (a): \( x \)-component of acceleration

We calculate the components of the forces and solve for \( a_x \).

1. \( F_{2x} = -22.1 \cos(46.1^\circ) \)
2. \( F_{3x} = 77.8 \cos(32.4^\circ) \)

Then, sum the x components:
\[
F_{\text{net},x} = 63 + F_{2x} + F_{3x}
\]
Finally, calculate:
\[
a_x = \frac{F_{\text{net},x}}{2.13}
\]

### Part (b): \( y \)-component of acceleration

Similar process for the y-components:

1. \( F_{2y} = 22.1 \sin(46.1^\circ) \)
2. \( F_{3y} = -77.8 \sin(32.4^\circ) \)

Then, sum the y components:
\[
F_{\text{net},y} = F_{2y} + F_{3y}
\]
And:
\[
a_y = \frac{F_{\text{net},y}}{2.13}
\]

---

Would you like me to continue with these specific calculations or clarify any particular steps?

The diagram shows all of the forces acting on a body of mass 2.13 kg. The three forces have magnitudes F1 = 63 N, F2 = 22.1 N, and F3 = 77.8 N, with directions as indicated in the diagram, where θ = 46.1 degrees and φ = 32.4 degrees. The dashed lines are parallel to the x and y axes. At t = 0, the body is moving at a speed of 8.33 m/s in the positive x direction. Part (a) What is the x component of the acceleration in m/s2? Part (b) What is the y component of the acceleration in m/s2? Part (c) What is the speed of the body, in m/s, at t = 8.3 s? Part (d) What is the magnitude of the displacement of the body, in meters, between t = 0 and 8.3 s?

Solve a physics problem involving a body of mass 2.13 kg under the influence of three forces with magnitudes F1 = 63 N, F2 = 22.1 N, and F3 = 77.8 N. This problem involves vector decomposition, Newton's laws of motion, and kinematics to find the acceleration components, speed, and displacement. Suitable for high school students in Grades 11-12.

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