Relative to the ground, a car has a velocity of 18.0 m/s, directed due north. Relative to this car, a truck has a velocity of 22.8 m/s, directed `52.1^(@)` south of east. Find the magnitude and direction of the truck's velocity relative to the ground.

Options

A. 4.8 m/s, `37.9^(@)` north of east
B. 22.8 m/s, `37.9^(@)` south of east
C. 14.0 m/s, due east
D. 20.4 m/s, `68.2^(@)` north of east

Answer

To solve the problem of finding the truck's velocity relative to the ground, we will follow these steps:

### Step 1: Define the velocities
1. **Velocity of the car (V_car)**: The car's velocity is given as 18.0 m/s directed due north. In vector form, this can be represented as:
\[ \mathbf{V}_{\text{car}} = 0 \hat{i} + 18.0 \hat{j} \quad \text{(where } \hat{i} \text{ is east and } \hat{j} \text{ is north)} \]

2. **Velocity of the truck relative to the car (V_truck/car)**: The truck's velocity is given as 22.8 m/s at an angle of 52.1 degrees south of east. We need to break this down into its components:
- The x-component (east-west) can be calculated using cosine:
\[ V_{x} = 22.8 \cos(52.1^\circ) \]
- The y-component (north-south) can be calculated using sine:
\[ V_{y} = -22.8 \sin(52.1^\circ) \quad \text{(negative because it's directed south)} \]

### Step 2: Calculate the components of the truck's velocity relative to the car
1. Calculate \( V_{x} \):
\[ V_{x} = 22.8 \cos(52.1^\circ) \approx 22.8 \times 0.6157 \approx 14.0 \text{ m/s} \]

2. Calculate \( V_{y} \):
\[ V_{y} = -22.8 \sin(52.1^\circ) \approx -22.8 \times 0.7880 \approx -18.0 \text{ m/s} \]

So, the velocity of the truck relative to the car can be expressed as:
\[ \mathbf{V}_{\text{truck/car}} = 14.0 \hat{i} - 18.0 \hat{j} \]

### Step 3: Find the truck's velocity relative to the ground
Using the formula:
\[ \mathbf{V}_{\text{truck}} = \mathbf{V}_{\text{truck/car}} + \mathbf{V}_{\text{car}} \]
Substituting the values:
\[ \mathbf{V}_{\text{truck}} = (14.0 \hat{i} - 18.0 \hat{j}) + (0 \hat{i} + 18.0 \hat{j}) = 14.0 \hat{i} + 0 \hat{j} \]

### Step 4: Determine the magnitude and direction of the truck's velocity relative to the ground
1. **Magnitude**:
\[ |\mathbf{V}_{\text{truck}}| = \sqrt{(14.0)^2 + (0)^2} = 14.0 \text{ m/s} \]

2. **Direction**: Since the y-component is zero and the x-component is positive, the direction is due east.

### Final Answer
The magnitude of the truck's velocity relative to the ground is **14.0 m/s**, and the direction is **due east**.

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