A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a right turn and travels 1.33 km east before making another right turn and then travels 1.45 km south to arrive at its destination. What is the magnitude and direction of the truck's displacement from the warehouse?

Options

A. 1.76 km, `40.8^(@)` north of east
B. 1.15 km, `59.8^(@)` north of east
C. 1.33 km, `30.2^(@)` north of east
D. 2.40 km, `45.0^(@)` north of east

Answer

To solve the problem of finding the magnitude and direction of the truck's displacement from the warehouse, we can follow these steps:

### Step 1: Define the Coordinate System
Assume the warehouse is at the origin of a coordinate system (0, 0). We will use the following directions:
- North corresponds to the positive y-axis.
- East corresponds to the positive x-axis.
- South corresponds to the negative y-axis.
- West corresponds to the negative x-axis.

### Step 2: Calculate the Position After Each Leg of the Journey
1. **First Leg (North)**: The truck travels 2.60 km north.
- Position after this leg: (0, 2.60)

2. **Second Leg (East)**: The truck makes a right turn and travels 1.33 km east.
- Position after this leg: (1.33, 2.60)

3. **Third Leg (South)**: The truck makes another right turn and travels 1.45 km south.
- Position after this leg: (1.33, 2.60 - 1.45) = (1.33, 1.15)

### Step 3: Determine the Displacement Vector
The displacement vector from the warehouse (origin) to the final position (1.33, 1.15) can be represented as:
- Displacement vector = (x, y) = (1.33, 1.15)

### Step 4: Calculate the Magnitude of the Displacement
The magnitude of the displacement can be calculated using the Pythagorean theorem:
\[ \text{Magnitude} = \sqrt{x^2 + y^2} = \sqrt{(1.33)^2 + (1.15)^2} \]
Calculating this:
\[ = \sqrt{1.7689 + 1.3225} = \sqrt{3.0914} \approx 1.76 \text{ km} \]

### Step 5: Calculate the Direction of the Displacement
To find the direction of the displacement, we can use the tangent function:
\[ \tan(\theta) = \frac{y}{x} = \frac{1.15}{1.33} \]
Calculating the angle:
\[ \theta = \tan^{-1}\left(\frac{1.15}{1.33}\right) \approx 40.7^\circ \]
This angle is measured from the east towards the north.

### Final Result
The magnitude of the truck's displacement from the warehouse is approximately **1.76 km**, and the direction is **40.7 degrees north of east**.

---