https://file.aiquickdraw.com/mathbot/file/1725986525749qrvnstsq.jpg

The question visible in the image is as follows:

"A disoriented physics professor drives 3.25 km north, then 2.20 km west, and then 1.50 km south. Find the magnitude and direction of the resultant displacement, using the component method and polygon method."

### Solution:

We are tasked with finding the resultant displacement vector after the professor moves in three different directions. We'll solve this using both the **component method** and **polygon method**.

#### 1. **Component Method**:
In this method, we break each movement into its north-south and east-west components, then sum them.

- First, let's define the directions:
  - **North** is positive along the y-axis.
  - **West** is negative along the x-axis.
  - **South** is negative along the y-axis.

Now, let's break down the displacements into x and y components:

- **Displacement 1 (3.25 km north)**:
  - x-component: 0 km (no movement east or west).
  - y-component: +3.25 km (positive since it's north).

- **Displacement 2 (2.20 km west)**:
  - x-component: -2.20 km (negative since it's west).
  - y-component: 0 km (no movement north or south).

- **Displacement 3 (1.50 km south)**:
  - x-component: 0 km (no movement east or west).
  - y-component: -1.50 km (negative since it's south).

#### Resultant Components:
- Sum of x-components = 0 + (-2.20) + 0 = **-2.20 km**
- Sum of y-components = 3.25 + 0 + (-1.50) = **1.75 km**

So, the total displacement is a vector with components:
- x = -2.20 km (west)
- y = 1.75 km (north)

#### Magnitude of the Resultant Displacement:
The magnitude \( R \) is found using the Pythagorean theorem:

\[
R = \sqrt{(x^2 + y^2)} = \sqrt{((-2.20)^2 + (1.75)^2)} = \sqrt{(4.84 + 3.06)} = \sqrt{7.90} \approx 2.81 \, \text{km}
\]

#### Direction of the Resultant Displacement:
The direction (θ) can be found using the tangent function:

\[
\theta = \tan^{-1} \left( \frac{y}{x} \right) = \tan^{-1} \left( \frac{1.75}{-2.20} \right)
\]

This gives:

\[
\theta \approx \tan^{-1} (-0.795) \approx -38.37^\circ
\]

Since the vector is in the second quadrant (northwest direction), the direction relative to the positive x-axis is:

\[
180^\circ - 38.37^\circ \approx 141.63^\circ \, \text{(counterclockwise from east)}.
\]

Thus, the direction is approximately **141.63° northwest**.

---

#### 2. **Polygon Method**:
In this method, we would visually represent the displacements as vectors and connect them head-to-tail to form a polygon. The resultant displacement is the vector that goes from the starting point to the end of the polygon. The result would be the same as obtained through the component method, i.e., 2.81 km at an angle of approximately 141.63°.

---

Let me know if you'd like further clarification, or have any additional questions!

### 5 Related Questions:
1. How do you find displacement using the graphical polygon method?
2. What is the difference between distance traveled and displacement?
3. How would the result change if the professor took a fourth displacement step eastward?
4. How can you calculate the resultant displacement for a three-dimensional movement?
5. What are common applications of vector addition in physics?

### Tip:
When working with vectors, always draw a diagram to visualize the displacements. It helps in understanding the direction and calculating components accurately.

A disoriented physics professor drives 3.25 km north, then 2.20 km west, and then 1.50 km south. Find the magnitude and direction of the resultant displacement, using component method and polygon method.

Solve a physics displacement problem where a professor drives north, west, and south. Learn how to find the resultant displacement magnitude and direction using vector addition, the Pythagorean theorem, and trigonometric functions.

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