A hiker’s trip consists of three segments. Path A is 8.0 km long heading 60 degrees north of east. Path B is 7.0 km long in a direction 30 degrees of north of west. Path C is 4.0 km long heading 70 degrees east of south. Find the displacement by a) adding the vectors(by components) and b) head to tail method.

To find the total displacement of the hiker, we can approach this problem using two methods:

### Part (a) Adding the Vectors by Components

Let's break down each path into its horizontal (x) and vertical (y) components.

1. **Path A: 8.0 km at 60° North of East**
   - \( A_x = 8.0 \times \cos(60°) \)
   - \( A_y = 8.0 \times \sin(60°) \)

2. **Path B: 7.0 km at 30° North of West**
   - \( B_x = 7.0 \times \cos(150°) \)  (Since North of West, \( \theta = 180° - 30° = 150° \))
   - \( B_y = 7.0 \times \sin(150°) \)

3. **Path C: 4.0 km at 70° East of South**
   - \( C_x = 4.0 \times \cos(290°) \) (Since East of South, \( \theta = 360° - 70° = 290° \))
   - \( C_y = 4.0 \times \sin(290°) \)

Let's calculate these components.

- For Path A:
  \[
  A_x = 8.0 \times \cos(60°) = 8.0 \times 0.5 = 4.0 \, \text{km}
  \]
  \[
  A_y = 8.0 \times \sin(60°) = 8.0 \times 0.866 = 6.93 \, \text{km}
  \]

- For Path B:
  \[
  B_x = 7.0 \times \cos(150°) = 7.0 \times (-0.866) = -6.06 \, \text{km}
  \]
  \[
  B_y = 7.0 \times \sin(150°) = 7.0 \times 0.5 = 3.5 \, \text{km}
  \]

- For Path C:
  \[
  C_x = 4.0 \times \cos(290°) = 4.0 \times 0.342 = 1.37 \, \text{km}
  \]
  \[
  C_y = 4.0 \times \sin(290°) = 4.0 \times (-0.94) = -3.76 \, \text{km}
  \]

Now, add up the components:

- Total x-component:
  \[
  R_x = A_x + B_x + C_x = 4.0 + (-6.06) + 1.37 = -0.69 \, \text{km}
  \]

- Total y-component:
  \[
  R_y = A_y + B_y + C_y = 6.93 + 3.5 + (-3.76) = 6.67 \, \text{km}
  \]

Finally, calculate the magnitude and direction of the displacement vector:

- Magnitude:
  \[
  R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-0.69)^2 + (6.67)^2} = \sqrt{0.48 + 44.49} = \sqrt{44.97} \approx 6.71 \, \text{km}
  \]

- Direction (angle north of east):
  \[
  \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{6.67}{-0.69}\right) \approx 95.9° \, \text{north of east}
  \]

### Part (b) Head to Tail Method

To find the displacement using the head-to-tail method:

1. Draw vector **A** from the origin.
2. From the head of **A**, draw vector **B**.
3. From the head of **B**, draw vector **C**.
4. The resultant vector **R** is the vector from the tail of **A** to the head of **C**.

Using the above components, the resultant vector can be represented with a magnitude of approximately **6.71 km** and a direction of **95.9° north of east**.

Would you like more details on any part of this problem or have any questions?

Here are some related questions to further explore:

1. How would the displacement change if Path C had a different angle?
2. What is the significance of breaking vectors into components in physics?
3. How does the head-to-tail method visually help in understanding vector addition?
4. What would happen to the displacement if the hiker followed the paths in reverse order?
5. How does vector addition differ when dealing with non-linear paths?

**Tip:** When dealing with vector problems, always double-check the angle used for each component to ensure accuracy.

Explore how to calculate a hiker's total displacement using vector addition and trigonometry. This problem involves breaking down vectors into components and using trigonometric functions to find magnitude and direction. Suitable for high school students.

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