a cave expolere is surveysing a cave. he follows a passage 100m straight east then 50m in a direction 30 drgee west of north then 150 m at 45 drgee west of south after a fourth unmesasured displacement he finds himselp back where he starated using a scal drawing determine the fourth displacement magnitude and direction 

This problem involves vector addition and determining the resultant displacement needed to bring the cave explorer back to the starting point. The displacements can be treated as vectors, and we will add them step by step. Since the explorer ends up where they started, the sum of all displacement vectors must equal zero.

### Step 1: Break down each displacement into its components

1. **First displacement:**
   - 100 m straight east. This only has an \( x \)-component (East-West direction).
   \[
   \text{Vector 1: } (100, 0) \text{ meters}
   \]

2. **Second displacement:**
   - 50 m at 30° west of north. We break this into \( x \)- and \( y \)-components.
   \[
   x_2 = 50 \sin(30^\circ) = 50 \times 0.5 = 25 \text{ m west} (-x)
   \]
   \[
   y_2 = 50 \cos(30^\circ) = 50 \times 0.866 = 43.3 \text{ m north} (+y)
   \]
   \[
   \text{Vector 2: } (-25, 43.3) \text{ meters}
   \]

3. **Third displacement:**
   - 150 m at 45° west of south. Again, we break this into \( x \)- and \( y \)-components.
   \[
   x_3 = 150 \sin(45^\circ) = 150 \times 0.707 = 106.1 \text{ m west} (-x)
   \]
   \[
   y_3 = 150 \cos(45^\circ) = 150 \times 0.707 = 106.1 \text{ m south} (-y)
   \]
   \[
   \text{Vector 3: } (-106.1, -106.1) \text{ meters}
   \]

### Step 2: Sum the known displacements

- \( x \)-components:
  \[
  x_{\text{total}} = 100 + (-25) + (-106.1) = 100 - 25 - 106.1 = -31.1 \text{ meters}
  \]

- \( y \)-components:
  \[
  y_{\text{total}} = 0 + 43.3 + (-106.1) = 43.3 - 106.1 = -62.8 \text{ meters}
  \]

### Step 3: Find the fourth displacement

Since the explorer ends up back at the starting point, the sum of all displacements must be zero. Therefore, the fourth displacement must cancel out the total \( x \)- and \( y \)-components found above.

- The fourth displacement's \( x \)-component must be \( +31.1 \) meters (to cancel \( -31.1 \)).
- The fourth displacement's \( y \)-component must be \( +62.8 \) meters (to cancel \( -62.8 \)).

### Step 4: Calculate the magnitude and direction of the fourth displacement

- **Magnitude**:
  \[
  d_4 = \sqrt{(31.1)^2 + (62.8)^2} = \sqrt{967.21 + 3948.84} = \sqrt{4916.05} \approx 70.1 \text{ meters}
  \]

- **Direction** (angle north of east):
  \[
  \theta = \tan^{-1}\left(\frac{62.8}{31.1}\right) = \tan^{-1}(2.02) \approx 63.4^\circ \text{ north of east}
  \]

### Final Answer:
The fourth displacement has a magnitude of **70.1 meters** and is directed at an angle of **63.4° north of east**.

Would you like further clarification or details?

Here are 5 related questions:
1. How would the problem change if the final position wasn’t the starting point?
2. What are the sine and cosine rules for solving vector components in other angles?
3. How do you interpret directions like "west of south" in vector terms?
4. Could we solve this problem graphically instead of analytically?
5. What is the difference between polar and rectangular coordinates in vectors?

**Tip:** Always label angles clearly when breaking vectors into components.

A cave explorer is surveying a cave. He follows a passage 100m straight east, then 50m in a direction 30 degrees west of north, then 150m at 45 degrees west of south. After a fourth unmeasured displacement, he finds himself back where he started. Using a scale drawing, determine the fourth displacement's magnitude and direction.

Explore how to calculate the magnitude and direction of a fourth displacement in a cave survey using vector addition and trigonometry. The problem breaks down into calculating x and y components of each movement and applying the Pythagorean theorem to find the final displacement. Suitable for high school students in Grades 9-12.

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