A child runs 50 m from a slide on a playground to a  lemonade stand which is on a bearing of 120°. From the lemonade stand, he runs to a icecream truck which is 30 m away on a bearing of 170°. Find the displacement of the child. (Hint: displacement is a vector that connects the initial and final position of moving object)
Question 13 options:

a)

73.00 m

b)

40 m

c)

80 m

d)

38.36 m

To find the displacement of the child, we'll break down the problem using vector analysis.

### Step 1: Represent the problem with vectors
1. **Vector from the slide to the lemonade stand:**
   - Magnitude: \( 50 \, \text{m} \)
   - Bearing: \( 120^\circ \)
   - This means the vector makes an angle of \( 120^\circ - 90^\circ = 30^\circ \) with the negative x-axis (assuming the positive x-axis is 0°).

   The components of this vector (using the fact that bearing angles are measured clockwise from the North):
   \[
   \text{Vector A}_x = 50 \times \cos(120^\circ) = 50 \times (-0.5) = -25 \, \text{m}
   \]
   \[
   \text{Vector A}_y = 50 \times \sin(120^\circ) = 50 \times \left(\frac{\sqrt{3}}{2}\right) \approx 50 \times 0.866 = 43.3 \, \text{m}
   \]

2. **Vector from the lemonade stand to the ice cream truck:**
   - Magnitude: \( 30 \, \text{m} \)
   - Bearing: \( 170^\circ \)
   - The angle with the positive x-axis would be \( 170^\circ - 90^\circ = 80^\circ \).

   The components of this vector are:
   \[
   \text{Vector B}_x = 30 \times \cos(170^\circ) = 30 \times (-0.985) \approx -29.55 \, \text{m}
   \]
   \[
   \text{Vector B}_y = 30 \times \sin(170^\circ) = 30 \times 0.174 \approx 5.22 \, \text{m}
   \]

### Step 2: Find the resultant vector
To find the displacement vector, we add the components of the two vectors:

- **Resultant x-component:**
  \[
  R_x = -25 + (-29.55) = -54.55 \, \text{m}
  \]

- **Resultant y-component:**
  \[
  R_y = 43.3 + 5.22 = 48.52 \, \text{m}
  \]

### Step 3: Calculate the magnitude of the resultant vector
The magnitude of the displacement is given by:
\[
\text{Displacement} = \sqrt{R_x^2 + R_y^2}
\]
\[
\text{Displacement} = \sqrt{(-54.55)^2 + (48.52)^2} \approx \sqrt{2973.7 + 2354.1} \approx \sqrt{5327.8} \approx 73.00 \, \text{m}
\]

### Final Answer:
The correct answer is:
**a) 73.00 m**

Would you like more details on any part of this solution or have any other questions? Here are some related questions you might find helpful:

1. How do you calculate the angle between two vectors?
2. What is the difference between distance and displacement?
3. How do you resolve a vector into its components?
4. How do bearings work in navigation?
5. What are some examples of vector addition in real life?

**Tip:** When dealing with vector problems, always break down vectors into their components and use trigonometry to handle bearings effectively.

Learn how to calculate the displacement of an object using vector analysis and trigonometry with a detailed step-by-step solution. This problem involves breaking down vectors into components, applying trigonometric functions to handle bearings, and calculating the magnitude of the resultant vector. Suitable for high school students.

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