Math Problem Statement

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

Solution

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: 50m50 \, \text{m}
    • Bearing: 120120^\circ
    • This means the vector makes an angle of 12090=30120^\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): Vector Ax=50×cos(120)=50×(0.5)=25m\text{Vector A}_x = 50 \times \cos(120^\circ) = 50 \times (-0.5) = -25 \, \text{m} Vector Ay=50×sin(120)=50×(32)50×0.866=43.3m\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: 30m30 \, \text{m}
    • Bearing: 170170^\circ
    • The angle with the positive x-axis would be 17090=80170^\circ - 90^\circ = 80^\circ.

    The components of this vector are: Vector Bx=30×cos(170)=30×(0.985)29.55m\text{Vector B}_x = 30 \times \cos(170^\circ) = 30 \times (-0.985) \approx -29.55 \, \text{m} Vector By=30×sin(170)=30×0.1745.22m\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: Rx=25+(29.55)=54.55mR_x = -25 + (-29.55) = -54.55 \, \text{m}

  • Resultant y-component: Ry=43.3+5.22=48.52mR_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: Displacement=Rx2+Ry2\text{Displacement} = \sqrt{R_x^2 + R_y^2} Displacement=(54.55)2+(48.52)22973.7+2354.15327.873.00m\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.

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Math Problem Analysis

Mathematical Concepts

Vector Analysis
Trigonometry

Formulas

Vector addition
Magnitude of a vector

Theorems

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Suitable Grade Level

High school