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A man covered following displacements. F...

A man covered following displacements. Find the net displacement of the person
(i) 25m west
(ii)` 20 m 30^(@)` east of south
(iii)`5sqrt(3)m` south
(iv) 10 m east
(v) `10 m 60^(@)` north of west

Text Solution

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The correct Answer is:
To find the net displacement of the person after covering the given displacements, we can break down each movement into its components along the x (east-west) and y (north-south) axes. ### Step-by-Step Solution: 1. **Define the Directions**: - East is positive x-direction. - West is negative x-direction. - North is positive y-direction. - South is negative y-direction. 2. **List the Displacements**: - (i) 25 m West - (ii) 20 m at 30° East of South - (iii) 5√3 m South - (iv) 10 m East - (v) 10 m at 60° North of West 3. **Convert Each Displacement to Components**: - **Displacement (i)**: - \( D_1 = -25 \hat{i} + 0 \hat{j} \) (25 m West) - **Displacement (ii)**: - \( D_2 = 20 \cos(30°) \hat{i} - 20 \sin(30°) \hat{j} \) - \( D_2 = 20 \left(\frac{\sqrt{3}}{2}\right) \hat{i} - 20 \left(\frac{1}{2}\right) \hat{j} \) - \( D_2 = 10\sqrt{3} \hat{i} - 10 \hat{j} \) - **Displacement (iii)**: - \( D_3 = 0 \hat{i} - 5\sqrt{3} \hat{j} \) (5√3 m South) - **Displacement (iv)**: - \( D_4 = 10 \hat{i} + 0 \hat{j} \) (10 m East) - **Displacement (v)**: - \( D_5 = -10 \cos(60°) \hat{i} + 10 \sin(60°) \hat{j} \) - \( D_5 = -10 \left(\frac{1}{2}\right) \hat{i} + 10 \left(\frac{\sqrt{3}}{2}\right) \hat{j} \) - \( D_5 = -5 \hat{i} + 5\sqrt{3} \hat{j} \) 4. **Sum the Components**: - **Total x-component**: \[ x_{total} = -25 + 10\sqrt{3} + 10 - 5 = -15 + 10\sqrt{3} \] - **Total y-component**: \[ y_{total} = -10 - 5\sqrt{3} + 5\sqrt{3} = -10 \] 5. **Calculate the Magnitude of the Net Displacement**: - The net displacement \( D \) can be calculated using the Pythagorean theorem: \[ D = \sqrt{(x_{total})^2 + (y_{total})^2} \] \[ D = \sqrt{(-15 + 10\sqrt{3})^2 + (-10)^2} \] \[ D = \sqrt{(15 - 10\sqrt{3})^2 + 100} \] \[ D = \sqrt{(225 - 300\sqrt{3} + 300) + 100} \] \[ D = \sqrt{625 - 300\sqrt{3}} \] 6. **Final Calculation**: - This expression can be simplified further if needed, but for practical purposes, we can evaluate it numerically to find the approximate value of the net displacement. ### Final Answer: The net displacement of the person is approximately **20 meters**.
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