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To find the total displacement, we need to break down each movement into its East-West (x) and North-South (y) components, then sum the components, and finally calculate the magnitude of the resultant

Physics

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To find the total displacement, we need to break down each movement into its East-West (x) and North-South (y) components, then sum the components, and finally calculate the magnitude of the resultant vector.

Let East be the positive x-direction and North be the positive y-direction.

  • Step 1: Resolve the first displacement into components. The man moves 6.0 m6.0 \text{ m} East.

    D1=(6.0 m,0 m)D_1 = (6.0 \text{ m}, 0 \text{ m})

    So, D1x=6.0 mD_{1x} = 6.0 \text{ m} and D1y=0 mD_{1y} = 0 \text{ m}.

  • Step 2: Resolve the second displacement into components. The man moves 10.0 m10.0 \text{ m} in the direction N30°E. This means 30∘30^\circ East of North. The angle with respect to the positive y-axis (North) is 30∘30^\circ. The x-component (East) is D2sin⁡(30∘)D_2 \sin(30^\circ). The y-component (North) is D2cos⁡(30∘)D_2 \cos(30^\circ).

    D2x=10.0 m×sin⁡(30∘)D_{2x} = 10.0 \text{ m} \times \sin(30^\circ) D2y=10.0 m×cos⁡(30∘)D_{2y} = 10.0 \text{ m} \times \cos(30^\circ)

    We know that sin⁡(30∘)=0.5\sin(30^\circ) = 0.5 and cos⁡(30∘)=32≈0.866\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866.

    D2x=10.0 m×0.5=5.0 mD_{2x} = 10.0 \text{ m} \times 0.5 = 5.0 \text{ m} D2y=10.0 m×32=53 m≈8.66 mD_{2y} = 10.0 \text{ m} \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ m} \approx 8.66 \text{ m}
  • Step 3: Sum the x and y components to find the total displacement. Total x-displacement (RxR_x):

    Rx=D1x+D2x=6.0 m+5.0 m=11.0 mR_x = D_{1x} + D_{2x} = 6.0 \text{ m} + 5.0 \text{ m} = 11.0 \text{ m}

    Total y-displacement (RyR_y):

    Ry=D1y+D2y=0 m+53 m=53 mR_y = D_{1y} + D_{2y} = 0 \text{ m} + 5\sqrt{3} \text{ m} = 5\sqrt{3} \text{ m}
  • Step 4: Calculate the magnitude of the resultant displacement. The magnitude of the resultant displacement RR is given by the Pythagorean theorem:

    R=Rx2+Ry2R = \sqrt{R_x^2 + R_y^2} R=(11.0 m)2+(53 m)2R = \sqrt{(11.0 \text{ m})^2 + (5\sqrt{3} \text{ m})^2} R=121 m2+(25×3) m2R = \sqrt{121 \text{ m}^2 + (25 \times 3) \text{ m}^2} R=121 m2+75 m2R = \sqrt{121 \text{ m}^2 + 75 \text{ m}^2} R=196 m2R = \sqrt{196 \text{ m}^2} R=14.0 mR = 14.0 \text{ m}

The man is 14.0 m14.0 \text{ m} from his starting point.

Comparing this with the given options:
A. 12.0 m
B. 17.0 m
C. 15.0 m
D. 14.0 m

The correct option is D.

The final answer is D. 14.0 m\boxed{\text{D. 14.0 m}}.

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