A person walks 6.0 m East, then 10.0 m N30°E. What is the magnitude of their total displacement?

• First displacement ($\vec{D_1}$): $6.0 \text{ m}$ East

  • $D_{1x} = 6.0 \text{ m}$
  • $D_{1y} = 0 \text{ m}$

• Second displacement ($\vec{D_2}$): $10.0 \text{ m}$ N30°E This means $30^\circ$ East of North. If we consider North as the y-axis and East as the x-axis, the angle with the y-axis is $30^\circ$.

  • The East (x) component is $D_2 \sin(30^\circ)$.

  • The North (y) component is $D_2 \cos(30^\circ)$.

  • $D_{2x} = 10.0 \text{ m} \times \sin(30^\circ) = 10.0 \text{ m} \times 0.5 = 5.0 \text{ m}$

  • $D_{2y} = 10.0 \text{ m} \times \cos(30^\circ) = 10.0 \text{ m} \times \frac{\sqrt{3}}{2} \approx 10.0 \text{ m} \times 0.866 = 8.66 \text{ m}$

• Total East component ($R_x$): $$R_x = D_{1x} + D_{2x} = 6.0 \text{ m} + 5.0 \text{ m} = 11.0 \text{ m}$$
• Total North component ($R_y$): $$R_y = D_{1y} + D_{2y} = 0 \text{ m} + 8.66 \text{ m} = 8.66 \text{ m}$$

Rounding to one decimal place, $R \approx 14.0 \text{ m}$.

Using exact values for $\sin(30^\circ)$ and $\cos(30^\circ)$:
$D_{2x} = 10 \times \frac{1}{2} = 5 \text{ m}$
$D_{2y} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ m}$

$R_x = 6 + 5 = 11 \text{ m}$
$R_y = 5\sqrt{3} \text{ m}$

$R = \sqrt{(11)^2 + (5\sqrt{3})^2}$
$R = \sqrt{121 + (25 \times 3)}$
$R = \sqrt{121 + 75}$
$R = \sqrt{196}$
$R = 14 \text{ m}$

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