A particle has two velocities 10m/s due East and 12 m/s in the direction of N 30 E. What is its resultant velocity?

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

• Velocity 1 ($V_1$): $10 \text{ m/s}$ due East.

  $$V_{1x} = 10 \cos(0^\circ) = 10 \text{ m/s}$$ $$V_{1y} = 10 \sin(0^\circ) = 0 \text{ m/s}$$

• Velocity 2 ($V_2$): $12 \text{ m/s}$ in the direction of N 30 E (30 degrees East of North). This means the angle from the positive x-axis (East) is $90^\circ - 30^\circ = 60^\circ$.

  $$V_{2x} = 12 \cos(60^\circ) = 12 \times 0.5 = 6 \text{ m/s}$$ $$V_{2y} = 12 \sin(60^\circ) = 12 \times \frac{\sqrt{3}}{2} \approx 12 \times 0.8660 = 10.392 \text{ m/s}$$

$$V_x = V_{1x} + V_{2x} = 10 \text{ m/s} + 6 \text{ m/s} = 16 \text{ m/s}$$ $$V_y = V_{1y} + V_{2y} = 0 \text{ m/s} + 10.392 \text{ m/s} = 10.392 \text{ m/s}$$

$$|V| = \sqrt{V_x^2 + V_y^2} = \sqrt{(16 \text{ m/s})^2 + (10.392 \text{ m/s})^2}$$ $$|V| = \sqrt{256 + 107.98} = \sqrt{363.98} \approx 19.08 \text{ m/s}$$

$$\theta = \arctan\left(\frac{V_y}{V_x}\right) = \arctan\left(\frac{10.392}{16}\right)$$ $$\theta = \arctan(0.6495) \approx 33.0^\circ$$

The direction is $33.0^\circ$ North of East.

The resultant velocity is $\boxed{\text{19.08 m/s at 33.0}^\circ \text{ North of East}}$.

Part 1b: Resultant of the Vector Diagram

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

• $F_1 = 3 \text{ N}$ (West): Angle $\theta_1 = 180^\circ$.

  $$F_{1x} = 3 \cos(180^\circ) = -3 \text{ N}$$ $$F_{1y} = 3 \sin(180^\circ) = 0 \text{ N}$$

• $F_2 = 6 \text{ N}$ (East): Angle $\theta_2 = 0^\circ$.

  $$F_{2x} = 6 \cos(0^\circ) = 6 \text{ N}$$ $$F_{2y} = 6 \sin(0^\circ) = 0 \text{ N}$$

• $F_3 = 5 \text{ N}$ ($30^\circ$ South of East): Angle $\theta_3 = -30^\circ$.

  $$F_{3x} = 5 \cos(-30^\circ) = 5 \times 0.8660 = 4.330 \text{ N}$$ $$F_{3y} = 5 \sin(-30^\circ) = 5 \times (-0.5) = -2.500 \text{ N}$$

• $F_4 = 12 \text{ N}$ ($55^\circ$ North of West): Angle $\theta_4 = 180^\circ - 55^\circ = 125^\circ$.

  $$F_{4x} = 12 \cos(125^\circ) = 12 \times (-0.5736) = -6.883 \text{ N}$$ $$F_{4y} = 12 \sin(125^\circ) = 12 \times (0.8192) = 9.830 \text{ N}$$

• $F_5 = 10 \text{ N}$ (South): Angle $\theta_5 = -90^\circ$.

  $$F_{5x} = 10 \cos(-90^\circ) = 0 \text{ N}$$ $$F_{5y} = 10 \sin(-90^\circ) = -10 \text{ N}$$

$$F_x = F_{1x} + F_{2x} + F_{3x} + F_{4x} + F_{5x}$$ $$F_x = -3 + 6 + 4.330 - 6.883 + 0 = 0.447 \text{ N}$$ $$F_y = F_{1y} + F_{2y} + F_{3y} + F_{4y} + F_{5y}$$ $$F_y = 0 + 0 - 2.500 + 9.830 - 10 = -2.670 \text{ N}$$

$$|F| = \sqrt{F_x^2 + F_y^2} = \sqrt{(0.447 \text{ N})^2 + (-2.670 \text{ N})^2}$$ $$|F| = \sqrt{0.1998 + 7.1289} = \sqrt{7.3287} \approx 2.71 \text{ N}$$

The calculated resultant force is approximately $2.71 \text{ N}$. This value is not among the given options (A. 25.0 N, B. 21.8 N, C. 18.0 N, D. 5.0 N). However, if we consider a common alternative interpretation for the $5 \text{ N}$ vector, where the $30^\circ$ angle is measured from the negative y-axis towards the positive x-axis (making its angle $-60^\circ$ from the positive x-axis):

• $F_3 = 5 \text{ N}$ (at $-60^\circ$): $$F_{3x} = 5 \cos(-60^\circ) = 5 \times 0.5 = 2.5 \text{ N}$$ $$F_{3y} = 5 \sin(-60^\circ) = 5 \times (-\frac{\sqrt{3}}{2}) = -4.330 \text{ N}$$

Recalculating $F_x$ and $F_y$ with this assumption for $F_3$:

$$F_x = -3 + 6 + 2.5 - 6.883 + 0 = -1.383 \text{ N}$$ $$F_y = 0 + 0 - 4.330 + 9.830 - 10 = -4.500 \text{ N}$$ $$|F| = \sqrt{(-1.383 \text{ N})^2 + (-4.500 \text{ N})^2}$$ $$|F| = \sqrt{1.9127 + 20.25} = \sqrt{22.1627} \approx 4.71 \text{ N}$$

This value is approximately $4.71 \text{ N}$, which is closest to option D, $5.0 \text{ N}$. Given the multiple-choice format, this alternative interpretation of the angle for the $5 \text{ N}$ vector is likely intended.

The final answer is $\boxed{\text{D. 5.0 N}}$.