1. **State the problem:** A car travels 150 m at 63° North of East, then 300 m at 34° South of West. We need to find the total displacement. 2. **Formula and approach:** Displacement is a vector sum of the two legs of the journey. We break each displacement into components along East (x-axis) and North (y-axis). 3. **First displacement components:** - Magnitude: 150 m - Direction: 63° North of East - $x_1 = 150 \cos 63^\circ$ - $y_1 = 150 \sin 63^\circ$ 4. **Calculate first displacement components:** - $x_1 = 150 \times 0.4540 = 68.1$ m - $y_1 = 150 \times 0.8910 = 133.7$ m 5. **Second displacement components:** - Magnitude: 300 m - Direction: 34° South of West - West is negative x, South is negative y - $x_2 = -300 \cos 34^\circ$ - $y_2 = -300 \sin 34^\circ$ 6. **Calculate second displacement components:** - $x_2 = -300 \times 0.8290 = -248.7$ m - $y_2 = -300 \times 0.5592 = -167.8$ m 7. **Total displacement components:** - $x = x_1 + x_2 = 68.1 - 248.7 = -180.6$ m - $y = y_1 + y_2 = 133.7 - 167.8 = -34.1$ m 8. **Magnitude of total displacement:** $$ D = \sqrt{x^2 + y^2} = \sqrt{(-180.6)^2 + (-34.1)^2} = \sqrt{32616.4 + 1162.8} = \sqrt{33779.2} = 183.7 \text{ m} $$ 9. **Direction of total displacement:** - Angle $\theta$ relative to East axis: $$ \theta = \tan^{-1} \left( \frac{y}{x} \right) = \tan^{-1} \left( \frac{-34.1}{-180.6} \right) = \tan^{-1} (0.1888) = 10.7^\circ $$ - Since both $x$ and $y$ are negative, the vector is in the third quadrant, so direction is $180^\circ + 10.7^\circ = 190.7^\circ$ from East, or equivalently 10.7° South of West. **Final answer:** The total displacement is approximately 184 m at 11° South of West.