Hiker Displacement 8D776D

1. **State the problem:** A hiker walks $80$ m East and then $60$ m at $30^\circ$ North of East. We need to find the resultant displacement vector. 2. **Formula and rules:** To find the resultant displacement using the component method, we break each vector into its $x$ (East-West) and $y$ (North-South) components and then add these components. 3. **Step 1: Components of the first vector** (80 m East): - $x_1 = 80$ m (since it is directly East) - $y_1 = 0$ m 4. **Step 2: Components of the second vector** (60 m at $30^\circ$ North of East): - $x_2 = 60 \cos 30^\circ = 60 \times \frac{\sqrt{3}}{2} = 30\sqrt{3}$ m - $y_2 = 60 \sin 30^\circ = 60 \times \frac{1}{2} = 30$ m 5. **Step 3: Sum the components:** - $x_R = x_1 + x_2 = 80 + 30\sqrt{3}$ m - $y_R = y_1 + y_2 = 0 + 30 = 30$ m 6. **Step 4: Calculate the magnitude of the resultant displacement:** $$ R = \sqrt{x_R^2 + y_R^2} = \sqrt{(80 + 30\sqrt{3})^2 + 30^2} $$ 7. **Step 5: Calculate the direction (angle) of the resultant displacement relative to East:** $$ \theta = \tan^{-1}\left(\frac{y_R}{x_R}\right) = \tan^{-1}\left(\frac{30}{80 + 30\sqrt{3}}\right) $$ 8. **Step 6: Numerical evaluation:** - Approximate $30\sqrt{3} \approx 30 \times 1.732 = 51.96$ - So, $x_R \approx 80 + 51.96 = 131.96$ m - $y_R = 30$ m - Magnitude: $$ R = \sqrt{131.96^2 + 30^2} = \sqrt{17411.5 + 900} = \sqrt{18311.5} \approx 135.4 \text{ m} $$ - Direction: $$ \theta = \tan^{-1}\left(\frac{30}{131.96}\right) = \tan^{-1}(0.2274) \approx 12.9^\circ $$ **Final answer:** The hiker's resultant displacement is approximately $135.4$ m at $12.9^\circ$ North of East.