1. **Problem statement:** A person walks 10 minutes in a direction 53° north of east, then 10 meters north, and finally 20 minutes in a direction 37° north of west. We need to find the magnitude and direction of the person's displacement. 2. **Assumptions and clarifications:** Since the problem mixes time and distance, we assume the walking speed is constant and convert times to distances proportionally. Let's denote the speed as $v$ meters per minute. 3. **Step 1: Calculate displacement vectors for each segment.** - First segment: 10 minutes at 53° north of east. Distance $d_1 = 10v$ meters. Components: $$x_1 = d_1 \cos 53^\circ = 10v \times 0.6018 = 6.018v$$ $$y_1 = d_1 \sin 53^\circ = 10v \times 0.7986 = 7.986v$$ - Second segment: 10 meters north. Components: $$x_2 = 0$$ $$y_2 = 10$$ - Third segment: 20 minutes at 37° north of west. Distance $d_3 = 20v$ meters. Components: $$x_3 = -d_3 \cos 37^\circ = -20v \times 0.7986 = -15.972v$$ $$y_3 = d_3 \sin 37^\circ = 20v \times 0.6018 = 12.036v$$ 4. **Step 2: Sum components to find total displacement vector.** $$x_{total} = x_1 + x_2 + x_3 = 6.018v + 0 - 15.972v = -9.954v$$ $$y_{total} = y_1 + y_2 + y_3 = 7.986v + 10 + 12.036v = (20.022v + 10)$$ 5. **Step 3: Calculate magnitude of displacement.** $$D = \sqrt{x_{total}^2 + y_{total}^2} = \sqrt{(-9.954v)^2 + (20.022v + 10)^2}$$ 6. **Step 4: Calculate direction of displacement.** Direction angle $\theta$ north of west: $$\theta = \tan^{-1} \left( \frac{y_{total}}{|x_{total}|} \right) = \tan^{-1} \left( \frac{20.022v + 10}{9.954v} \right)$$ 7. **Step 5: Interpretation:** Without a given speed $v$, the displacement depends on $v$. If speed is known, substitute to find numeric values. **Summary:** The displacement vector components are $$x = -9.954v$$ and $$y = 20.022v + 10$$. Magnitude and direction depend on $v$ as shown. If speed $v$ is given, plug in to get final numeric answers.