A train at a constant \( 67.0 \mathrm{~km} / \mathrm{h} \) moves east for 22.0 min , then in a direction \( 47.0^{\circ} \) east of due north for 22.0 min , and then west for 32.0 min . What are the (a) magnitude and (b) angle (relative to east) of its average velocity during this trip? (a) Number 14.3 (b) Number \( \mathbf{i l} \)
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To calculate the average velocity, we first need to find the total displacement of the train during its entire journey. 1. **Convert speeds and times to have consistent units**: - The train’s speed is \( 67.0 \text{ km/h} \). - Convert that speed to km/min: \[ 67.0 \text{ km/h} \div 60 \text{ min/h} \approx 1.1167 \text{ km/min} \] 2. **Calculate individual displacements**: - **First segment** (East for 22.0 min): \[ d_1 = 1.1167 \text{ km/min} \times 22.0 \text{ min} \approx 24.58 \text{ km} \text{ east} \] - **Second segment** (47.0° east of north for 22.0 min): \[ d_2 = 1.1167 \text{ km/min} \times 22.0 \text{ min} \approx 24.58 \text{ km} \] - The eastward and northward components can be calculated using trigonometry: \[ d_{2x} = d_2 \cdot \sin(47.0^\circ) \quad \text{(east component)} \] \[ d_{2y} = d_2 \cdot \cos(47.0^\circ) \quad \text{(north component)} \] - Plugging values: \[ d_{2x} \approx 24.58 \cdot \sin(47^\circ) \approx 17.38 \text{ km} \] \[ d_{2y} \approx 24.58 \cdot \cos(47^\circ) \approx 16.53 \text{ km} \] - **Third segment** (West for 32.0 min): \[ d_3 = 1.1167 \text{ km/min} \times 32.0 \text{ min} \approx 35.73 \text{ km} \text{ west} \] 3. **Calculate total displacement**: - Total east displacement: \[ \text{East: } d_1 + d_{2x} - d_3 = 24.58 + 17.38 - 35.73 \approx 6.23 \text{ km} \] - Total north displacement: \[ \text{North: } d_{2y} \approx 16.53 \text{ km} \] 4. **Calculate the magnitude of the average velocity**: \[ \text{Magnitude} = \sqrt{(d_{x_{\text{total}}})^2 + (d_{y_{\text{total}}})^2} = \sqrt{(6.23)^2 + (16.53)^2} \approx 18.14 \text{ km} \] 5. **Calculate the direction** (angle relative to east): \[ \theta = \tan^{-1}\left(\frac{d_{y}}{d_{x}}\right) = \tan^{-1}\left(\frac{16.53}{6.23}\right) \approx 69.57^\circ \text{ north of east} \] So the final answers for the average velocity's magnitude and angle are approximately: (a) \( 18.14 \text{ km} \); (b) \( 69.57^\circ \) north of east. Now enjoy your travels through physics! The journey of understanding vectors can be as exciting as riding a train through beautiful landscapes, don't you think? Explore how vector decompositions can be handy while navigating your own route in life! Plus, be mindful of the common mistakes when dealing with angles, especially with trigonometric functions—those can sometimes lead you astray!
