1. **Problem statement:** A suspension bracket in a train carriage inclines at an angle of $5^\circ$ to the vertical during the first minute and at $5^\circ$ in the opposite direction during the last minute of a 5-minute journey. It remains vertical during the middle part. We need to find the train's acceleration and total distance traveled. 2. **Understanding the problem:** The bracket inclines due to acceleration. The angle $\theta$ relates to acceleration $a$ by the formula for a pendulum in accelerating frame: $$\tan(\theta) = \frac{a}{g}$$ where $g = 9.8$ m/s$^2$ is acceleration due to gravity. 3. **Calculate acceleration:** Given $\theta = 5^\circ$, $$a = g \tan(5^\circ) = 9.8 \times \tan(5^\circ)$$ Using $\tan(5^\circ) \approx 0.0875$, $$a = 9.8 \times 0.0875 = 0.8575 \text{ m/s}^2$$ 4. **Analyze motion:** - First minute: acceleration $a = 0.8575$ m/s$^2$ - Middle 3 minutes: no acceleration (bracket vertical) - Last minute: acceleration $a = -0.8575$ m/s$^2$ (opposite direction) 5. **Calculate velocity at end of first minute:** $$v = u + at = 0 + 0.8575 \times 60 = 51.45 \text{ m/s}$$ 6. **Calculate distance traveled in first minute:** $$s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 0.8575 \times 60^2 = 1547.1 \text{ m}$$ 7. **Middle 3 minutes:** constant velocity $v = 51.45$ m/s $$s_2 = vt = 51.45 \times 180 = 9261 \text{ m}$$ 8. **Last minute:** deceleration $a = -0.8575$ m/s$^2$, initial velocity $51.45$ m/s Distance: $$s_3 = vt + \frac{1}{2}at^2 = 51.45 \times 60 + \frac{1}{2} \times (-0.8575) \times 60^2 = 1547.1 \text{ m}$$ 9. **Total distance traveled:** $$s = s_1 + s_2 + s_3 = 1547.1 + 9261 + 1547.1 = 12355.2 \text{ m}$$ **Final answers:** - Acceleration magnitude: $0.8575$ m/s$^2$ - Total distance traveled: $12355.2$ m