To find how long it takes for a train with a mass of \(2.1 \times 10^6\) kg, moving at 12 m/s, to stop when a constant braking force of \(3.7 \times 10^5\) N is applied: 1. **Calculate deceleration**: \[ a = \frac{F}{m} = \frac{3.7 \times 10^5 \, \text{N}}{2.1 \times 10^6 \, \text{kg}} \approx -0.176 \, \text{m/s}^2 \] 2. **Calculate stopping time**: \[ t = \frac{v_i}{|a|} = \frac{12 \, \text{m/s}}{0.176 \, \text{m/s}^2} \approx 68.18 \, \text{s} \] 3. **Calculate stopping distance**: \[ d = v_i t \frac{1}{2}at^2 = 12 \times 68.18 \frac{1}{2} \times (-0.176) \times (68.18)^2 \approx 409.24 \, \text{m} \] So, the train takes approximately 68.18 seconds to stop and travels about 409.24 meters during this time.