1. **Stating the problem:** A train moves along a straight line with uniform acceleration. Initially, its speed is 20 km/h. When it passes the next kilometer post, its speed is 30 km/h. We need to find the speed at which the train passes the next kilometer post and the time taken to do so. 2. **Given data:** - Initial speed, $u = 20$ km/h - Final speed after passing the next kilometer post, $v = 30$ km/h - Distance between posts, $s = 1$ km 3. **Convert speeds to m/s:** - $u = 20 \times \frac{1000}{3600} = \frac{50}{9}$ m/s - $v = 30 \times \frac{1000}{3600} = \frac{25}{3}$ m/s 4. **Formula used:** - Using the equation of motion with uniform acceleration: $$v^2 = u^2 + 2as$$ where $a$ is acceleration. 5. **Calculate acceleration $a$:** $$a = \frac{v^2 - u^2}{2s}$$ Substitute values: $$a = \frac{\left(\frac{25}{3}\right)^2 - \left(\frac{50}{9}\right)^2}{2 \times 1000}$$ Calculate numerator: $$\left(\frac{25}{3}\right)^2 = \frac{625}{9}, \quad \left(\frac{50}{9}\right)^2 = \frac{2500}{81}$$ Find common denominator 81: $$\frac{625}{9} = \frac{625 \times 9}{81} = \frac{5625}{81}$$ So, $$a = \frac{\frac{5625}{81} - \frac{2500}{81}}{2000} = \frac{\frac{3125}{81}}{2000} = \frac{3125}{81 \times 2000} = \frac{3125}{162000} \approx 0.01929 \text{ m/s}^2$$ 6. **Calculate time $t$ to pass the next post:** Using formula: $$v = u + at$$ Rearranged: $$t = \frac{v - u}{a}$$ Substitute values: $$t = \frac{\frac{25}{3} - \frac{50}{9}}{0.01929} = \frac{\frac{75}{9} - \frac{50}{9}}{0.01929} = \frac{\frac{25}{9}}{0.01929} = \frac{2.7778}{0.01929} \approx 144.0 \text{ seconds}$$ 7. **Final answers:** - Speed at next kilometer post: 30 km/h (given) - Time taken to reach next post: approximately 144 seconds or 2 minutes 24 seconds. This means the train accelerates uniformly from 20 km/h to 30 km/h over 1 km in about 2 minutes 24 seconds.