1. **Problem statement:** Roy is in the middle of a 200 m train moving at 4 km/h. Max is 125 m behind the train, running at 6 km/h towards it. Roy moves towards the last door at 1 km/h to help Max board. Find the time taken for Max to board the train. 2. **Convert speeds to m/s:** - Train speed: $4 \text{ km/h} = \frac{4 \times 1000}{3600} = \frac{10}{9} \text{ m/s}$ - Max speed: $6 \text{ km/h} = \frac{6 \times 1000}{3600} = \frac{5}{3} \text{ m/s}$ - Roy speed relative to ground: $1 \text{ km/h} = \frac{1000}{3600} = \frac{5}{18} \text{ m/s}$ 3. **Positions and distances:** - Train length = 200 m, Roy is in the middle, so 100 m from either end. - Max is 125 m behind the train's rear. 4. **Relative speeds:** - Max approaches the train at $\frac{5}{3} \text{ m/s}$. - Train moves forward at $\frac{10}{9} \text{ m/s}$. - Roy moves towards the rear door at $\frac{5}{18} \text{ m/s}$ relative to ground. 5. **Calculate time for Max to reach the rear of the train:** - Max's speed relative to train rear = Max speed - train speed = $\frac{5}{3} - \frac{10}{9} = \frac{15}{9} - \frac{10}{9} = \frac{5}{9} \text{ m/s}$ - Distance Max must cover to reach rear = 125 m - Time for Max to reach rear = $\frac{125}{5/9} = 125 \times \frac{9}{5} = 225 \text{ seconds}$ 6. **Calculate Roy's movement towards rear door:** - Roy is initially 100 m from rear door inside the train. - Roy moves towards rear at $\frac{5}{18} \text{ m/s}$ - Time for Roy to reach rear door = $\frac{100}{5/18} = 100 \times \frac{18}{5} = 360 \text{ seconds}$ 7. **Interpretation:** - Max reaches rear of train in 225 seconds. - Roy reaches rear door in 360 seconds. 8. **Since Max arrives earlier, Roy must wait for Max to board. But Roy cannot pull Max until Roy reaches rear door. So, Max waits for Roy.** 9. **Time taken for Max to board = time for Roy to reach rear door = 360 seconds = 6 minutes.** **Final answer:** Max boards the train after 6 minutes.