Physics (12)

The car is 550 meters behind the train when it regains a velocity of 25 m/s [W]. The train maintained its constant velocity of 25 m/s [W] while the car was momentarily stopped, allowing the train to move ahead at 25 m/s [W]. Since the car decelerates to and accelerates from a zero velocity, the displacement of the car relative to the train from the beginning to the end of this period of zero velocity is calculated by multiplying the average velocity of the car (0 m/s) by the amount of time the car was at rest (45 s). This gives a total displacement of 0 m/s * 45 s, or 0 meters. Since the difference in velocities of the train and car remains constant throughout the time the car is motionless and accelerating, the total distance between the train and the car is equal to the displacement of the car relative to the train at the end of this period of acceleration (25 m/s * 45 s), which is 550 meters.