A freight train passed the station at a constant speed of 18 km / h. 2 minutes later, a passenger train began to move

A freight train passed the station at a constant speed of 18 km / h. 2 minutes later, a passenger train began to move from the same station with an acceleration of 0.5 m / s2. How long and at what distance from the station the passenger train will catch up with the freight train

To solve this problem, we compose the coordinate equations for both trains:
Since the passenger departed only after 2 minutes, then
Xt = 600 + 5t
Xп = 0.25t ^ 2
Since the passenger will catch up with the commodity, then Xt = For, then
600 + 5t = 0.25t ^ 2, we get:
t = 60c, do not take the negative root
This means that the distance from the station (the coordinate of the meeting) is equal to: X = 900m