A train pulls away from a station with a constant acceleration of 0.4 m/s2. A passenger arrives at a point next to the track 6 s after the end of the train has passed the very same point. What is the slowest constant speed at which she can run and still catch the train? (changed 6.4 s to 6 s and .41 m/s2 to .4)
The Attempt at a Solution
So I got the distance and the velocity of the train when the passenger shows up which is 7.2+2.4t+.2t2.
For the speed of the passenger I got V(t-6). I set them equal to each other and got V(t-6)=7.2+2.4t+.2t2 I am not sure where to go after this. I saw some other forums with the same questions, but I didn't really understand what to do, any help or hints as to what to do now would be appreciated.