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## Homework Statement

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)

## Homework Equations

V=V

_{0}+at

X=X

_{0}+V

_{0}t+.5at

^{2}

## 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+.2t

^{2}.

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+.2t

^{2}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.

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