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## Homework Statement

A train pulls away from a station with a constant acceleration of 0.41 m/s

^{2}. A passenger arrives at a point next to the track 6.4 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?

## Homework Equations

x = 1/2*a*t

^{2}

x = v*t

## The Attempt at a Solution

I think the 0.41 m/s

^{2}belongs in the first equation, and gives a graph of position of the end of the train vs time. The 6.4 seconds can be used in the second equation, turing it into x = v*(t-6.4) to graph the motion of the girl running to catch up to the train. I can find the derivative of the first equation to find the slope that the line of the second equation must match, but I always get strange results when I try to tell that line to also go through the point (6.4,0). All I need is the value of v in the second equation, but I always end up eliminating it or getting strange results when simplifying.