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A Train pulls away froma station with a constant acceleration of .4m/s^2. A passenger arrives at the track 6.0s after the end of the train has passed the very same point. what is the slowest constant speed at which she can run and catch the train?

I am rather confused on what to do but i have some idea

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distance of train in x(6s) = 1/2 * (.4m/s^2) * (6s)^2 = 7.2m

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So the person is 7.2m away when she gets there

But how do create 2 equations and set them equal to eachother?

Im thinking it might be as simple as expressing the train as a parabolic equation which intercepts at 7.2, and the linear line starts at 0, but how do i express accelleration in an equation that i can graph??

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# Homework Help: Constant Acceleration of a train

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