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# 5min this seems to be a reasonable time our velocity

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Unformatted text preview: takes: t= t= 4v2 ∆x 2 v1f 4 (45mph) (3.5miles) 2 (60mph) t = 0.175hrs t = 10.5min This seems to be a reasonable time. Our velocity when we pass the freight train: v = at 2 v1f 4v2 ∆x 2 2∆x v1f 2v2 v= 1 v = 2v2 v= v = 2 (45mph) v = 90mph While 90 mph seems high, it is reasonable considering the train is assumed to be undergoing constant acceleration. to solve for acceleration: a1 = 2 2 vf −vi 2∆x It is important to remember that vf is the velocity of our train at the end of our ∆x, 3.5 miles, which is not the point where the pass. Substituting this into our time equation: t= 2v2 2∆x 2 2 1 vf −vi or, knowing that t= , vi is zero for our train and clearing up the ambiguity of using vf : 4vi ∆x 2 v1f To fi...
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## This document was uploaded on 02/13/2014.

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