## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Let x be the length of one train car; $x = \frac{1}{2}at^2$ $x = \frac{1}{2}a(5.0~s)^2$ $x = (12.5~s^2)~a$ We can find the time for the train to travel a distance of $7x$; $7x = \frac{1}{2}at^2$ $(7)(12.5~s^2)~a = \frac{1}{2}at^2$ $175~s^2 = t^2$ $t = 13.2~s$ The back of the seventh car passes after 13.2 seconds.