Answer
The back of the seventh car passes after 13.2 seconds.
Work Step by Step
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.
