Answer
The speed of the last car will be 21 m/s.
Work Step by Step
$v_0 = 0 ~m/s$
$x_0 = 0 ~m$
To find the acceleration, we can use this equation:
$v^2 = v_0^2 + 2a(x-x_0)$
$a = \frac{v^2}{2x} = \frac{18^2}{2\times180} = 0.9 ~m/s^2$
We need to find the speed of the train when the front of the train has travelled a distance of 255 meters.
$v^2 = v_0^2 + 2a(x-x_0) = 2\times0.9\times255 = 459$
$v = \sqrt{459} = 21 ~m/s$
Note that the answer is rounded off to two significant digits.
