## University Physics with Modern Physics (14th Edition)

We can break up the distance into three parts called $x_1, x_2$ and $x_3$. $x_1 = \frac{1}{2}at^2 = \frac{1}{2}(1.60~m/s^2)(14.0~s)^2$ $x_1 = 156.8~m$ If the train accelerates at a rate of $a = 1.60~m/s^2$ for a time of 14.0 s, the velocity is $v = (1.60~m/s^2)(14.0~s)$ which is $v = 22.4 ~m/s$. $x_2 = vt = (22.4~m/s)(70.0~s)$ $x_2 = 1568~m$ $x_3 = \frac{v^2-v_0^2}{2a} = \frac{0-(22.4~m/s)^2}{(2)(-3.50~m/s^2)}$ $x_3 = 71.7~m$ The total distance $d$ is the sum of $x_1, x_2,$ and $x_3$. $d = 156.8~m + 1568~m + 71.7~m = 1800~m$ The total distance is 1800 meters.