## Solution:

### Part A

We are given

$v_0=22.0\:m/s$

$a=-0.150\:m/s^2$

$x=210\:m$

We are required to solve for time, t. So we have

$x=v_0t+\frac{1}{2}at^2$

$\left(210\:m\right)=\left(22.0\:m/s\right)t+\frac{1}{2}\left(-0.150\:m/s^2\right)t^2$

If we rearrange the terms into a general quadratic equation, we have

$0.075t^2-22t+210=0$

Solve for t using the quadratic formula. We are given $a=0.075;\:b=-22t;\:c=210$

$\displaystyle t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle t=\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\left(0.075\right)\left(210\right)}}{2\left(0.075\right)}$

$t=9.88\:s\:\text{and}\:t=283.46\:s$

Discard t=283.46 s as it can be seen from the problem that this is not a feasible solution. So, we have

$t=9.88\:s$

### Part B

We have the same given values from part a. So we have

$v^2=v_0^2+2ax$

$v=\sqrt{v_0^2+2ax}$

$v=\sqrt{\left(22.0\:m/s\right)^2+2\left(-0.150\:m/s^2\right)\left(210\:m\right)}$

$v=20.6\:m/s$

### Part C

We are given the same values as in the previous parts, except that for the value of x since we should incorporate the length of the train. For the distance, x, we have

$x=210\:m+130\:m$

$x=340\:m$

Therefore, we have

$x=v_ot+\frac{1}{2}at^2$

You can solve this problem by the method presented in part A. But we can also rearrange the equation to solve for t, using also quadratic equation. We have

$\displaystyle t=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}$

$\displaystyle t=\frac{-\left(22.0\:m/s\right)\pm \sqrt{\left(22.0\:m/s\right)^2+2\left(-0.150\:m/s^2\right)\left(340\:m\right)}}{-0.150\:m/s^2}$

$t=16.4\:s$

### Part D

We have the same given values.

$v^2=v_0^2+2ax$

$v=\sqrt{\left(22.0\:m/s\right)^2+2\left(-0.150\:m/s^2\right)\left(340\:m\right)}$

$v=19.5\:m/s$