Answer
(a) Yes, the cows nearby will witness a collision at $t=22.5s$.
(b) The collison occurs at $538m$.
(c) See the sketch below.
Work Step by Step
Let the origin of our system be at the initial position of the front of the passenger train when the brakes are applied.
Let
$x_{p,i}$=initial position of the front of the passenger train=$0$
$x_{p,f}$=final position of the front of the passenger train at time t
$x_{f,i}$=initial position of the caboose of the freight train=$200m$
$x_{f,f}$=final position of the caboose of the freight train at time t
$v_{p,i}$=initial velocity of the passenger train=$25.0\frac{m}{s}$
$v_{f,i}$=initial velocity of the caboose of the freight train=$15.0\frac{m}{s}$
$a_{p}$=acceleration of the passenger train=$-0.100\frac{m}{s^2}$
$a_{f}$=acceleration of the freight train=$0$
Since the motion is uniform, we can apply the kinematics equation for constant acceleration.
For the two trains to collide, the front of the passenger train must be at the same position as the caboose of the freight train at some time t ,i.e,
$x_{p,f}=x_{f,f}$
$=>x_{p,i}+v_{p,i}t+\frac{1}{2}a_{p}t^2=x_{f,i}+v_{f,i}t+\frac{1}{2}a_{f}t^2$ [kinematics equation]
$=>25.0t-\frac{1}{2}\times0.100t^2=200+15.0t$
$=>\frac{0.100}{2}t^2-10.0t+200=0$
Using the quadratic formula, we get
$t=\frac{10.0+-\sqrt ((-10.0)^2-4\times\frac{0.100}{2}\times200)}{2\times\frac{0.100}{2}}=177.5s$ or $22.54s$
(a) Yes, the cows nearby will witness a collision at $t=22.5s$
(b) Now, we calculate the distance the front of the passenger train travels before collision.
$x_{p,f}=x_{p,i}+v_{p,i}t+\frac{1}{2}a_{p}t^2=25.0\times22.54-\frac{0.100}{2}\times22.54^2=538m$
The passenger train travels a distance of $538m$ after the brakes were applied from the origin of our system before collison ,i.e, the collison occurs at $538m$.
(c) The sketch of the position of the front of the passenger train and the back of the freight train i.e, $x$ vs $t$ is shown below.
