Answer
$460.3\space m$
Work Step by Step
Please see the attached image first.
First of all, let's convert $68.5\space km/h$ into $m/s$, by using the following simple conversion factors which we already know.
$$68.5\space km/h=(\frac{68.5\space km}{h})(\frac{1000\space m}{km})(\frac{h}{3600\space s})=19\space m/s$$
Now, let's apply equation $S=\frac{(U+V)}{2}t$ to train 1 as follows.
$\rightarrow S=\frac{(U+V)}{2}t$
Now plug the known values into this equation.
$S=\frac{(19\space m/s+0)}{2}\times48.3\space s$
$S=458.85\space m$
Distance between trains at the = S+1.45 m = 458.85 m+ 1.45 m = 460.3 m
beginning to brake
