# Prove the postulate below?

## The drivers of a train moving at a speed ${v}_{1}$ sight another train at a distance $d$ ahead of them on the same track, moving with slower speed ${v}_{2}$. 1st driver pulles the brake to create a retardation $r$. Show that if $d > {\left({v}_{1} - {v}_{2}\right)}^{2} / \left(2 r\right)$, there will be no collision.

May 7, 2017

Using the equation of motion: $s = {v}_{i} t + \frac{1}{2} a {t}^{2}$

Using the position of the first train at $t = 0$ as the Origin, the respective displacement of the trains at time $t$ is:

${p}_{1} \left(t\right) = {v}_{1} t - \frac{1}{2} r {t}^{2}$

${p}_{2} \left(t\right) = d + {v}_{2} t$

${p}_{1} \left(t\right) = {p}_{2} \left(t\right) \implies {v}_{1} t - \frac{1}{2} r {t}^{2} = d + {v}_{2} t$

$\frac{r}{2} {t}^{2} + \left({v}_{2} - {v}_{1}\right) t + d = 0$
In the quadratic equation, we force the discriminant (ie ${b}^{2} - 4 a c$) to be negative to ensure there are no real solutions,
${b}^{2} < 4 a c \implies {\left({v}_{2} - {v}_{1}\right)}^{2} < 4 \frac{r}{2} d$
$\implies d > {\left({v}_{2} - {v}_{1}\right)}^{2} / \left(2 r\right)$