Converting the gangsters' speed to ms$^{-1}$, $108$ km/h $=108/3.6$ ms$^{-1} = 30$ ms$^{-1}$.

The distance the cop travelled in time $t$ s is $\frac12at^2 = t^2$ metres, and the distance the gangsters travelled in time $t$ s is $30t$ metres. The time when they met the second time is
$$\begin{align*}t^2 =&amp; 30t\\t=&amp;30\text{ seconds}\end{align*}$$

Then the distance they have travelled in $30$ seconds is $900$ metres, and this is before they arrive Indiana.

Solve for time, given distance and acceleration
Consider this a followup question of this one

In the classical schwarszchild solution with an eternal black hole, the user falls through the event horizon in finite local time, but this event does take place for distant observers in the infinite future. As Leonard Susskind explains, measurements of objects near the horizon are subject to large uncertainties, since all the radiation used to see them is largely red-shifted and received in increasingly large intervals in the future. This is why in some still vaguely defined sense, the effective event horizon grows, even if no matter has actually yet crossed the original event horizon (the stable event horizon after the black hole collapse period)

But when the event horizon radiates, the event horizon must recede faster than any infalling observer. Any infalling matter will probably be still stir-fried by the hawking radiation over the eons the evaporation takes (which in the observer frame is intensely blue shifted). The only matter that ever gets to be inside the formal event horizon (not the effective horizon) before the final evaporation is the matter that was already in the original collapsing star, around which the event horizon formed

Is there still any room to say otherwise? can still be argued that matter will fall in the event horizon in finite time for distant observers (faster than the evaporation will shrink the event horizon)?
