so hmm recall your d = rt, or distance = rate * time
so.. .first train leaves at 2pm, and the second one leaves 2 hours later
at 6pm, the faster train overtakes the first train
alrite... now, overtaking, means, they meet first, then the faster one keeps on rolling
now, when they meet, the distance they've covered is "d"
though the train leaving at 4pm has been going for 2hrs, it covered "d" distance
whilst the first train who left at 2pm, has covered also the same "d" distance
now, we know the faster train is 48 miles per hour faster
so, if the first train has a speed rate of "r", the second one has a speed rate of " r + 48 "
thus [tex]\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{first train}&d&r&4\\
\textit{second train}&d&r+48&2
\end{array}\\\\
-----------------------------\\\\
\begin{cases}
d=4r\\
d=2(r+48)\\
-------\\
4r=2(r+48)
\end{cases}[/tex]
solve for "r"
