Question

Andre and Morgan are both headed to Florida for vacation. Andre will drive 50 miles per hour. Morgan will drive 45 miles per hour but will leave 2 hours earlier than Andre. How many hours will Morgan have been driving before Andre catches up with her?

Answer 1

Let 't' be the time at which Andre catches up with Morgan.

We know by definition of distance that:

$d=s\cdot t$

where 's' is the speed and 't' is the time.

In this case, the distance that Morgan travels in time t is:

$d=45t$

Since Andre starts 2 hours late, we have that in his case, the expression would be:

$d=50(t-2)$

Equating both expressions and solving for t, we get the following:

$\begin{gathered} 45t=50(t-2) \\ \Rightarrow45t=50t-100 \\ \Rightarrow45t-50t=-100 \\ \Rightarrow-5t=-100 \\ \Rightarrow t=\frac{-100}{-5}=20 \\ t=20 \end{gathered}$

therefore, Morgan would have driven for 20 hours before Andre catches up to him