Question 127779
The way I do this is to put a stopwatch on them
both at the moment Dick takes off and Ernesto
is already flying.
When Dick takes off, Ernesto has already gone
(distance = rate x time) {{{60*1 = 60}}}miles
Now start the stopwatch and let {{{t}}}=
the time elapsed until they meet. Note that
they are now {{{280 - 60 = 220}}}miles apart
{{{50t + 60t = 280 - 60}}}
{{{110t = 220}}}
{{{t = 2}}}hrs
Ernesto has been in the air for 1 hour, so
{{{1 + 2 = 3}}} and
It takes him 3 hours to meet Dick