Question 280099
Train A has a 15 minute head start over B if you consider
the station as the starting point. In that time, A has gone
{{{.25*60 = 15}}} miles.
Assume I have a stopwatch. I'll start it when B reaches the station
If I stop it when B catches A, both trains will have been moving
for the same length of time, which I'll call {{{t}}}
given:
{{{v[a] = 60}}} mi/hr
{{{v[b] = 80}}} mi/hr
{{{d[b] = d[a] + 15}}} mi
----------------
Now I can write equations for both trains
{{{d[a] = 60t}}}
{{{d[a] + 15 = 80t}}}
By substitution,
{{{60t + 15 = 80t}}}
{{{20t = 15}}}
{{{t = 3/4}}} hr
Train B will catch train A in 45 min, or at 8:25 + 45 = 9:05
check answer:
{{{d[a] = 60t}}}
{{{d[a] = 60*(3/4)}}}
{{{d[a] = 45}}} mi
and
{{{d[a] + 15 = 80t}}}
{{{d[a] + 15 = 80*(3/4)}}}
{{{d[a] + 15 = 60}}}
{{{d[a] = 45}}} mi
OK