Question 1078342
Let time start when the second plane takes off.
At that point the first plane is traveling at ({{{0}}},{{{100+200t}}}) where {{{t}}} is the time in hours from 0630AM.
The second plane is traveling at ({{{170t}}},{{{0}}}). 
The first component of the distance function is east positive, west negative.
The second component of the distance function is north positive, south negative.
We can calculate the distance between the first and second plane using the distance formula,
{{{D^2=(170t-0)^2+(0-(100+200t))^2}}}
{{{D^2=28900t^2+40000t^2+40000t+10000}}}
{{{D^2=68900t^2+40000t+10000}}}
Find t when D=500,
{{{500^2=68900t^2+40000t+10000}}}
{{{250000=68900t^2+40000t+10000}}}
{{{68900t^2+40000t-240000=0}}}
{{{689t^2+400t-2400=0}}}
Using the quadratic formula,
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-400 +- sqrt( 400^2-4*689*(-2400) ))/(2*689) }}}
{{{x = (-400 +- 40sqrt(4234))/(2*689) }}}
{{{x = (-200 +- 20sqrt(4234))/(689) }}}
Only the positive solution makes sense here,
{{{x = (-200 + 20sqrt(4234))/(689) }}}
or approximately,
{{{x=1.5985}}}{{{hrs}}}
1 hr 36 minutes from 0630AM which would be 0806AM.