Question 389493
(a)
Set {{{t = 0}}}
(4,3) and (3,-5)
(b)
For velocity vectors:
hiker:
{{{sqrt(3^2 + 2^2)}}} at 180 degrees + arc tan(2/3)
{{{sqrt(13)}}} at {{{33.69}}} degrees
rescue party:
{{{sqrt(1^2 + 4^2)}}} at 180 degrees - arc tan(4)
{{{sqrt(5)}}} at {{{104.04}}} degrees
(c)
I need a general formula for the distance between them
It would be square root of(x2 - x1)^2 + (y2 - y1)^2)
{{{sqrt((3 - t - 4 + 3t)^2 + (-5 + 4t - 3 + 2t)^2)}}}
{{{sqrt((-1 + 2t)^2 + (-8 + 6t)^2)}}}
The 1st term is zero when {{{t = 1/2}}}, then the distance 
would be {{{sqrt((-5)^2) = 5}}}
The 2nd term is zero when {{{t = 8/6}}}
Then distance = {{{5/3}}}
I think they come the closest in 1 hr 20 min and
they are 1.67 km apart.
This is kind of a guess, I would get another opinion, too