By Pythagorean Theorem:
$s = \sqrt{(100 - 40t)^2 + (200 - 40t)^2}$
Set ds/dt = 0
$\dfrac{ds}{dt} = \dfrac{2(100 - 40t)(-40) + 2(200 - 40t)(-40)}{2\sqrt{(100 - 40t)^2 + (200 - 40t)^2}} = 0$
$2(100 - 40t)(-40) + 2(200 - 40t)(-40) = 0$
$(100 - 40t) + (200 - 40t) = 0$
$300 - 80t = 0$
$t = 3.75 \, \text{ hrs}$
$t = 3 \, \text{ hrs} \, 45 \, \text{ min}$
Time: 3:45 PM answer
Minimum distance will occur at t = 3.75,
$s_{min} = \sqrt{[\,100 - 40(3.75)\,]^2 + [\,200 - 40(3.75)\,]^2}$
$s_{min} = 70.71 \, \text{ miles}$ answer