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*