Solution 23 above shows that the distance s at any time after noon is given by

$s = \sqrt{(45 - 60t)^2 + (30 + 40t)^2}$

$\dfrac{ds}{dt} = \dfrac{2(45 - 60t)(-60) + 2(30 + 40t)(40)}{2\sqrt{(45 - 60t)^2 + (30 + 40t)^2}}$ *See Solution 22*

$\dfrac{ds}{dt} = \dfrac{-60(45 - 60t) + 40(30 + 40t)}{\sqrt{(45 - 60t)^2 + (30 + 40t)^2}}$

$\dfrac{ds}{dt} = \dfrac{-2700 + 3600t + 1200 + 1600t}{\sqrt{(45 - 60t)^2 + (30 + 40t)^2}}$

$\dfrac{ds}{dt} = \dfrac{5200t - 1500}{\sqrt{(45 - 60t)^2 + (30 + 40t)^2}}$

(a) at 12:15 PM, t = 15/60 = 0.25 hr

$\dfrac{ds}{dt} = \dfrac{5200(0.25) - 1500}{\sqrt{[45 - 60(0.25)]^2 + [30 + 40(0.25)]^2}}$

$\dfrac{ds}{dt} = -4 \, \text{ mi/hr}$ *answer*

(b) at 12:45 PM, t = 45/60 = 0.75 hr

$\dfrac{ds}{dt} = \dfrac{5200(0.75) - 1500}{\sqrt{[45 - 60(0.75)]^2 + [30 + 40(0.75)]^2}}$

$\dfrac{ds}{dt} = 40 \, \text{ mi/hr}$ *answer*