From Solution 20,
$s = \sqrt{(65 - 100t)^2 + 900}$ at any time after noon.
From Solution 19,
$\dfrac{ds}{dt} = \dfrac{-100(65 - 100t)}{\sqrt{(65 - 100t)^2 + 900}}$
(a) at 12:15 PM, t = 15/60 = 0.25 hr
$\dfrac{ds}{dt} = \dfrac{-100[ \, 65 - 100(0.25) \, ]}{\sqrt{[ \, 65 - 100(0.25) \, ]^2 + 900}}$
$\dfrac{ds}{dt} = -80 \, \text{ mi/hr}$ answer
(b) at 12:30 PM, t = 30/60 = 0.5 hr
$\dfrac{ds}{dt} = \dfrac{-100[ \, 65 - 100(0.5) \, ]}{\sqrt{[ \, 65 - 100(0.5) \, ]^2 + 900}}$
$\dfrac{ds}{dt} = -44.72 \, \text{ mi/hr}$ answer
(c) at 1:15 PM, t = 1 + 15/60 = 1.25 hr
$\dfrac{ds}{dt} = \dfrac{-100[ \, 65 - 100(1.25) \, ]}{\sqrt{[ \, 65 - 100(1.25) \, ]^2 + 900}}$
$\dfrac{ds}{dt} = 89.44 \, \text{ mi/hr}$ answer