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*