37-38 How fast a ship leaving from its starting point

$s = \sqrt{100t^2 - 200t + 400}$

$\dfrac{ds}{dt} = \dfrac{200t - 200}{2\sqrt{100t^2 - 200t + 400}}$

$\dfrac{ds}{dt} = \dfrac{100t - 100}{\sqrt{100t^2 - 200t + 400}}$
 

after 6 hrs from start, t = 6 - 2 = 4 hrs
$\dfrac{ds}{dt} = \dfrac{100(4) - 100}{\sqrt{100(4^2) - 200(4) + 400}}$

$\dfrac{ds}{dt} = \dfrac{100(4) - 100}{\sqrt{100(4^2) - 200(4) + 400}}$

$\dfrac{ds}{dt} = 8.66 \, \text{ mi/hr}$           answer

$s = \sqrt{100t^2 + 200t + 400}$

$\dfrac{ds}{dt} = \dfrac{200t + 200}{2\sqrt{100t^2 + 200t + 400}}$

$\dfrac{ds}{dt} = \dfrac{100t + 100}{\sqrt{100t^2 + 200t + 400}}$
 

after 6 hrs from start, t = 6 - 2 = 4 hrs
$\dfrac{ds}{dt} = \dfrac{100(4) + 100}{\sqrt{100(4^2) + 200(4) + 400}}$

$\dfrac{ds}{dt} = 9.45 \, \text{ mi/hr}$           answer