31-32 Train in an elevated track and car in perpendicular road

$s = \sqrt{x^2 + 900}$
 

$s = \sqrt{(1300t^2 - 2400t + 1600) + 900}$

$s = \sqrt{1300t^2 - 2400t + 2500}$

$\dfrac{ds}{dt} = \dfrac{2600t - 2400}{2\sqrt{1300t^2 - 2400t + 2500}}$

$\dfrac{ds}{dt} = \dfrac{1300t - 1200}{\sqrt{1300t^2 - 2400t + 2500}}$
 

after 1 sec, t = 1
$\dfrac{ds}{dt} = \dfrac{1300(1) - 1200}{\sqrt{1300(1^2) - 2400(1) + 2500}}$

$\dfrac{ds}{dt} = 2.67 \, \text{ ft/sec}$           answer

the train and the car are nearest together if ds/dt = 0
$\dfrac{1300t - 1200}{\sqrt{1300t^2 - 2400t + 2500}} = 0$

$1300t - 1200 = 0$

$t = 12/13 \,\, \text{ sec}$

$t = 0.923 \, \text{ sec}$           answer