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

Problem 31
An elevated train on a track 30 ft above the ground crosses a street at the rate of 20 ft/sec at the instant that a car, approaching at the rate of 30 ft/sec, is 40 ft up the street. Find how fast the train and the car separating 1 second later.

Solution 31

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From the isometric box:
$s = \sqrt{x^2 + 30^2}$

$s = \sqrt{x^2 + 900}$

where:
x² = (20t)² + (40 - 30t)²
x² = 400t² + 1600 - 2400t + 900t²
x² = 1300t² - 2400t + 1600

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

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

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

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

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

Problem 32
In Problem 31, find when the train and the car are nearest together.

Solution 32

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From Solution 31,
$\dfrac{ds}{dt} = \dfrac{1300t - 1200}{\sqrt{1300t^2 - 2400t + 2500}}$

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

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