# 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**

$s = \sqrt{x^2 + 30^2}$

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

x

^{2}= (20t)

^{2}+ (40 - 30t)

^{2}

x

^{2}= 400t

^{2}+ 1600 - 2400t + 900t

^{2}

x

^{2}= 1300t

^{2}- 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**

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

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*