19-21 Two cars driving in parallel roads

Problem 19
One city A, is 30 mi north and 55 mi east of another city, B. At noon, a car starts west from A at 40 mi/hr, at 12:10 PM, another car starts east from B at 60 mi/hr. Find, in two ways, when the cars will be nearest together.

Solution 19

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1st Solution (Specific):
The figure to the right shows the position of the cars when they are nearest to each other.

$40t + 60 \left( t - \dfrac{10}{60} \right) = 55$

$100t = 65$

$t = 0.65 \,\, \text{ hr }$

$t = 39 \,\, \text{ min }$

Time: 12:39 PM answer

2nd Solution (General):
From the figure shown in the right:
$s = \sqrt{x^2 + 30^2}$

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

where:
$x = 55 - 40t - 60(t - \frac{10}{60})$

$x = 65 - 100t$

Thus,
$s = \sqrt{(65 - 100t)^2 + 900}$

$\dfrac{ds}{dt} = \dfrac{2(65 - 100t)(-100)}{2\sqrt{(65 - 100t)^2 + 900}}$

$\dfrac{ds}{dt} = \dfrac{-100(65 - 100t)}{\sqrt{(65 - 100t)^2 + 900}}$

when ds/dt = 0
$\dfrac{-100(65 - 100t)}{\sqrt{(65 - 100t)^2 + 900}} = 0$

$-100(65 - 100t) = 0$

$100t = 65$

$t = 0.65 \,\, \text{ hr }$

$t = 39 \,\, \text{ min }$

Time: 12:39 PM answer

Problem 20
For the condition of Problem 19, draw the appropriate figures for times before 12:39 PM and after that time. Show that in terms of time after noon, the formula for distance between the two cars (one formula associated with each figure) are equivalent.

Solution 20

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For time before 12:39 PM, see the figure in the general solution of Solution 20.
$x = 65 - 100t$

$s = \sqrt{x^2 + 30^2} = \sqrt{(65 - 100t)^2 + 900}$

For time after 12:39 PM, there are three conditions that worth noting. Each are thoroughly illustrated below.

First condition: (after 12:39 PM but before 1:05 PM)

$[60(t - 10/60) - x] + 40t = 55$

$x = 60t - 10 + 40t - 55$

$x = -(65 - 100t)$

$x^2 = (65 - 100t)^2$

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

$s = \sqrt{(65 - 100t)^2 + 900}$ okay

Second condition: (after 1:05 PM but before 1:22:30 PM)

$[60(t - 10/60) - x] + 40t = 55$

$x = 60t - 10 + 40t - 55$

$x = -(65 - 100t)$

$x^2 = (65 - 100t)^2$

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

$s = \sqrt{(65 - 100t)^2 + 900}$ okay

Third condition: (after 1:22:30 PM)

$40t - [x - 60(t - 10/60) ] = 55$

$x = 55 - 40t - 60t + 10$

$x = 65 - 100t$

$x^2 = (65 - 100t)^2$

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

$s = \sqrt{(65 - 100t)^2 + 900}$ okay

Problem 21
For Problem 19, compute the time-rate of change of the distance between the cars at (a) 12:15 PM; (b) 12:30 PM; (c) 1:15 PM

Solution 21

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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

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