28-29 Time Rates: Two cars driving on roads that intersects at 60 degree

Problem 28
At noon a car drives from A (Fig. 48) toward C at 60 miles per hour. Another car starting from B at the same time drives toward A at 30 miles per hour. If AB = 42 miles, find when the cars will be nearest each other.

Solution 28

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By cosine law:
$s^2 = (60t)^2 + (42 - 30t)^2 - 2(60t)(42 - 30t) \cos 60^\circ$

$s^2 = 3600t^2 + (1764 - 2520t + 900t^2) - (2520t - 1800t^2)$

$s^2 = 6300t^2 - 5040t + 1764$

$s = \sqrt{6300t^2 - 5040t + 1764}$

$\dfrac{ds}{dt} = \dfrac{12600t - 5040}{2\sqrt{6300t^2 - 5040t + 1764}} = 0$

$12600t - 5040 = 0$

$t = 2/5 \, \text{ hr}$

$t = 24 \, \text{ min}$

Time = 12:24 PM answer

Problem 29
Solve Problem 28 if the car from B leaves at noon but the car from A leaves at 12:07 PM.

Solution 29

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By cosine law:
$s^2 = [ \, 60(t - 7/60) \, ]^2 + (42 - 30t)^2 - 2[ \, 60(t - 7/60) \, ](42 - 30t) \cos 60^\circ$

$s^2 = (3600t^2 - 840t + 49) + (1764 - 2520t + 900t^2) - (-1800t^2 + 2730t - 294)$

$s^2 = 6300t^2 - 6090t + 2107$

$s = \sqrt{6300t^2 - 6090t + 2107}$

$\dfrac{ds}{dt} = \dfrac{12600t - 6090}{2\sqrt{6300t^2 - 6090t + 2107}} = 0$

$12600t - 6090 = 0$

$t = 29/60 \, \text{ hr}$

$t = 29 \, \text{ min}$

Time: 12:29 PM answer

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