52-53 Two cars traveling from the same point but going to different directions

Problem 52
A car drives south at 20 mi/hr. Another car, starting from the same point at the same time and traveling 40 mi/hr, goes east for 30 minutes then turns north. Find the rate of rotation of the line joining the cars (a) 1 hour after the start; (b) at the time the second car makes its turn.

Solution 52

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A car drives north, another car drives south

$\tan \theta = \dfrac{20t + 40(t - 0.5)}{20}$

$\tan \theta = t + 2(t - 0.5)$

$\tan \theta = 3t - 1$

$\theta = \arctan (3t - 1)$

$\dfrac{d\theta}{dt} = \dfrac{3}{1 + (3t - 1)^2}$

(a) 1 hour after the start, t = 1
$\dfrac{d\theta}{dt} = \dfrac{3}{1 + [ \, 3(1) - 1 \, ]^2}$

$\dfrac{d\theta}{dt} = \dfrac{3}{1 + 4}$

$\dfrac{d\theta}{dt} = 0.6 \, \text{ rad/hr}$ answer

(b) At the time the second car makes its turn, t = 0.5
$\dfrac{d\theta}{dt} = \dfrac{3}{1 + [ \, 3(0.5) - 1 \, ]^2}$

$\dfrac{d\theta}{dt} = \dfrac{3}{1 + 0.25}$

$\dfrac{d\theta}{dt} = 2.4 \, \text{ rad/hr }$ answer

Problem 53
Prove that the results in Problem 52 are independent of the speed of the cars, if the second car travels twice as fast as the first car.

Solution 53

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Let r = rate or speed of first car
2r = rate or speed of the second car

$\tan \theta = \dfrac{rt + 2r(t - 0.5)}{0.5(2r}$

$\tan \theta = t + 2(t - 0.5)$

$\tan \theta = 3t - 1$

$\theta = \arctan (3t - 1)$

The angle θ above is independent from rate of cars, r. answer

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