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

$\tan \theta = 3t - 1$

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

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

(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

$\dfrac{d\theta}{dt} = \dfrac{3}{1 + [ \, 3(0.5) - 1 \, ]^2}$

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

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

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