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

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

$\tan \theta = 3t - 1$

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

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

(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

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

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

$\tan \theta = 3t - 1$

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

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