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

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

$\tan \theta = 3t - 1$

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

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

(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

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

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

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

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