Answer
(a) The time increases by a factor of two.
(b) The distance increases by a factor of four.
Work Step by Step
Let $t_1$ be the original time required to stop.
$a=\frac{speed_1}{t_1}$. Rearrange the equation:
$t_1=\frac{speed_1}{a}$. Notice that $a$ is the magnitude of the acceleration.
Let $\Delta x_1$ be the original distance to stop.
$v^2=v_0^2+2a\Delta x_1$ (equation 2-12, page 35). But, $v=v_f=0$.
Rearrange the equation:
$\Delta x_1=-\frac{v_0^2}{2a}$. If the car is moving in the positive direction ($\Delta x\gt0$ and $v_0\gt0$) and decelerating, then $a\lt0$. So, $\Delta x\gt0$. But, considering $a$ as the magnitude of the deceleration and $v_0=speed_1$:
$\Delta x_1=\frac{(speed_1)^2}{2a}$
(a) If $speed_2=2speed_1$:
$t_2=\frac{speed_2}{a}=\frac{2speed_1}{a}=2t_1$
(b) $\Delta x_2=\frac{(speed_2)^2}{2a}=\frac{(2speed_1)^2}{2a}=4\frac{(speed_1)^2}{2a}=4\Delta x_1$
