#### Answer

(a) The shortest stopping distance is 52.5 meters.
(b) We should drive at a speed of 16.0 m/s.

#### Work Step by Step

(a) We can find the magnitude of deceleration.
$F_f = ma$
$mg~\mu_k = ma$
$a = g~\mu_k = (9.80~m/s^2)(0.80)$
$a = 7.84~m/s^2$
We can find the stopping distance.
$x = \frac{v^2-v_0^2}{2a} = \frac{0-(28.7~m/s)^2}{(2)(-7.84~m/s^2)}$
$x = 52.5~m$
The shortest stopping distance is 52.5 meters.
(b) We can find the magnitude of deceleration.
$F_f = ma$
$mg~\mu_k = ma$
$a = g~\mu_k = (9.80~m/s^2)(0.25)$
$a = 2.45~m/s^2$
We can find the initial speed $v_0$.
$v^2 = v_0^2 +2ax$
$v_0^2 = 0-2ax = -2ax$
$v_0 = \sqrt{-(2)(-2.45~m/s^2)(52.5~m)}$
$v_0 = 16.0~m/s$
We should drive at a speed of 16.0 m/s.