S 2=29m
以汽车初速度方向为正方向，设质量为 m 的汽车受到的阻力与其对路面压力之间的比例系数为 μ ，在平直公路上紧急制动的加速度为 a 1。
根据牛顿第二定律 － μmg = ma 1·················································································· ①
根据匀减变速直线运动规律 a 1＝ ·········································································· ②
联立解得 μ ＝＝0.8
设汽车沿坡角为 θ 的斜坡向下运动时，紧急制动的加速度为 a 2，制动距离为 S 2。
根据牛顿第二定律 mg sin θ － μmg cos θ ＝ ma 2······································································ ③
根据匀变速直线运动规律 S 2＝··············································································· ④
由题意知tan θ =0.1，则 sin θ ＝≈0.1 cos θ ＝≈1
联立解得 S 2="29m" ··································································································· ⑤
说明：若用其他方法求解，参照参考答案给分。
；28.57m