题目内容
（分）一汽车行驶时遇到紧急情况，驾驶员迅速正确地使用制动器在最短距离内将车停住，称为紧急制动，设此过程中使汽车减速的阻力与汽车对地面的压力成正比，其比例系数只与路面有关。已知该车以72km/h的速度在平直公路上行驶，紧急制动距离为25m；若在相同的路面上，该车以相同的速率在坡度（斜坡的竖直高度和水平距离之比称为坡度）为1：10的斜坡上向下运动，则紧急制动距离将变为多少？（g=10m/s2，结果保留两位有效数字）
S2=29m
解析:
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a1。
根据牛顿第二定律 －μmg=ma1 ·················································································· ①
根据匀减变速直线运动规律 a1＝ ·········································································· 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a2，制动距离为S2。
根据牛顿第二定律 mgsinθ－μmgcosθ＝ma2······································································ ③
根据匀变速直线运动规律 S2＝··············································································· ④
由题意知tanθ=0.1，则 sinθ＝≈0.1 cosθ＝≈1
联立解得 S2=29m ··································································································· ⑤
说明：若用其他方法求解，参照参考答案给分。
；28.57m
