Question

（分）一汽車行駛時遇到緊急情況，駕駛員迅速正確地使用制動器在最短距離內將車停住，稱為緊急制動，設此過程中使汽車減速的阻力與汽車對地面的壓力成正比，其比例系數只與路面有關。已知該車以72km/h的速度在平直公路上行駛，緊急制動距離為25m；若在相同的路面上，該車以相同的速率在坡度（斜坡的豎直高度和水平距離之比稱為坡度）為1：10的斜坡上向下運動，則緊急制動距離將變為多少？（g=10m/s²，結果保留兩位有效數字）

Answer 1

  S₂=29m

解析:

以汽車初速度方向為正方向，設質量為m的汽車受到的阻力與其對路面壓力之間的比例系數為μ，在平直公路上緊急制動的加速度為a₁。

根據牛頓第二定律  －μmg=ma₁  ·················································································· ①

根據勻減變速直線運動規律  a₁＝ ·········································································· ②

聯立解得  μ＝＝0.8

設汽車沿坡角為θ的斜坡向下運動時，緊急制動的加速度為a₂，制動距離為S₂。

根據牛頓第二定律  mgsinθ－μmgcosθ＝ma₂······································································ ③

根據勻變速直線運動規律  S₂＝··············································································· ④

由題意知tanθ=0.1，則  sinθ＝≈0.1        cosθ＝≈1

聯立解得  S₂=29m ··································································································· ⑤

說明：若用其他方法求解，參照參考答案給分。

；28.57m