Question

甲、乙、两辆车行驶在平行公路上,车速分别为6m/s,8m/s,.当甲乙丙三车依次相距5m时

Answer 1

设乙的加速度为a2

依题意知：

乙刹车后位移S2+5米≤甲刹车后位移S1

由Vt=V1-at=0得

t=V1/a1=6m/s÷1m/s²=6s

∴S1=V1t-1/2a1t²=6m/s×6s-1/2×1m/s²×6²=18m

S2=V2t-1/2a2t²=8m/s×6s-1/2×a2×6²=-18a2+36

∴-18a2+36+5≤18

解得a2≤23/18m/s²

∴乙车做减速运动的加速度至少是23/18m/s²