| 解:(1)由题意得: s=0.7×11+0.08×11 2 =17.38≈17.4m(8分) (2)设志愿者饮酒后的反应时间为t 1 ,则t 1 ×17+0.08×17 2 =46 t l ≈1.35s. 当v=11m/s时,s=t l ×11+0.08×11 2 =24.53. ∴24.53﹣17.38≈7.2(m) 答:刹车距离将比未饮酒时增加7.2m (3)为防止“追尾”当车速为17m/s时,刹车距离必须小于40m, ∴t×17+0.08×17 2 <40 解得t<0.993(s) 答:反应时间不超过0.99s. |