问火车紧急刹车的车速为v(t)=10-t+108/t+2,则刹车后行驶的距离约为

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解：v(t)=10-t+108/(t+2)=0得，t=16(-8舍去）S=∫（0---16）[10-t+108/(t+2)]dt=10t-1/2t²+108ln(t+2)|0---8 =80-32+108ln10-108ln2 =48+108ln5