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共回答了17个问题采纳率:100% 举报
1)
就是相遇问题(追及)
设t秒后摩托车追上公共汽车
则,在t秒内,公共汽车匀速运动,行驶S1 = vt0 + vt = 8 + 4t
摩托车匀加速运动,行驶S2 = at²/2 = 3 × t²/2
t秒时刚好追上,即S1 = S2
8 + 4t = 3 × t²/2
t = 4 s
2)
S2 = at²/2 = 3 × t²/2 = 3×4²/2 m = 24m
(也可以S1 = 8+4×4m = 24m)
3)
当摩托车速度小于汽车速度时,两车距离一直增加
故,当两车速度相同时,两车距离最短
摩托车速度V = at
at = 4 t = 4/3 s
此时S = 8+4×4/3 m = 40/3m
1年前
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