2nd problem
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感觉相遇在中点的时候,trajectory 应该刚好是a quarter sphere.
是不是啊。
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这里好冷清,再出两个数学/物理题,活跃一下气氛 大树下   (455 bytes , 947reads )
我的解法-第二题 大树下   (1061 bytes , 412reads )
我的解法-第一题 大树下   (1157 bytes , 368reads )
so any hint to Q1? 1300cc   (0 bytes , 228reads )
fianlly got it. only A contribute to the progress between A,B. 1300cc   (12 bytes , 238reads )
第一题,正方形的话爬了1米,正三角形的话是2/3 shwan   (77 bytes , 400reads )
cant figure out q1. 1300cc   (6 bytes , 260reads )
第一题你是对的 大树下   (87 bytes , 391reads )
小虫匀加速?加速度是0.01?那就是51? shwan   (0 bytes , 309reads )
2nd problem 1300cc   (71 bytes , 423reads )
... 大树下   (60 bytes , 318reads )
i mean 1D. a quarter circle maybe is preferred. 1300cc   (15 bytes , 263reads )
老兄你在乱猜是吧 大树下   (16 bytes , 298reads )
the trajectory of each worm should be a whirling line. (my guess) 1300cc   (224 bytes , 270reads )
1。永远不能相遇,或者题目给的不对 1300cc   (438 bytes , 468reads )
第二题:条件足够了 大树下   (980 bytes , 479reads )
think in a 2nd way 1300cc   (343 bytes , 506reads )
I don't know how do you derive that and what is n stand for 大树下   (346 bytes , 374reads )
while. we need formula to prove that. think e.g the worm speed is a function 1300cc   (135 bytes , 275reads )
speed converge to v1+v2 大树下   (880 bytes , 400reads )
i solve the integration wrongly 1300cc   (125 bytes , 226reads )
actually 1300cc   (483 bytes , 259reads )
approaching v1 + v2. 大树下   (0 bytes , 324reads )
so it will never catch up. 1300cc   (153 bytes , 257reads )
第一题:如果你觉得永远不能相遇请给出理由 大树下   (151 bytes , 340reads )
你确定? 掬水掇月   (121 bytes , 332reads )
上贴删除。。。 掬水掇月   (56 bytes , 310reads )
很久以前都做过了.. hash   (0 bytes , 266reads )
这两道题都有好几种解决方法:有些简单,有些复杂 大树下   (92 bytes , 254reads )