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Physics 高手請進
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最佳解答:
a) 1/10 = 0.1 s b) Average speed in the first interval = 0.054/0.1 = 0.54 m/s Average speed in the last interval = 0.341/0.1 = 3.41 m/s c) In each time interval, the average speed is supposed to happen at the middle instant and so, the time instant for which the 0.54 m/s and 3.41 m/s happen should be at t = 0.05 s and t = 0.35 s So the acceleration of the ball is (3.41 - 0.54)/(0.35 - 0.05) = 9.57 m/s2 d) With v = u + at for u = 0, t = 2 and a = 9.57, we have v = 19.1 m/s When t = 1 s, v = 9.57 m/s, so applying v2 - u2 = 2as with v = 19.1, u = 9.57 and a = 9.57: s = 14.35 m which is the distance covered in the last second.
其他解答:
這裏可以幫到你 http://actioni.w7.c361.net/yahoo.com.hk/hk/auction/178987536
Physics 高手請進
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圖片參考：http://imgcld.yimg.com/8/n/HA00039250/o/701004260124813873387870.jpg please help, thanks!!! please help, thanks!!! 更新: http://yfrog.com/jk2222sdj最佳解答:
a) 1/10 = 0.1 s b) Average speed in the first interval = 0.054/0.1 = 0.54 m/s Average speed in the last interval = 0.341/0.1 = 3.41 m/s c) In each time interval, the average speed is supposed to happen at the middle instant and so, the time instant for which the 0.54 m/s and 3.41 m/s happen should be at t = 0.05 s and t = 0.35 s So the acceleration of the ball is (3.41 - 0.54)/(0.35 - 0.05) = 9.57 m/s2 d) With v = u + at for u = 0, t = 2 and a = 9.57, we have v = 19.1 m/s When t = 1 s, v = 9.57 m/s, so applying v2 - u2 = 2as with v = 19.1, u = 9.57 and a = 9.57: s = 14.35 m which is the distance covered in the last second.
其他解答:
這裏可以幫到你 http://actioni.w7.c361.net/yahoo.com.hk/hk/auction/178987536
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