Peter hears 'the please stand back from the doors' warning when he heads for an MTR train. At this time, he is walking at 0.8 ms^-1 and is 15m away from the train. He immediately accelerates at 2.5 ms^-2 to chase but the train doors close at 3 s after the warning announcement.
Is he able to catch the train?
Physics motion
2011-11-13 12:50 am
回答 (1)
2011-11-13 1:13 am
✔ 最佳答案
Initial velocity of Peter, u = 0.8 ms^-1Acceleration of Peter, a = 2.5 ms^-2
In 3 s, he can travel,
By s = ut + 1/2 at^2
s = (0.8)(3) + 1/2 (2.5)(3)^2
= 13.65 m < 15 m
He can travel 13.65 m in 3 s, but he was 15 m away from the train.
So, he is not able to catch the train.
參考: Prof. Physics
收錄日期: 2021-04-20 11:51:16
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111112000051KK00576