[~~ Phy] Kinematics

[匿名]

2007-01-08 7:35 pm

Motor car P travelling at 30km/h on straight road is overtaken by another car Q at 40 km/h. P immediately accelerates after travelling a distance of 0.5 km and overtakes second car Q which has kept its speed constant.

a.) Find the time taken.
b.) Find the greatest distance between the cars at this time.

回答 (1)

用戶4366

2007-01-09 7:17 am

✔ 最佳答案

因為上次答了你的問題之後，你取消了題目，白費了我的心機和時間，這次，我雖然已經計出了答案，但沒有興趣再打出來給你了。

2007-01-08 23:22:22 補充：
a) 45sb) 250m

2007-01-11 18:55:46 補充：
30 km/h = 30000/3600 = 8.33 m/s40 km/h = 40000/3600 = 11.11 m/s500 m = 11.11tt = 45 sS=ut + 0.5at^2500 = (8.33)(45) + 0.5a(45)^2 gives a = 0.1236 m/s^2

2007-01-11 18:56:29 補充：
Let when t=0, position of car P = Xp and position of car Q = XqXp = 8.33t (0.5)(0.1236)t^2Xq = 11.11t

2007-01-11 18:56:57 補充：
Difference = Xq-Xp = 2.78t - 0.0618t^2Since this is a quadratic equaton, the optium point must at the mid point of roots, that is22.5s,Xq-Xp|max = 2.78(22.5) - (0.0618)(22.5)^2 = 31.26m

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