簡單phy-0-

2007-01-28 1:21 am
一輛汽車以 30km h-1 的速度行。司機踏下掣動器,汽車向前行了12m後停下。
若車初速為 60km h-1 ,停車距離是多少??
更新1:

用錯formula -=-用左 s=[(u+v)/2]*t

回答 (1)

2007-01-28 2:05 am
✔ 最佳答案
u= 30 km/h = 8.333 m/s, v=0 m/s, s = 12m, a=?

use v^2 = u^2 + 2a.s
0 = 8.333^2 + 2a(12)
giving a = -2.893 m/s2

Now, u = 60 km/h= 16.667 m/s, v = 0 m/s, a=2.893 m/s2, s=?
use again v^2 = u^2+2.a.s
0 = 16.667^2 +2.(-2.893)s
solve for s gives s = 48 m


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