同上一個問題 == 現在小學老是都這樣出的唷??? 甲、乙兩車作直線運動，其速率分別為50m/s 與30m/s，若兩車以相同之減速度踩煞車而終致停止， 試求煞車距離比？ 【Ans：25：9】?
回答 (2)
2017-03-15 11:19 am
Use equation of motion: v^2 = u^2 + 2a.s
For car 1: 0 = 50^2 + 2a(s1)
i.e. s1 = -50^2/2a
For car 2: 0 = 30^2 + 2a(s2)
i.e. s2 = -30^2/2a
where s1 and s2 are the braking distances for car 1 and car 2 respectively, and a is their common deceleration.
Hence, s1/s2 = (50/30)^2 = 25/9
or s1:s2 = 25:9
For car 1: 0 = 50^2 + 2a(s1)
i.e. s1 = -50^2/2a
For car 2: 0 = 30^2 + 2a(s2)
i.e. s2 = -30^2/2a
where s1 and s2 are the braking distances for car 1 and car 2 respectively, and a is their common deceleration.
Hence, s1/s2 = (50/30)^2 = 25/9
or s1:s2 = 25:9
2017-03-14 1:43 pm
Sol
設甲車之減速度為xm/s，乙車之減速度為ym/s
時間=tSec
50/x=30/y
30x=50y
3x=5y=15p
x=5p，y=3p
甲車煞車距離=(1/2)(xt)^2
乙車煞車距離=(1/2)(yt)^2
甲車煞車距離：乙車煞車距離
=x^2：y^2
=25p^2：9p^2
=25：9
設甲車之減速度為xm/s，乙車之減速度為ym/s
時間=tSec
50/x=30/y
30x=50y
3x=5y=15p
x=5p，y=3p
甲車煞車距離=(1/2)(xt)^2
乙車煞車距離=(1/2)(yt)^2
甲車煞車距離：乙車煞車距離
=x^2：y^2
=25p^2：9p^2
=25：9
收錄日期: 2021-04-11 21:36:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170313153129AAW7azL