一名運動員參加了一項100m短跑比賽。如圖(請看意見欄)為他的速度時間關係圖。於時間t=0,比賽開始。於t=T時,運動員衝過終點。運動員加速時的加速度為2.16m/s^2及他完成比賽的時間為14.65 s。
如果該運動員下次比賽時的加速度和最大速度均提高5%,如果運動員的反應時間依然是0.3s,問他在100m短跑比賽中會快多少秒?
物理:問他在100m短跑比賽中會快多少秒
2013-12-30 2:15 am
回答 (2)
2013-12-30 4:57 am
✔ 最佳答案
New acceleration = (1+5/100) x 2.16 m/s^2 = 2.268 m/s^2New maximum speed = (1+5/100) x 8 m/s = 8.4 m/s
Time needed to reach maximum speed
= 8.4/2.268 s = 3.704 s
Distance travelled during this time
= mean speed x time taken = (8.4/2) x 3.704 m = 15.56 m
Remaining distance to go = (100 - 15.56) m = 84.44 m
Time needed to cover this distance = 84.44/8.4 s = 10.05 s
Therefore, total time taken
= (0.3 + 3.704 + 10.05) s = 14.06 s
Difference in time = (14.65 - 14.06) s = 0.59 s
2013-12-30 2:15 am
圖:photo.pchome.com.tw/forever2015/138831204559
收錄日期: 2021-04-11 20:14:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131229000051KK00139