運動員以10ms^-1沿筆直跑道跑步。突然有人走到他前方25m處
若他在那人面前停下,他的減速度最少是多少???
設運動員反應時間是0.2s。
Answer：2.17ms^-2
計極個Answer都係2
[力學] 運動2一問
2007-02-27 11:48 pm
回答 (4)
2007-02-28 1:51 am
✔ 最佳答案
佢反應時間是0.2s所以佢開始減速ge距離只係25-(10x0.2)=23m 因為有0.2s時間冇減速, 仲係10ms^-1
唔知時間, 所以用v^2-u^2=2as
開始速度係10ms^-1, 最後速度係 0ms^-1, 距離係23m
So, 0^2-10^2=2(23)a
-100=46a
a=-100/46
a=2.173913043
a=2.17ms^-2 cor. to 3 sig.fig.
因為答案係減速度, so唔駛負號, ans係2.17ms^-2
參考: 自己
2007-03-01 2:30 am
好答案
2007-02-28 12:05 am
The distance traveled in the reaction time is 0.2 X 10 = 2m. There are 23m to go.
By v^s = u^2 + 2as
0= 100 + 46a
So, Decelaration is 2.17ms-2
By v^s = u^2 + 2as
0= 100 + 46a
So, Decelaration is 2.17ms-2
2007-02-28 12:00 am
distance between the runner and the man when the runner started to decellerate:
25-10*0.2=23m
decelleration of the runner:
v*v-u*u=2as
0-(10)(10)=2a(23)
a= -2.174
25-10*0.2=23m
decelleration of the runner:
v*v-u*u=2as
0-(10)(10)=2a(23)
a= -2.174
參考: myselve
收錄日期: 2021-04-12 20:19:54
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