偉傑以1.4m/s的平均速率步行5km，再以平均速率v步行10km。若塔在整個旅程的平均速率是2m/s,v是多少?
回答 (2)
2016-10-29 2:48 am
可是他的答案顯示是=1.12m/s
2016-10-28 11:43 pm
Sol
T1=5km/(1.4m/s)=5000m/(1.4m/s)=(25000/7)s
T2=10km/(vm/s)=10000s/v
2m/s=(5km+10km)/(T1+T2)
2m/s=15000m/[(25000/7)s+10000s/v]
2=15000/[(25000/7)+10000/v]
2=15000*7v/(25000v+70000)
2=15*7v/(25v+70)
50v+140=105v
55v=140
v=140/55=28/11
T1=5km/(1.4m/s)=5000m/(1.4m/s)=(25000/7)s
T2=10km/(vm/s)=10000s/v
2m/s=(5km+10km)/(T1+T2)
2m/s=15000m/[(25000/7)s+10000s/v]
2=15000/[(25000/7)+10000/v]
2=15000*7v/(25000v+70000)
2=15*7v/(25v+70)
50v+140=105v
55v=140
v=140/55=28/11
收錄日期: 2021-04-30 21:54:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161028091824AAabunL