1)甲車平均速率48公里每小時,乙車平均速率36公里每小時,乙車先行半小時,請問幾小時後甲車可追及乙車?
2)甲地和乙地相距800公里,甲火車以50 公里每小時的速度由甲地駛向乙地, 而乙火車則以75公里每小時由乙地駛向甲地,若兩車同時出發,幾小時後會相遇?
2題數學題 - 速率......請幫忙! PLEASE! ( 5+5=10分)
2007-03-04 8:19 am
回答 (3)
2007-03-04 8:25 am
✔ 最佳答案
1. 36*0.5/(48-36)=18/12
=3/2
2. 800/(50+75)
=800/125
=32/5
參考: me
2007-03-04 9:05 am
1: 速率 (V) = 距離 (S) / 時間 (t)
距離 (S) = 速率 (V) x 時間 (t)
甲車: S甲 = 48 x t
乙車: S乙 = 36 x (t + 0.5hr)
S甲 = S乙
So, 48 x t = 36 x (t + 0.5hr)
48T = 36t + 36 x 0.5
48T - 36t = 18
12T = 18
t = 18 / 12
t = 1.5hr
2.
甲車: S甲 = 50 x t
乙車: S乙 = 75 x t
S甲 + S乙 = 800
So, 50t + 75t = 800
125t = 800
t = 800 / 125
t = 6.4 hr
距離 (S) = 速率 (V) x 時間 (t)
甲車: S甲 = 48 x t
乙車: S乙 = 36 x (t + 0.5hr)
S甲 = S乙
So, 48 x t = 36 x (t + 0.5hr)
48T = 36t + 36 x 0.5
48T - 36t = 18
12T = 18
t = 18 / 12
t = 1.5hr
2.
甲車: S甲 = 50 x t
乙車: S乙 = 75 x t
S甲 + S乙 = 800
So, 50t + 75t = 800
125t = 800
t = 800 / 125
t = 6.4 hr
參考: Me
2007-03-04 8:30 am
1) let x hour be the time :
0.5 * 36 +36x =48x
18=12x
x=1.5
the ans. is 1.5 hours
2)let x hour be the time
50x + 75 x = 800
125x = 800
x = 6.4
the ans is 6.4 hours
0.5 * 36 +36x =48x
18=12x
x=1.5
the ans. is 1.5 hours
2)let x hour be the time
50x + 75 x = 800
125x = 800
x = 6.4
the ans is 6.4 hours
收錄日期: 2021-04-13 00:06:25
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