A car travelling at 110kmh^-1 has its speed reduced to 38km^-1 whilst
travelling over a distance of 200m.Find the value of the retardation(assuumed uniform)and the time taken.Find also how much further the car will travel with the same retardation before coming to rest.
另外仲有一個問題 further呢個字係phy裡面點解??
F.4 PHY
2008-12-26 6:23 pm
回答 (3)
2008-12-26 6:56 pm
✔ 最佳答案
V=(38*1000)/3600=10.6ms-2U=(110*1000)/3600=30.6ms-2
S=200m
V2-U2=2AS
10.62-30.62=2A(200)
A=-2.06ms-2
V=U+AT
0=10.6+(-2.06)T
T=5.15s
Further解更遠的.係E條問題到,姐係問,部車仲要行幾遠先可以靜止.
V2-U2=2AS
02-10.62=2(-2.06)S
S=27.3M
The car will travel 27.3m further before coming to rest.
2008-12-26 10:57:30 補充：
以上2位均沒有埋會further的意思.
參考: ME
2008-12-26 6:55 pm
Sol: Given: u=110km/h=110x10^3/3600 ms^-1
...................v=38km/h=38x10^3/3600 ms^-1
...................s=200m
...................a=? t=? s=?
Find the retardation (a=??) ???
..By the formula: v^2 = u^2 + 2as
..[(38x10^3)/3600]^2 = [(110x10^3)/3600]^2 + 2a(200)
.................................a=-2.056 ms^(-2) ( Corr to 4 sig.fig)
Find the time taken (t=??) ???
.....By the formula: v = u + at
.....[(38x10^3)/3600] = [(110x10^3)/3600] + ( -2.056 )t
................................t = [(110x10^3 - 38x10^3)/3600](1/2.056)
.................................t= 9.730 s ( Corr to 4 sig.fig)
ind also how much further the car will travel with the same retardation before coming to rest (s=??) ???
By the formula: v^2 = u^2 + 2as
........................(0)^2= [(110x10^3)/3600]^2 + 2(-2.056)s
......................... .....s= 227.1m ( Corr to 4 sig.fig)
2008-12-26 11:21:05 補充：
v^2 = u^2 + 2as
(0)^2= [(110x10^3)/3600]^2 + 2(-2.056)s
.......s= 227.1m ( Corr to 4 sig.fig)
......s = 227.1 - 200
......s = 27.1 m (The Answer)
The required further distance travelled is 27.1m.
...................v=38km/h=38x10^3/3600 ms^-1
...................s=200m
...................a=? t=? s=?
Find the retardation (a=??) ???
..By the formula: v^2 = u^2 + 2as
..[(38x10^3)/3600]^2 = [(110x10^3)/3600]^2 + 2a(200)
.................................a=-2.056 ms^(-2) ( Corr to 4 sig.fig)
Find the time taken (t=??) ???
.....By the formula: v = u + at
.....[(38x10^3)/3600] = [(110x10^3)/3600] + ( -2.056 )t
................................t = [(110x10^3 - 38x10^3)/3600](1/2.056)
.................................t= 9.730 s ( Corr to 4 sig.fig)
ind also how much further the car will travel with the same retardation before coming to rest (s=??) ???
By the formula: v^2 = u^2 + 2as
........................(0)^2= [(110x10^3)/3600]^2 + 2(-2.056)s
......................... .....s= 227.1m ( Corr to 4 sig.fig)
2008-12-26 11:21:05 補充：
v^2 = u^2 + 2as
(0)^2= [(110x10^3)/3600]^2 + 2(-2.056)s
.......s= 227.1m ( Corr to 4 sig.fig)
......s = 227.1 - 200
......s = 27.1 m (The Answer)
The required further distance travelled is 27.1m.
參考: 本人對”物理”的基本知識, 希望本人的解答幫助到各下解決”物理功課”疑難, 請投本人一票, 謝謝!!, 今次的錯誤真是不好意思, 本人下次會小心地答題目, 但本人仍希望各下能投下本人一票, 謝謝!!
2008-12-26 6:34 pm
v^2-u^2=2as
u=30.55ms^-1 v=10.56ms^-1 a=? s=200
(10.56)^2-(30.55)^2=2a(200)
a=-2.05ms^-2
time taken:v=u+at
10.56=30.55+(-2.05t)
t=9.75
v^2-u^2=2as
0^2-30.55^2=2(-2.05)s
s=227.6m
u=30.55ms^-1 v=10.56ms^-1 a=? s=200
(10.56)^2-(30.55)^2=2a(200)
a=-2.05ms^-2
time taken:v=u+at
10.56=30.55+(-2.05t)
t=9.75
v^2-u^2=2as
0^2-30.55^2=2(-2.05)s
s=227.6m
參考: 希望冇錯
收錄日期: 2021-04-25 23:15:47
原文連結 [永久失效]:
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