## [數學]等加速度

### [數學]等加速度

yll

yll

[hide:27a9c86c51]設加速度a，總時間t
0.5at2-0.5a(t-3)2=0.5at2*0.75

scsnake

[hide:3a5e76077d]u=0, sinitial=0

a=dv/dt
v=at

v=ds/dt
s=(1/2)at2

When t=T-3, then s=(3/4)S, t=T, then s=S

(3/4)S=(1/2)a(T-3)2 ---(1)
S=(1/2)aT2 ---(2)

(2)/(1):
4/3=[T/(T-3)]2

T=6(2-√3)s (Rejected) or 6(2+√3)s

So, T=6(2+√3)s[/hide:3a5e76077d]

kevin

[hide:0197a21093]Given : t = 3s , displacement = x/4 ( let 路程 is x ) , initial velocity = 0.

Using equation of motion " s = ut + 1/2 *at2 "

x/4 = 0*3 + 1/2 * a(3)2

we find the 等加速度 is x/18.

Using equation again. " s = ut + 1/2 *at2 "

Put uniform acceleration = x/18.

x = 0*t + 1/2 * x/18 * t2
t2 = 36/x * x
t = 6

So 他花了 6s 完成這個比賽[/hide:0197a21093]
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E.T

In H.K, it is F.4 ch.10 physics ( speed, velocity and acceleration )
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E.T

i don't think so ~ the question 提供的資料 is enough to answer.
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E.T

http://hk.user.auctions.yahoo.com/hk/show/auctions?userID=ice_stella

stelny

E.T 寫到:In H.K, it is F.4 ch.10 physics ( speed, velocity and acceleration )

......也許是我的physics 不太好吧....這課是剛教完的(ch.10)...真失敗
....約有半個月沒翻過physics 書了...

Maybe you are very clever , no need to see phy. book = ="

No need to do revision ..............
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E.T

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Let's go to discuss ~*

E.T

[hide:791e166700]設起點速度為0 ，到餘下1/4路程那點的速度是y, 到終點的速度是x，全程長z
a是加速，t是全程所需時間

v=u+at
x=0+at
x=at

x=y+3a
at=y+3a
y=(t-3)a

(0+x)t/2=z
at^2=2z

(x+y)3/2=z/4
(x+y)3=z/2
(2t-3)a(3)(4)=at^2
12(2t-3)=t^2
t^2-24t+36=0
t= 24 + sqrt(576-144)/2
t= 12 + 6sqrt3

My phy is very bad, so ...
[/hide:791e166700]

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wa... who's correct ar

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[hide:38f48e1be0]Ans:12+6√3[/hide:38f48e1be0]

yll

I and siuhochung 1000