phy問題天天都多

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有架車從a點出發到c點我想問呢 .. 點解搵displacement個角果時 ..係搵角c而唔係角a 既 ?

我想問點知架 =] thx

abc666

某些會考solution很多錯的
如果只有u,v,a找s的話應用v^2-u^2=2as
或先v=u+at再s =ut+1/2at^2也不會當錯
但這條問題直接用s =ut+1/2at^2解釋嚴格來說是錯的
我估這條應該是mc的一些問哪些statement correct/not correct吧
如果在非mc問題(不過好像不會這樣問),只答bys =ut+1/2at^2實沒了一分

小霞的fans

原帖由 sunnychy 於 10/1/2009 08:08 PM 發表

                               
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你不可以代u = 0 , 因為如果u = 0的話, 個物件已經係停左
而且你用錯式  (成件事都同時間t 無關)

一件野減速減得慢(decelaration細) 當然要行長d ge路先停得低
正確要用ge式應該係v^2 - u^2 = 2as

...

我諗我明啦 =]
但係我睇會考solution佢都係用 s =ut+1/2at^2 ..

sunnychy

原帖由 小霞的fans 於 11/1/2009 03:49 PM 發表

                               
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咁例如我代u=0 ,, t=2 ,, a=-2
另1個u=0 ,, t=2 ,,a=-4

咁1個就係s=0(2)+1/2(-2)(2)^2
另1個就係s=0(2)+1/2(-4)(2)^2

第1條個答案係-4 .. 咁s姐係4啦..
第2條個答案係-8 ,, 咁s姐係8啦,,

唔係decele ...

你不可以代u = 0 , 因為如果u = 0的話, 個物件已經係停左
而且你用錯式  (成件事都同時間t 無關)

一件野減速減得慢(decelaration細) 當然要行長d ge路先停得低
正確要用ge式應該係v^2 - u^2 = 2as

一開始有某個速度u
停的時候, 速度= 0 (v=0)
有個decelaration a
s就係distance

我代一開始的速度v 為 4

如果 a = -2

0^2 - 4^2 = 2(-2)s
-16 = -4s
s = 4

4米先停低


如果 a = -4

0^2 - 4^2 = 2(-4)s
-16 = -8s
s = 2

2米就停左

小霞的fans

原帖由 超夢夢的夢境 於 10/1/2009 07:34 PM 發表

                               
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Let W be the weight of the block
Resolve the weight into components along the plane and perpendicular to the plane.
Component along the plane = W sin a (downwards along the plane)
Component perpe ...

咁例如我代u=0 ,, t=2 ,, a=-2
另1個u=0 ,, t=2 ,,a=-4

咁1個就係s=0(2)+1/2(-2)(2)^2
另1個就係s=0(2)+1/2(-4)(2)^2

第1條個答案係-4 .. 咁s姐係4啦..
第2條個答案係-8 ,, 咁s姐係8啦,,

唔係deceleration 比較大果個s比較大咩?

我諗錯左d咩 ?

超夢夢的夢境

原帖由 小霞的fans 於 11/1/2009 11:07 AM 發表

                               
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a block is placed on aroughplane inclined at an angle a to the horizontal.a force P parallel to the inclined plane is applied to the block so that it moves up the plane at a constant speed

點解weight of the block 會 = P/sin a = slope of P/sin a 既?

Let W be the weight of the block
Resolve the weight into components along the plane and perpendicular to the plane.
Component along the plane = W sin a (downwards along the plane)
Component perpendicular to the plane = W cos a (balanced by the normal reaction)
Friction = k * W cos a, where k is the coefficient of kinetic friction (downwards along the plane)

Since the block moves at a constant speed
P = W sin a + k * W cos a
沒 coefficient of kinetic friction 的話,問題無解
那個是 smooth plane 吧...

原帖由 小霞的fans 於 11/1/2009 11:07 AM 發表

                               
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仲有點解by s=ut+1/2at^2

smallerdecelerationrequires a longer stopping distance既 ?

唔係應該shorter stopping distance 咩 ? 代數入去我係咁計到..

Let the deceleration be -d where d is positive
v = u - dt
When v = 0
t = u/d

s = ut - 1/2 * dt^2
s = (u^2 - 1/2 * u^2)/d
s = (u^2)/2d

s increases as d decreases
i.e.smaller deceleration requires a longer stopping distance
忘了考慮 t 增加吧

小霞的fans

唔該哂*
又有題新啦 ..

a block is placed on a rough plane inclined at an angle a to the horizontal.a force P parallel to the inclined plane is applied to the block so that it moves up the plane at a constant speed

點解weight of the block 會 = P/sin a = slope of P/sin a 既?

仲有點解bys=ut+1/2at^2

smaller deceleration requires a longer stopping distance既 ?

唔係應該shorter stopping distance 咩 ? 代數入去我係咁計到..

[ 本帖最後由 小霞的fans 於 10/1/2009 06:10 PM 編輯 ]

abc666

應該因為direction用S x度E表達

加個point D在左下個 (ABCD form 長方形)(自行幻想)
角c=角DAC=x

Royal

若c到b是北方
那A到C的角是S c E