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[問題] Phy kinematics 一問

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Phy kinematics 一問

E條點做?正確答案係b,c,d



   

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sorry,補返張圖

圖係E度
附件: 您所在的用戶組無法下載或查看附件

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(a) is incorrect
a = dv/dt = 2*i + 2*t*j
By Pyth's Theorem and putting t = 1,
|a| = [ (2)^2  +(2*1)^2 ]^(1/2) = 8^(1/2) m/s^2
(b) is correct
By Pyth's Theorem,
|v| = [ (2*t)^2 + (t^2)^2 ]^(1/2) = [ 4*(t^2) + t^4 ]^(1/2) = t * [ (4 + t^2)^(1/2) ]
tangential acceleration = d|v|/dt
By differentiating |v| with respect to t,
d|v|/dt = [ (4 + t^2)^(1/2) ] + t * (1/2) * (2*t) * [ (4 + t^2)^(-1/2) ]
By putting t = 1,
d|v|/dt = [ (4 + 1^2)^(1/2) ] + 1 * (1/2) * (2*1) * [ (4 + 1^2)^(-1/2) ] = 5^(1/2) + 5^(-1/2) = 6/[5^(1/2)] m/s^2
(c) is correct
By Pyth's Theorem,
acceleration^2 = (tangential acceleration)^2 + (radial acceleration)^2
(radial acceleration)^2 = acceleration^2 - (tangential acceleration)^2
By (a), acceleration = 8^(1/2) m/s^2
By (b), tangential acceleration = 6/[5^(1/2)] m/s^2
(radial acceleration)^2 = 8 - 36/5 = (40 - 36)/5 = 4/5
radial acceleration = 2/[5^(1/2)] m/s^2
(d) is correct
radial acceleration = (|v|^2) / radius of curvature
radius of curvature = (|v|^2) / radial acceleration
By (b), |v| = t * [ (4 + t^2)^(1/2) ]
(|v|^2) = (t^2) * (4 + t^2)
By putting t = 1,
(|v|^2) = (1^2) * (4 + 1^2) = 5
By (c), radial acceleration = 2/[5^(1/2)] m/s^2
radius of curvature = (|v|^2) / radial acceleration = 5 / { 2 / [5^(1/2)] } = 5 * [ 5^(1/2) ] / 2 m
2017 May/June
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so if I want to know what the tangential acceleration is,I need to find out the magnitude of v first,then differentiate v with respct to time?

And I just need to differentiate vector v and subsitute t=1 to get the acceleration?



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Yes. By definition, tangential acceleration = d|v|/dt .
係呀,呢個係定義,即係話tangential acceleration係喺當時嗰一剎那,向住目前方向H加速度。
|v| stands for magnitude of velocity facing at the direction of the vector v at that time.
|v| 係喺當時嗰一剎那,向住目前方向H速度有幾快。
Of course, acceleration is rate of change of velocity, that means a = dv/dt .
acceleration(加速度)就當然啦,acceleration係速度改變H率,當然要d咗佢啦。
ps:
呢個其實唔係physics,
而係applied mathematics入面Hmechanics再入面Hkinematics。

[ 本帖最後由 斧頭 於 2016-9-21 11:09 PM 編輯 ]
2017 May/June
CIE GCE:AL Maths,AL Further Maths
Edexcel IAL:AL Phy
2017 Oct/Nov
CIE GCE:AL Chinese
2018 Jan
Edexcel IAL:AL Chem
2018 Apr
HKDSE:Eng,物理,化學
逃離香港 服務英國 拯救世界

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明白曬

明曬啦,多謝你



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