=.= 發問於 科學及數學其他 - 科學 · 8 年前

# physics position and motion

point a :。

point b :。

point c :。

point a至point b 係14.5cm point b至point c係24.1cm

the flashing light of the stroscobe flashes at a fixed frequency of 10Hz

ans:9.6ms^-2

14.5/100=0*0.1+1/2 a*0.1^2

a=29ms^2

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=0.*0.1+1/2 *9.6*0.1^2

=0.048m-->唔係14.5cm 點解???

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=0+9.6*0.1

=0.96ms^1

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### 2 個解答

• 8 年前
最愛解答

你不可以這樣計，因為你無法得知個Ball是在何時何地開始Fall freely from rest，也無法得知個Ball在Point A的速度

但你在Sub數時，已假設個Ball是在Point A fall freely from rest，也就是事先假設個Ball在Point A的速度是零

不過題目沒有指明這個條件，而且驗算時這個速度也不是零，因此你的方法會出錯

正確的方法是用兩組s與t(A至B 與 A至C)，此時兩組的a與u都相同(因此不能用B至C)，可以聯立兩組 s=ut+1/2*at^2 來求得u與a：

0.145 = u*0.1 + 1/2*a*0.1^2

0.145+0.241 = u*0.2 + 1/2*a*0.2^2

即 0.145 = 0.1u + 0.005a -(1)

0.386 = 0.2u + 0.02a -(2)

(2)-2*(1), 0.386 - 0.145*2 = (0.2u + 0.02a) - 2(0.1u + 0.005a)

0.096 = 0.2u + 0.02a - 0.2u - 0.01a = 0.01a

得出正解a = 9.6 m/s^2

2013-06-16 11:15:10 補充：

以上即acceleration due to gravity = 9.6 m/s^2

4*(1)-(2), 0.145*4-0.386 = 4(0.1u+0.005a) - (0.2u+0.02a)

0.194 = 0.4u+0.02a-0.2u-0.02a = 0.2u

u = 0.97 m/s

Ball在Point A的速度是0.97 m/s(向下)

2013-06-16 11:20:07 補充：

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• 8 年前

Stage 1 (From Point A to Point B)

By s = ut + (1/2)at^2

0.145 = u(1/10) + (1/2)a(1/10)^2

14.5 = 10u + (1/2)a

29 = 20u + a..........(1)

Stage 2 (Whole Motion, from Point A to Point C)

Again, by s = ut+(1/2)at^2

(0.145+0.241) = u[(1/10)(2)] + (1/2)(a)[(1/10)(2)]^2

0.386 = u/5 + a/50

19.3 = 10u + a.............(2)

(1) - (2) :

29 - 19.3 = 10u

u = 0.97ms^-1

Sub u =0.97 to (1)

a = 9.6ms^-2//

2013-06-17 22:23:47 補充：

As above calculation has stated, the initial speed of the ball (at Point A) IS NOT 0ms^-1. So you cannot find the correct answer.