F.3 Physics Questions

以下我用hr代表hour﹐這些純屬個人喜好
1
(a)
From the graph
initial speed
=20 m/s
=20 (m/s)(1km/1000m)(3600 s/1 hr)
=72 km/hr
1 這是最保險的做法﹐雖然會做得慢些
2 題目已經給了初速﹐不用計算
(b)
Thr area under the graph
=area of rectangle+ area of triangle
=20*0.5+20*(2.5-0.5)/2
=10+20
=30 m
Its physical meaning is:
The total traveling distance of the car from the driver saw a dog until the car stopped moving.
1 由s=∫v dt 可知v-t曲線圖中所包圍的面積是距離。
2 又v-t曲線圖中所包圍的面積不是位移。
3 在這裡題目沒有問減速度﹐所以不用計算。
2
We notice that if the car pass through the speed limit sign at 50km/h, then the deceleration of the car should be at the minimum value.
We use v^2-u^2=2as
Here u= 70km/h, v=50km/h, s=30m=30/1000 km=0.03 km
50^2-70^2=2a(0.03)
a=-40000 kmhr^-2 (minus sign means decelerate)
Since
-40000 kmhr^-2
=-40000 (km/hr^2)(1000 m/1 km)(1 hr^2/3600*3600 s^2)
=-3.0864 ms^-2
So the minimum deceleration of the car is -3.0864 ms^-2
理論上可以首先全部換成 m/s 再做﹐不過因為題目有2個原始數據都是km/hr 所以若果直接用km/hr (注意30m轉成km都是完整數字﹐沒有捨入誤差) 計算出來的答案會準確些。

第1條問題
a)
v = u + at
0 = u + 0.5a
u = -0.5a

i.e. if the acceleration is -2km/h2, initial speed is 1km/h
it is in negative side coz it is deceleration

b)
the area under the graph can be found by considering the area of a triangle, while the area of a triangle can be computed by the formula (base * height / 2)

the area under the velocity-time graph is the displacement of the car

第2條問題
v2 = u2 + 2as
502 = 702 + 2a(30/1000)
2500 = 4900 + 6a/100
-2400 = 6a/100
6a = -240000
a = -40000
i.e. the minimum deceleration is -40000km/h2

2007-04-06 16:19:45 補充：
the acceleration can be found by finding the slope of the graph

2007-04-06 17:15:19 補充：
to convert -40000km/h^{-2} to SI unit (ms^{-2})1km = 1000m1hour = 3600sso -40000 * 1000/ 3600 / 3600 = -3.086ms^{-2}

2007-04-06 17:17:15 補充：
a = slope = (0 - 20) / (2.5 - 0.5) = -20/2 = -10so u = -0.5 * - 10 = 5ms^{-1}area = 20*0.5 + (2.5 - 0.5)*20/2= 30m

1. I think Q1 has miss one figure.
so I can't solve it.



2.Based on v^2 - u^2 = 2as,
(v= final speed,u= initial speed, a= acceleration, s=displacement)

(50/3.6)^2 - (70/3.6)^2 = 2(30)
a= -3.09 m/s^2
Therefore the minimum deceleration of the car is 3.09m/s^2

2007-04-06 19:03:50 補充：
1a.有graph就唔同哂 :P As we can see in the graph,the initial speed is 20m/s,20m/s = 20 x 1km/1000m x 3600s/1hour= 20 x 3.6km/hour= 72 km/hour

2007-04-06 19:04:06 補充：
b.the area under the graph is (0.5 2.5)s x 20m/s /2 =3s x 20m/s /2=30mTherefore the physical meaning of the graph is "the distance of stopping the car".