F.3 Physics Questions

第1條問題

A driver of moving car suddenly saw a dog running across the road. He applied the brake 0.5s later and the car decelerated uniformly to rest.

(a) Find the initial speed of the car in km/h.

(b) Find the area under the graph and state its physical meaning

第2條問題

A car is moving at a velocity of 70km/h. The driver then sees a 50km/h speed limit sign at a distance of 30m ahead. In order not to exceed the speed limit, find the minimum deceleration of the car.

3 Answers

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  • 1 decade ago
    Favorite Answer

    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。

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  • 阿一
    Lv 7
    1 decade ago

    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

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  • 1 decade ago

    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".

    Source(s): my physical knowledge
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