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Chapter Three Motion in a Plane 第三章平面上的運動

3-1 Displacement As a Vector Quantity 位移即向量

In the following three cases, the displacement is 7.5 km[NE], even though the actual path taken and the total distance travelled are different. Also, if the trip is reversed, beginning at B and arriving, by any path, at A, the displacement is reversed. It is 7.5 km[SW].

甚至實際的路徑和總位移不同

，但下列三種情況，位移均是7.5 km[NE]。假如行程是倒著走，起點是B ，經由其它任何路徑，達終點A ，整個位移是倒轉的，位移是7.5 km[SW]。

W E

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The actual displacement between any two points may be determined by drawing a scale diagram with a ruler and a protractor, or by using simple trigonometry. The following example illustrates this.

任何兩點之間實際的位移可用比例尺和分度器畫簡圖予以判斷，或者使用簡單的三角學計算之。以下列的範例說明:

Example 範例

Find the displacement of an airplane that flies 5.0 km due east and then turns and flies 7.0 km due north.

求飛機的位移，此飛機向正東飛行5.0Km，然後轉向向正北飛行7.0 Km.

Solution: 解題

1.Scale Drawing Solution: 以比例尺作圖解題

Using a scale of 1 cm=2.0 km and with directions as indicated, a scale

diagram of the plane's motion is drawn.

使用1cm=2.0km的比例尺且指出方向，將飛機飛行的比例簡圖畫出。

Using a ruler and the scale, the magnitude of the plane's displacement is found to be 8.5 km.

使用尺和比例尺，找出飛機位移的大小是8.5km。

Using the protractor and measuring the angle between the plane's

displacement and due east, its direction is found to be [E55o N].

使用分度器，測量飛機的位移和正東方之間的角度，找出它的方向是[E55o N]。

Thus, ?d = 8.5 km [E55o N]

W E

S 5.0 km

2.Mathematical Solution: 以數學解題

Draw a rough sketch representing the plane's motion. Using the Pythagorean theorem to find the magnitude of its displacement, we have 畫一簡圖描述飛機的運動，使用畢氏定理求其位移的大小，得到

|?d|2 = (5.0 km)2 + (7.0 km)2 = 74 km2

|?d| = 8.6 km

W E 7.0 km

5.0 km

Also, if we call the angle between the plan's displacement and due east θ, we have

同時，假如我們稱飛機位移和正東方之間的夾角為θ，得到

7.0 km

tan θ = = 1.4

5.0 km

θ = 54o

Thus, ?d = 8.6 km [E54o N]

3-2 Velocity as a Vector Quantity 速率即向量

The difference between velocity and speed 速率和速度的差異

Speed is the rate of change of distance with time, without regard to direction. Velocity is the rate of change of displacement with time, and it includes an expression of direction. Speed is represented by the symbol v and velocity is

速度是距離和時間的改變，不須考慮到方向。速率是位移和時間的改變，且包含方向。速度記號是v而速率是v。

Average Velocity involves the total displacement and the total time required for the displacement as follow:

平均速率含總位移和位移所須的時間，如下:

V av =

Instantaneous velocity: To determine the instantaneous velocity, the slope of the d-t graph must be determined at one point of time. An instant in time means that the time interval decreases or approaches the value of zero.

瞬時速率: 判定瞬時速率，必須於時間的一點中，求d-t圖表中斜率。

V inst =lim

?t → 0 ?t

Example 範例

A clock has a second hand that is 12 cm long. Find each of the following. 時鐘秒針長12 cm ，求下列各題

(a) the average speed of the tip of the second hand

(b) its instantaneous velocity as it passes the 6 and the 9 on the clock face (c) its average velocity in moving form the 3 to the 12 on the clock face

Solution 解題

(a) the tip of the second hand makes one complete revolution in 60 s.

秒針繞一圈費時60s.

?d 2π(12 cm) V av = = ?t 60 s

= 1.3 cm/s

(b) Since the speed is constant, |v | = v = 1.3 cm/s

Therefore, v 6 = 1.3 cm/s[left]

V 9= 1.3 cm/s[up]

(d) The displacement of the tip of the second hand in moving from the 3 to 12 may be found from the following diagram:

秒針位移，從3點移動到12點，由下列簡圖得到

scale: 1 cm = 5 cm

|?d |2 = (12cm)

2 + (12cm)2 = 288 cm 2 |?d | = 17 ?d = 17 cm[left 45o up]

Therefore, V av = 17 cm/45 s = 0.38 cm/s [left 45o up]

Practice 練習

1.A police cruiser chasing a speeding motorist travelled 60 km[s], then 35 km[NE], and finally 50 km[w].

一輛警車追逐一超速駕駛，行駛60 km[s], 然後再35km[NE]，最後行駛50 km[W]。

(a)Calculate the total displacement of the cruiser.

(b)I f the chase took 1.3 h, what was the cruiser's (I) average speed and (ii)

average velocity, for the trip.

2.An express bus travels directly from A-town to B-ville. A local bus also links these two towns, but it goes west 30 km from A-town to C-city, then 30 km south to D-ville, and finally 12 km west to E-town and 30 km northeast to B-ville.

一快車從A-town直達行駛到B-ville。一本地的公車亦連接這兩鎮，但它從A-town到C-city向西行駛30km，然後向南行駛30km到D-ville，最後向西12km到E-town且東北30km到B-ville。

(a)What is the shortest distance from A-town to B-ville?

(b)In what direction does the express bus travel?

(c)If the express bus takes 0.45 h to go from A-town to B-ville, and the

local bus takes 3.0h, calculate the average speed and the average velocity for each bus.

3.A hiker walks 10.0 km[NE], 5.0 km[W], and then 2.0 km[S] in 2.5 h.

一健行者，在2.5h內，行走10.0 km[NE]、5.0 km[W]、而後2.0 km[S]。

(a)What is the hiker's displacement?

(b)In what direction must the hiker aet out, in order to return by the most

direct route to the starting point?

(c)If the hiker walks at a constant speed for the entire trip and returns by

the most direct route, how long will the total walk take?

Answer: 1) 43 km[S36o W]; 1.1 x 102 km/h, 33 km/h[S 36o W]

2) 23 km, [W24o S]; 51 km/h, 51 km/h[W24o S]; 34 km/s,

7.7 km/h[W24o S]

3) 5.5 km[E38o N],[S22o W], 3.3 h

3-3 The Addition and Subtraction of Vectors 向量的加減

1.The resultant: Sum of a number of vectors of a particular type is that single vector that would have the same effect as all the original vectors taken together.

合力: 一種特殊種類向量的總合，它是一種單一向量，與所有原來向量的組合具有相同的效力。

2.Polygon Method: 多邊形法

This method for finding the resultant R of several vectors (A, B, and C) consists in beginning at any convenient point and drawing ( to scale and in the proper directions ) each vector arrow in turn. They may be taken in any order of succession: A + B + C = C + A + B = R. The tail end of each arrow is positioned at the tip end of the preceding one, as shown in Figure.

此方法是求許多向量的合力，此合力的組成是始於任何合適的點且依序連續畫出每一向量的箭頭：。每一向量的尾端落在前一向量的尖端，如圖示。

Start

3.Parallelogram Method 平行四邊形法

Parallelogram method for adding two vectors: The resultant of two vectors acting at any angle may be represented by the diagonal of a parallelogram. The two vectors are drawn as the sides of the parallelogram and the resultant is its diagonal, as shown in Figure. The direction of the resultant is away from the origin of the two vectors.

平行四邊形法用來將兩個向量相加：形成任何角度兩向量的合力，可用平形四邊形的對角線表示。兩向量是平行四邊的兩邊，而合力是對角線，如圖示。合力的方向是遠離原來的兩個向量。

Resultant 合力

4. Subtraction of Vectors: 向量的減法

To subtract a vector B from a vector A, reverse the direction of B and add individually to vector A, that is, A - B = A + ( -B).

從A向量減去B向量，顛倒B向量的方向且個別加在A向量上，即A - B = A + ( -B)。

B - B

A B

A - B

3-4 The Multiplication of a Vector by a Scale

向量的乘法使用比例尺

Vectors may be multiplied together in two distinct ways. The vector dot product, written as A?B, produces a result that is not a vector, but a scalar quantity whose magnitude is AB cosθ, where θis the angle between A and B, the vector cross product, written A × B, produces a result that is a vector of magnitude AB sinθ, in a direction perpendicular to the plane of A and B, given by a right-hand rule.

向量相乘以二種方式，向量乘號寫成A?B，產生的合力，不是向量，是數量，它的大小是AB cosθ，θ是向量A和B的夾角，向量相乘，寫作A× B，產生AB sinθ向量，與AB平面成垂直，由右手定則求出。

In general, the product of a scalar K and a vector A is a vector whose magnitude is K|A|, whose units are the product of the units of K and A, and whose direction is that of the vector A. Division of a vector A by a scalar K is treated as if it were multiplication by the scalar 1/k.

大體來講，數量K和向量A的乘積是向量，它的大小是K|A|，單位是K和A單位的乘積，它的方向是向量A的方向。向量A除以數量K，就好像向量乘以1/K數量。

Graphically, the product is a vector parallel to A, but K times as long.

圖形來講，乘積是向量平行A，長度乘以K。

Where k=3 where k =1/3

3-5 The Components of a Vector

向量的分量

A component of a vector is its effective value in a given direction. For example, the x-component of a displacement is the displacement parallel to the x-axis caused by the given displacement. A vector in two dimensions may be solved into two component vectors acting along any two mutually perpendicular directions. Figure shows the vector R and its

and R y, which have magnitudes

在一已知的方向中，向量的分量是有效值，例如，位移x的分量是由已知的位移引出平行x軸的位移。在二度空間的向量能分解成兩個分量，此兩個分量的方向是順著互相垂直的方向。圖顯示出向量R和它的分量R x和R y，分量有大小，如下:

| R x| = | R| cos θand | R y| = | R| sin θ

or equivalently 同樣地

R x = R cos θand R y = Rsin θ

Example 範例

1.The displacement of an airplane from its starting point is 100 km[E30o N]. Determine the components of its displacement in the easterly and northerly directions.

一飛機從起點起算位移100 km[E30o N]，求位移東和北方向的分量。Graphical Solution: Scale 1 cm = 20 km

x(east)

Mathematical Solution:

|?d x| = |?d x| cos θ = 100 km (cos 30o) = 100 km (0.8660)

= 86.6 km

|?d y| = |?d| sin θ = 100 km (sin 30o) = 100 km (0.5000)

= 50.0 km

Therefore, the components of a displacement of 100 km[E30o N] are

86.6 km[E] and 50.0 km[N].

因此，位移100km[E30o N]的分量是86.6 km[E] and 50.0 km[N]。

2.The easterly and northerly components of a car's velocity are 24 m/s and 30 m/s, respectively. In what direction and with what speed is the car moving? In other words, what is the car's velocity?

一汽車向東和向北的分量分別是24 m/s和30 m/s，試問此汽車以多少速度和方向行駛？換句話說，汽車的速率是多少？

Graphical Solution:

[E]

24 m/s

Mathematical Solution:

|V y|30 m/s

tan θ = = = 1.25 θ = 51o

|V x|24 m/s

|V|2 = |V x|2 + |V y|2 = (24 m/s)2 + (30 m/s)2

= 1476 m2/s2

|V| = 38 m/s

Therefore, the car is moving in a direction 51o to the north of due east, with a speed of 38 m/s ( V = 38 m/s [E51o N]).

因此，汽車以38 m/s速度朝51o正東北方行駛。

Example 範例

1.A boat sails in a straight line 20 km[N30o E]. What are the components of its displacement to the north and east?

船以20 km[N30o E]直線航行，求船向北和東位移的分量？

2.A cannon fires a cannonball with a speed of 100 m/s at an angle of 20o above the horizontal. What are the horizontal and vertical components of the initial velocity of the cannonball?

一加農砲以水平夾角20o速度100 m/s射出砲彈，求其初速率的水平和垂直分量？