Problem 4 - 物理學系- 東海大學

Fundamentals of Physics
Halliday & Resnic
【CH04】Two- and three- Dimensional Motion
Homework of Chapter 4:
5, 9, 13, 15, 17, 25, 27, 31, 35, 37, 41, 45
東海大學物理系
【Problem 4-5】
An ion’s position vector is initially r = 5 i− 6j+ 2k
, and 10s later is r = − 2 i+ 8j−2k, all in
meters. In unit-vector notation, what is its vavg
during the 10s?
一個離子初始位置 r = 5 i− 6j+ 2k
,10 秒後位置 r =− 2 i+ 8j−2k,單位都是米,問在 10 秒
期間,平均速度 vavg ?
( − 2i+ 8j−2 k) −(5i− 6j+ 2 k)
: vavg
=
= ( − 0.7 i+ 1.4j−0.4 k) m/ s
10s
【Problem 4-9】
Figure 4-33 gives the path of a squirrel moving about on level ground, from point A (at time t=0), to
point B (at t=5min), C (at t=10min), and finally D (at t=15min). Consider the average velocities of
the squirrel from point A to be each of the other three points. Of them, what are the (a) magnitude
and (b) angle of the one with the least magnitude and the (c) magnitude and (d) angle of the one
with the greatest magnitude?
(圖 4-33)
一隻松鼠在平地上的移動路徑如圖 4-33,從 A 點(在時間 t = 0),B 點(在時間 t = 5 分鐘),
C 點(在時間 t = 10 分鐘),最後 D 點(在時間 t = 15 分鐘)。考慮松鼠從 A 點到其他三個
點的平均速度。其中,最小的速度與角度為何?為大的速度與角度為何?
:各點座標與時間如下:
A(15, − 15)( t = 0) ,
B(30, − 45)( t = 5min) ,
C(20, − 15)( t = 10min) ,
D(45,45)( t = 15min)
2 2
33.5m
AB = (30 − 15) + ( −45 −( − 15)) = 33.5m
, vav( AB) = = 0.11 m/ s
300s
Fundamentals of Physics
Halliday & Resnic
2 2
5m
AC = (20− 15) + ( −15 −( − 15)) = 5m
, vav( AC) = = 0.0083 m/ s
600s
2 2
67.1m
AD = (45 − 15) + (45 −( − 15)) = 67.1m
, vav( AD) = = 0.075 m/ s
900s
5m
(a) 最小 vav( AC) = = 0.0083 m/ s
600s
(b) 角度是 0 度
33.5m
(c) 最大 vav ( AB) = = 0.11 m/ s
300s
−1 −60
0
(d) 角度 tan = − 63
30
東海大學物理系
【Problem 4-13】
The position r 3 4
of a particle moving in an xy plane is given by r = (2t − 5 t) i+ (6−7 t ) j,
with r
in meters and t in sections. In unit-vector notation, calculate (a) r , (b) v , and (c) a for t = 2s.
(d) what is the angle between the positive direction of the x axis and a line tangent to the particle’s
path at t = 2s?
3 4
一個粒子在 xy 平面上移動的位置 r = (2t − 5 t) i+ (6−7 t ) j,
r 單位為米,t 單位為秒,計算
t = 2s,(a)
r ,(b) v ,和 (c) a ,(d)問粒子在 t = 2s時路徑方向和
x 軸的夾角?
:(a) rt ( = 2 s) = 6 i−106j
dr 2 3
(b) vt ( = 2 s) = = (6t −5) i− 28t j= 19 i−224j dt
dv
2
(c) at ( = 2 s) = = (12 ti ) − (84 t) j= 24 i−336j dt
−1 −224
(d) tan =− 85.15
19
【Problem 4-15】
2
2
A car is propelled over an xy plane with acceleration components ax= 4 m/ s and ay=− 2 m/ s .
Its initial velocity has components v0x= 8 m/ s and v0y= 12 m/ s.
In unit-vector notation, what is
the velocity of the cart when it reaches its greatest y coordinate?
2
2
一台車子在 xy 平面上,加速度分量為 a = 4 m/ s 和 a =− 2 m/ s ,初速度分量分別為
v0x= 8 m/ s和
v0y= 12 m/ s,當車子到最大的
y 座標時,其速度為何?
a = 4 i−2 j( m/ s )
v
0 = 8 i+ 12 j( m/ s)
1
y = 12 t+ ( − 2) t = 12t−
t
2
: 2
2 2
x
y

Fundamentals of Physics
Halliday & Resnic
當 y = ymax
時,對上式微分, 0 12 2t
v = 8+ 4(6) = 32( m/ s)
x
v y = 0
當 y = ymax
時, v = 32 i( m/ s)
= − ⇒ t = 6s
東海大學物理系
【Problem 4-17】
A particle leaves the origin with an initial velocity v = 3 i ( m/ s)
and a constant acceleration
2
a =−1 i−0.5 j ( m/ s ) . When it reaches its maximum x coordinate, what are is (a) velocity and (b)
position vector?
和等加速度 2
一個粒子以初速度 v = 3 i ( m/ s)
座標時,問(a)速度(b)位置向量?
a =−1 i−0.5 j ( m/ s ) 離開原點,當它達到最大 x
1 2 1 2
:(a) x = 3 t+ ( − 1) t = 3t−
t
2 2
微分後, 0= 3− t , t = 3
v ( 0.5)(3) ( / ) 1.5
y = − j m s =− j( m/ s)
1 2 1
(b) r = v0t+ at
2
= (3 i)(3) + ( −1i−0.5 j)(3
) = 4.5i−2.25 j(
m)
2
2
【Problem 4-25】
A dart is thrown horizontally with an initial speed of 10m/s toward point P, the bull’s-eye on a dart
board. It hits at point Q on the rim, vertically below P, 0.19s later. (a) what is the distance PQ? (b)
How far away from the dart board is the dart released?
一飛鏢以水平 10m/s 射向標靶上的的靶心 P 點。0.19 秒後射中 P 點下方的 Q 點。(a) PQ 兩
點距離為何?(b) 飛鏢射出時,與標靶距離多遠?
1 2 2
:(a) PQ= (9.8 m/ s )(0.19 s) = 0.18m
2
(b) x = (10 m/ s)(0.19 s) = 1.9m
【Problem 4-27】
0
A certain airplane has a speed of 290km/h and diving at an angle of θ = 30 below the horizontal
when the pilot releases a radar decoy (Fig 4-37). The horizontal distance between the release point
Fundamentals of Physics
Halliday & Resnic
東海大學物理系
and the point where the decoy strikes the ground is d = 700m.
(a) How long is the decoy in the air?
(b) How high was the release point?
某飛機以時速 290 公里,水平往下 30 度俯衝飛行,飛行員釋放出雷達誘餌(圖 4-37)。釋
放點與著地點水平距離為 700 公尺。(a)誘餌在空中多久時間?(b)被釋放時的高度?
(圖 4-37)
0 0
:飛機初速度 v = 290cos30 i− 290sin 30 j = 251.1 i−145 j(
km/ h)
= 69.75i−40.3 j(
m/ s)
s 700m
(a) s = v0xt ⇒ t = = = 10s
v0x69.75 m/ s
1 2
1
2
(b) 0 − y = v0yt− gt = ( −40.3)(10) − (9.8)(10 ) =− 893m
2
2
y = 893m
【Problem 4-31】
0
A plane, diving with constant speed at an angle of 53 with the vertical, releases a projectile at an
altitude of 730m. The projectile hits the ground 5s after release. (a) What is the speed of the plane?
(b) Hw far does the projectile travel horizontally during its flight? What are the (c) horizontal and (d)
vertical components of its velocity just before striking the ground?
一架飛機,以與垂直角度 53 度俯衝在 730 公尺高度時,投擲一枚炸彈,炸彈在 5 秒後落地。
(a)飛機速度為何?(b)炸彈飛行水平距離為何?炸彈著地時速度的(c)水平分量與(d)垂直分
量為何?
0 1
2
:(a) 0 − 730 m= ( −v0cos53 )(5 s) − (9.8)(5 s)
2
v0= 202 m/ s
0
(b) (202 m/ s)(sin 53 )(5 s) = 806.6m
0
(c) vx= (202 m/ s)sin 53 = 161.3( m/ s)
0 2
v = ( −202 m/ s)cos53 − (9.8 m/ s )(5 s) =− 170.6( m/ s)
y
【Problem 4-35】
A projectile’s launch speed is five times its speed at maximum height. Find launch angle θ 0 .

Fundamentals of Physics
Halliday & Resnic
一子彈的啟動速度,是它在最高點速度的 5 倍,問啟動時的角度?
v = v θ = v,
v = 0
:最高點時, x 0cos 0
啟動速度 v0= 5v
5vcosθ 0 = v
− 1
θ0
= cos = 78.5
5
1 0
y
東海大學物理系
【Problem 4-37】
A ball is shot from the ground into the air . At a height of 9.1m, its velocity is
v = (7.6 i+ 6.1 j)
m/ s,
with i horizontal and j upward. (a) To what maximum height does the
ball rise? (b) What total horizontal distance does the ball travel? What are the (c) magnitude and (d)
angle (below the horizontal) of the ball’s velocity just before it hits the ground?
一顆球從地面射向空中,在高度 9.1m 時,速度為 v = (7.6 i+ 6.1 j)
m/ s(
i 為水平, j 為向上),
(a)球達到的最大高度?(b)球旅行的總水平距離?球撞到地面時,其速度(c)大小(d)
和水平的夾角為何?
2 2
: v = v0− 2gs
2 2 2
(6.1 m/ s) = ( v m/ s) − 2(9.8 m/ s )(9.1 m)
0 y
v0y= 14.7 m/ s
2 2 2
(a) (0) = (14.7 m/ s) − 2(9.8 m/ s )( Hm)
H = 11m
1
2v 2
0 y 2(14.7)
(b) 0 = v0yt− gt ⇒ t = = = 3s
2
g 9.8
R = v0 xt
= (7.6)(3) = 22.8m
(c) v3x= 7.6
i
v3y = v0y − gt = 14.7 − (9.8)(3) =− 14.7
2 2
v3 = (7.6) + ( − 14.7) = 16.5 m/ s
−1 −14.7
0
(d) θ = tan =− 63
7.6
【Problem 4-41】
In Fig. 4-39, a ball is thrown leftward from the left edge of the roof, at height h above the ground.
0
The ball hits the ground 1.5s later, at distance d=25m from the building and at angle θ = 60 with
the horizontal. (a) Find h. (Hint: one way is to reverse the motion, as if on videotape.) What are the
(b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown? (d)
Is the angle above or below the horizontal?
Fundamentals of Physics
Halliday & Resnic
(圖 4-39)
東海大學物理系
圖 4-39,一球從屋頂的左邊緣被往左方丟出,離地面高度 h。1.5 秒後,球在距離建築物 d =
0
25 公尺,角度 θ = 60 的地方撞到地面,。(a)求出 h(提示:一種方法是往回復運動去想,
就好像錄影帶倒帶播放。球被丟出時的(b)水平速度與(c)角度?(d)球是被往上丟或往下丟?
0
: 25 m= ( v0cos60 )(1.5 s)
v0= 33.3 m/ s
0 1
2
(a) h= ( v0sin 60 )(1.5) − (9.8)(1.5) = 32.2m
2
0
(b) vx = v0x = 33.3cos 60 = 16.7 m/ s
0
v = v0− gy = 33.3sin 60 − (9.8)(1.5) = 14.1 m/ s
y y
2 2
v= (16.7) + (14.1) = 21.9 m/ s
− 14.1
(c) θ = tan = 40.2
16.7
(d) 往下
1 0
【Problem 4-45】
Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect
to knock it into the water (Fig 4-43). Although the fish sees the insect along straight-line path at
angle φ and distance d, a drop must be launched at a different angle θ 0 if its parabolic path is to
0
intersect the insect. If φ = 36 , d = 0.9m,
and the launch speed is 3.56 /
for the drop to be at the top of the parabolic path when it reaches the insect?
(圖 4-43)
m s, what 0
θ is required
射手魚在樹枝上的水滴發現了一隻昆蟲,於是噴出水滴要將昆蟲射下水(圖 4-43)。雖然魚
是沿著直線路徑的角度φ 和距離 d 看到這隻昆蟲,但水滴必須從 θ 0 角度沿著拋物線路徑射到
0
昆蟲。如果, φ = 36 , d = 0.9m,發射速度為
3.56 m/ s, θ 0 必須在何種角度下,水滴才能
射到昆蟲?
:魚和昆蟲距離
φ
y = dsin φ = (0.9)sin 36 =
0.53m
0
x = dcos = (0.9)cos36 = 0.73m
0

Fundamentals of Physics
Halliday & Resnic
2 2
水滴運動狀況: 0 = ( v0sin θ ) − 2gymax
( v0
sin θ )
y = ymax
=
2g
−1 2gy −1
2(9.8)(0.53)
0
θ0
= sin = sin =
64.2
2 2
v (3.56 m/ s)
0
2
東海大學物理系