Lecture 03 20210929 - ecourse2.ccu.edu.tw
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Transcript of Lecture 03 20210929 - ecourse2.ccu.edu.tw
A. M2L2TB. ML2TC. ML2T-1
D. M-1L2T
M2L
2T
ML2
T
ML2
T‐1
M‐1L
2T
0% 0%0%0%
The energy of a photon can be calculated with the following equation: E=hꞏv, where E is the energy, v is the frequency of the photon (1/s), h is the Planck constant. Which of the following is the dimension of h?
A. M2L2TB. ML2TC. ML2T-1
D. M-1L2T
M2L
2T
ML2
T
ML2
T‐1
M‐1L
2T
0% 0%0%0%
The energy of a photon can be calculated with the following equation: E=hꞏv, where E is the energy, v is the frequency of the photon (1/s), h is the Planck constant. Which of the following is the dimension of h?
Ans: C
At which of the following position the car has largest magnitude of instantaneous speed?
At which of the following position the car has largest magnitude of instantaneous speed?
Ans: D
Chapter 2 Kinematics in One Dimension
Reference Frame
Displacement, Velocity, and Acceleration
Differentiation and Integration
( ) ( ) ( )x t v t a t
( ) ( ) ( )a t v t x t? ?
??
xxdxd cossin
CCU Physics 2 - 6
cos sind x xdx
1n
nd x n xd x
Important Differentiation Formulas
2
1 ( ) ( )(2) ( )/ ( ) ( ) ( )( )
d df x dg xf x g x g x f xdx g x dx dx
2 2. . sin 2 sin cosde g x x x x x xdx
( ) ( )(1) ( ) ( ) ( ) ( )d dg x df xf x g x f x g xdx dx dx
2
1. . sin / cos sinde g x x x x xdx x
CCU Physics 2 - 7
Practice
?21 22 xxdxd
?212 xxdxd
Practice
122121 222222 xdxdxx
dxdxxx
dxd
xxxx 2221 22 xx 64 3
21122
1
21
222
2
xdxdxx
dxdx
x
xxdxd
122
21 2
2
xxxx
142
1 22
xx
x
Proof ( Option)
0
( ) ( ) ( ) ( )( ) ( ) limx
d f x x g x x f x g xf x g xdx x
0
0
( ) ( ) ( ) ( )lim
( ) ( ) ( ) ( )lim
x
x
f x x g x x f x x g xx
f x x g x f x g xx
( ) ( )( ) ( )dg x df xf x g xdx dx
( ) ( )( ) ( ) ( ) ( )d dg x df xf x g x f x g xdx dx dx
(1)
CCU Physics 2 - 9
?tan
dxxd
2
1 ( ) ( )(2) ( )/ ( ) ( ) ( )( )
d df x dg xf x g x g x f xdx g x dx dx
x
dxdxx
dxdx
xcossinsincos
cos1
2
xx
dxd
dxxd
cossintan
xxxxx
sinsincoscoscos
12
xxx
222 sincos
cos1
x2cos1
x2sec
2
2 2sin( )sin( ) 2 cos( )d d u dxx x xdx du dx
( ( )) ( ) ( )(3) df u x df u du xdx du dx
3 232 2 2( 1)1 6 ( 1)d du d xx x xdx du dx
Chain Rule
3
3 4(sin )sin 3sin cosd du d xx x xdx du dx
(i)
(ii)
(iii)
Proof ( Option)
0
( ( )) ( ) ( )limx
df u x f x x f xdx x
( ) ( )df u du xdu dx
(3) ( ( )) ( ) ( )df u x df u du xdx du dx
0
( ) ( ) ( ) ( )lim( ) ( )x
f x x f x u x x u xu x x u x x
0
( ) ( )limx
f u u f u uu x
CCU Physics 2 - 10
Practice
?12 xdxd
Practice
dxdy
dydyx
dxd
2/1
2 1 )1( 2 xy
dxxdy 1
21 2
2/1
xx
21
121
2
12
xx
Proof ( Option)
(2) 2
1 ( ) ( )( ) / ( ) ( ) ( )( )
d df x dg xf x g x g x f xdx g x dx dx
1 1 ( )( ) / ( ) ( )( ) ( )
d d df xf x g x f xdx dx g x g x dx
2
1 ( ) 1 ( )( )( ) ( )
dg x df xf xg x dx g x dx
2
1 ( ) ( )( ) ( )( )
df x dg xg x f xg x dx dx
CCU Physics 2 - 11
log ?ad xdx
?xd bdx
logay x
0log ( ) log ( )lim a a
xx x xdy
dx x
01lim log ( )x a
x xx x
01lim log (1 )x a
x xx x x
log 1 logaa
d x edx x
1lim (1 ) 2.718N
NeN
yx a log ?ad xdx
01lim log (1 )
xx
x ax
x x
;
log lne x x 1 1ln log loge ed dx x edx dx x x
Natural logarithm
1lim(1 ) 2.718N
Ne
N
1(1 )N
N
N
3.0
2.8
2.6
2.4
2.2
2.0
100
101
102
103
104
105
2 3
1 ... ...2! 3! !
nx x x xe x
n
( 1)x
2 3
1 ... ...2! 3! !
nx x x xe x
n
0
limx x x
x
x
d b bbdx x
y= bx
ln ln
0
( ) ( )limb x x b x
x
e ex
x lnx ln
0
1limb
b
x
eex
2
x ln
0
( x ln )(1 x ln .... 12limb
x
bbe
x
2
x
0
(ln )lim ln x2x
bb b
x lnb b xln b b
lnx x xd e e e edx
2 ?xd edx
2
?xd edx
lnx x xd e e e edx
2 22x xd e edx
2 2
2x xd e x edx
xxdxd cossin
CCU Physics 2 - 6
cos sind x xdx
1n
nd x n xd x
log 1 logaa
d x edx x
log lne x x
1lnd xdx x
( ) ( ) ( )x t v t a t
( ) ( ) ( )a t v t x t? ?
dtdx
dtdv
CCU Physics 2 - 2P. 20
Motion of Constant Velocity (等速度)
0)( vtv
0 ( )f ix v t t it ft
0v
v(t) x(t)
consttxtv
)(
Displacement is the area under the velocity vs. time plot.
0( ) ( ) ( )f i f ix t x t v t t
0 ( )f i f ix x v t t Or,
)(tv
it ft
f i
f i
v va
t t
2
2i f i f iax v t t t t
212
f ii f i f i
f i
v vv t t t t
t t
2f i
i f i f i
v vv t t t t
2f i
f i
v vt t
Motion of Constant Acceleration (等速度)
)(tv
it ft
f i
f i
v va
t t
2
2i f i f iax v t t t t
In fact,
212
f ii f i f i
f i
v vv t t t t
t t
2f i
i f i f i
v vv t t t t
2f i
f i
v vt t
Motion of Constant Acceleration (等速度)
)(tv
at bt?x
General Motion
it ft
)(tv
1v2v
t
tvtvtvx N 21
Nxxxx 21
1x
2xNx
Division:
nn
xvt
,n nx v t
n nn n
x x v t
0, ( )dxt v v tdt
( ) ,f
i
t
f i tx x v t dt
it ft
)(tv
1v2v
t
( )f
i
t
f i tx x v t dt
0, 1(N )
lim ( )f
i
N t
n tt nx v t v t dt
( )f
i
t
tx v t dt
Next,
f ix x x
( )f
i
t
f i tx x v t dt since
and ,d xvd t
In general, if )()( xgdx
xdf
)()()()( ab
x
x
x
xxfxfdx
dxxdfdxxg b
a
b
a
( )f
i
t
tv t dt is an inverse operation of .d x
d t
, ( ) ( )i
t
i tor x t x v t dt
therefore
For example, if we know x(t) = t3, then
2( )( ) 3dx tv t tdt
Therefore, if v(t) =3t2, then according to
0( ) (0) ( ) dt
tx t x v t
We see that 3 2
0( ) (0) 0 3
tx t x t t dt
2 3
03
tt dt t
( )f
i
t
tv t dt is an inverse operation of .d x
d t
1
1
nnd t t
dt n
1 11
1 1 1
b n nnb n b aa
a
t ttt dtn n n
How to calculate ?( )b
a
t
tf t dt
CCU Physics 2 - 22
nn
xnx
dxd
1
1 b
a
nb
a
n
nxdxx
1
1
1lnd xdx x
1 ln
b b
aadx x
x
xx eedxd
b
a
xb
a
x edxe xx
dxd cossin b
a
b
axdxx sin)(cos
xxdxd sincos
b
a
b
axdxx cos)(sin
Table
CCU Physics 2 - 24
(x>0)
