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C. 2si..
. s..
cos $
$
sin $$
98%
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( ) ( ) ( )x t v t a t
( ) ( ) ( )a t v t x t? ?
dtdx
dtdv
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
nn
xnx
dxd
1
1 b
a
nb
a
n
nxdxx
1
1
1lnd xdx x
1 ln
b b
aadx xx
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)
Change of variable for Integration( ) ( ) ( ) ( )du x g x du x g x dx dudx
Example: 2( ) ,u x x2( ) ( ) 2 2du x d x x du x dx
dx dx
2( ) sin( ),f x x2 2
22
2 2 2cos( )2 cos( ) sin( ) sin( ) sin( )b b b
aa ax x dx u du u b a
Application for integration:2( ) ,u x x
sintancos
b b
a a
xx dx dxx
( ) cos ,u x x ( ) sin ,du x x dx
cos cos
coscos
sintan lncos
b b b b
aa a a
xdx dux dx ux u
ln cos ln cosb a coslncosba
22
0
xe xdx2
2
0
1 22
xe xdx 2( )u x x
( ) 2du x xdx
2du x dx
4
0
12
ue du 4 4
0
1 1 ( 1)2 2
ue e
2. Find the answer for the following.2
1
1( )2 5
t
tb dt
t
3 5 2
0( ) xc e dx
0( ) sin(3 )
2a x dx
(d)6
5
32 9
dxx
2. Find the answer for the following.2
1
1( )2 5
t
tb dt
t
3 5 2
0( ) xc e dx
0( ) sin(3 )
2a x dx
(d)6
5
32 9
dxx
52102
3 ;2 3
1sin(3 ) sin2 3
1 5 1cos co2
0s( )3 2
duu x dx
x dx udu
22
11
2
1
1 1 1(2 ) ln 2
1 2 5
5
ln
2 2 5 2
2 2 5
tt
tt
d t tt
tt
2 13
13
2
1 ( )5
5 215
15
u
u x
dx du
du
e e
e
162
5
12
65
3 2 9
3 (
3 3 3
2 9)2 1 / 2
xx
x dx
x
( ) ( ) ( )a t v t x t
For general 1D motion,
0( ) (0)
(0)
tv t v adt
v a t
0
2
( ) (0) ( )
1(0) (0)2
tx t x v t dt
x v t at
t
xa) it speeds up all the timeb) it slows down all the timec) it moves at constant
velocityd) sometimes it speeds up and
sometimes it slows downe) not really sure
The graph of position vs. time for a car is given
below. What can you say about the velocity of the
car over time?
PRS 2-6
t
xa) it speeds up all the timeb) it slows down all the timec) it moves at constant
velocityd) sometimes it speeds up and
sometimes it slows downe) not really sure
The graph of position vs. time for a car is given
below. What can you say about the velocity of the
car over time?
bPRS 2-6
You toss a ball straight up in the air and catch it again. Right
after it leaves your hand and before you catch it, which of the above plots represents the v vs. t graph for this motion? (Assume your y-axis is pointing up).
v
td
v
tb
v
tc
v
ta
PRS 2- 7
You toss a ball straight up in the air and catch it again. Right
after it leaves your hand and before you catch it, which of the above plots represents the v vs. t graph for this motion? (Assume your y-axis is pointing up).
v
td
v
tb
v
tc
v
ta
dPRS 2- 7
v
ta
v
tb
v
tcv
td
You drop a very bouncy rubber ball. It falls, and then it hits
the floor and bounces right back up to you. Which of the
following represents the v vs. t graph for this motion?
PRS 2-8
v
ta
v
tb
v
tcv
td
You drop a very bouncy rubber ball. It falls, and then it hits
the floor and bounces right back up to you. Which of the
following represents the v vs. t graph for this motion?
dPRS 2-8
2-3 Instantaneous Velocity
A jet engine moves along an experimental track (which we call the xaxis) as shown. We will treat the engine as if it were a particle. Its position as a function of time is given by the equation x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m. (a) Determine the displacement of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s. (b) Determine the average velocity during this time interval. (c) Determine the magnitude of the instantaneous velocity at t = 5.00 s.
2-32 2 2
1 1 (2.10 m /s )(3.00 s) 2.80 mx At B
2 22 (2.10 m /s )(5.00 s) 2.80 m 55.3 mx
21.7 m
2x At B
22 2 2(2.10 m /s )(5.00 s) 21.0 m /sv At
2 1 55.3 m 21.7 m 33.6 mx x
2 1
2 1
33.6 m 16.8 m /s2.00 s
x x xvt t t
( )a
( )b
2( ) 2dx dv At B Atdt dt
( )c
1 2 ; ( 1) n ndx dvv n A t a n n A tdt dt
( ) nx t A t
Average acceleration (平均加速度)
f i
f i
v vvat t t
2
20lim ( )t
v dv d dx d xat d t d t d t d t
Instantaneous acceleration (瞬時加速度 )
Example :
CCU Physics 2 - 15P. 24
The position of a electron is plotted a function of time in Fig. . In the plot five position are selected. At which position(s) does the electron has maximum acceleration in the positive x-direction?
-1.0
-0.5
0.0
0.5
6543210
Posi
tion
Time (s)
(A) (B) (C) (D) (E) (F) (G) x
t
The position of a electron is plotted a function of time in Fig. . In the plot five position are selected. At which position(s) does the electron has maximum acceleration in the positive x-direction?
-1.0
-0.5
0.0
0.5
6543210
Posi
tion
Time (s)
(A) (B) (C) (D) (E) (F) (G) x
t
Ans B
( ) ( ) ( )x t v t a t
( ) ( ) ( )a t v t x t? ?
dtdx
dtdv
A time-varying acceleration problem.
An experimental vehicle starts from rest (v0 = 0) at t = 0 and accelerates at a rate given by a(t) = (7.00 m/s3)t. What is (a) its velocity and (b) its displacement 2.00 s later?
0( ) (0)
tv t v a dt
23 3
00
( ) (7.00 m / s ) (7.00 m / s )2
tt tv t t dt
2
3 3 2(7.00 m /s ) 0 (3.50 m /s )2t t
3 22.00 s, (3.50 m /s )(2.00 s) 14.0 m /st v At
( )a
t t
35
30
25
20
15
10
5
0
543210
80
60
40
20
0
543210
a(t) v(t)
0( ) (0)
tx t x v dt
2.00 s 3 2
0( ) (3.50 m / s )x t t dt
2.00 s33
0
(3.50 / ) 9.33 m3tm s
2.00 s, 14.0 m /st v At
( )b
9.33 mx and
a(t)
t
v(t)
t
35
30
25
20
15
10
5
0
543210
80
60
40
20
0
543210
140
120
100
80
60
40
20
0
543210
t
x(t)
Example : The acceleration of a particle can be expressed as a function of time below.
, 0 2(t)=
2, 2t t
at
Find the velocity of the particle as a function of time, given that v = 0 at t =0.
Example : The acceleration of a particle can be expressed as a function of time below.
, 0 2(t)=
2, 2t t
at
Find the velocity of the particle as a function of time, given that v = 0 at t =0.
5
4
3
2
1
0
543210
t
a(t)
, 0 2(t)=
2, 2t t
at
Solution: For 0≤ t <2,
0( ) (0) (t)dt
tv t v a
0( ) 0 dt
tv t t
2 2
0
( )2 2
tt tv t
5
4
3
2
1
0
543210
t
a(t)
For t ≥2, 2
0 0 2( ) ( ) ( ) ( )
t tv t a t dt a t dt a t dt
22
220
2 2 2 2 (2 4) 2 22
t tt dt t t t 2 2 , 0 2
(t)=2 2, 2t t
vt t
Therefore,
5
4
3
2
1
0
543210
8
6
4
2
0
543210
t
a(t)
v(t)
t
