Lecture 04 Chap 2&3

Lecture 04Chap 2&3

10/3/2021

甘宏志, 物理館 416 室 , phyhck@ccu.edu.tw

C. 2si..

. s..

cos $

$

sin $$

98%

2%0%0%

( ) ( ) ( )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