ch02 motion - eunil wonparticle.korea.ac.kr/class/2006/phys183-01/ch02_med.pdfa youngster throws a...

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Eunil Won Dept. of Physics, Korea University 1 Ch02 Motion

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Eunil Won Dept. of Physics, Korea University 1

Ch02 Motion

Eunil Won Dept. of Physics, Korea University 2

Displacement

position: x

1x 2x

displacement 12 xxx −≡∆

Eunil Won Dept. of Physics, Korea University 3

Eunil Won Dept. of Physics, Korea University 4

Average speed, instantaneous speed

tx

ttxxv

∆∆=

−−≡

12

12

dtdx

ttxttxv

t=

∆−∆+≡

∞→∆

)()(lim

Eunil Won Dept. of Physics, Korea University 5

Acceleration

2

2

limdt

xddtdx

dtd

dtdv

tva

t=⎟

⎠⎞

⎜⎝⎛==

∆∆≡

∞→∆

Eunil Won Dept. of Physics, Korea University 6

Eunil Won Dept. of Physics, Korea University 7

Examples 2.1/2.2/2.3

Eastminkm40.0

min15East km0.6 ==v Eastkm/h 24

hmin60

minkm40.0 =×=v

Westsh

km0.6Wests15

km/h 90⋅

==∆∆=

tva

Westm/s67.1km1

m10s3600

h1sh

km0.6 23

=××⋅

=a

38smin 4s1078.2m/s4.5

m1500 2 ⇒×===vlt

Eunil Won Dept. of Physics, Korea University 8

Examples 2.4/2.5/2.6

20

m/s08

m/s00

.avt

=

==

?s75.0

==

vt constant=v

20

m/s2

??

−=

==

avt

velocity

Time elapsed

Distance run

m/s0.6)75.0)(0.8(00 =+=+= atvv

s0.32

0.600 =−−=−=

avvt

m0.9)0.3)(0.2()0.3)(0.6( 2212

21

0 =−+=+= attvs

Eunil Won Dept. of Physics, Korea University 9

Example 2.7

24

20

2

m/s1028.1)400)(16(20

2

×=+=

+= asvv

A constant accelerating vehicle (a=16 m/s2) is started from rest. What is the instantaneous velocity after traveling 400 m?

Eunil Won Dept. of Physics, Korea University 10

Constant accelerating motion

.consdtdva ==

∫ ∫ ∫== dtaadtdv

dtdxv =

∫ ∫ ∫ +== dtatvvdtdx )( 0

221

00 attvxx ++=

atvv += 0

Eunil Won Dept. of Physics, Korea University 11

Acceleration of gravity2m/s8.9−=g

Eunil Won Dept. of Physics, Korea University 12

Example 2.8

m/s100 =v

m10.5)8.9(2)10(0

2

222

02

20

2

=−

−=−=

+=

avvs

asvv

Distance traveled upward

Time elapsed

s02.18.9

1000 =−−=−=

avvt

Eunil Won Dept. of Physics, Korea University 13

Example 2.9

s50.38.9

)0.60(2

2

0

2/1

2212

21

0

=⎥⎦⎤

⎢⎣⎡

−−=

=

+=+=

ast

atattvs

m1.44)0.3)(8.9(

m6.19)0.2)(8.9(

m9.4)0.1)(8.9(

0

2212

321

3

2212

221

2

2212

121

1

2212

21

0

−=−==

−=−==

−=−==

+=+=

ats

ats

ats

atattvs

A rock is dropped into a 60.0 m deep well.

(a) How long will it take the rock to reach bottom?

(b) Calculate the position of the rock 1.0, 2.0, and 3.0 sec after its release.

Eunil Won Dept. of Physics, Korea University 14

Example 2.10

2

2

20

m/s7.34

)50)(8.9(2)15(

2

−=

−−++=

+= asvv

2

2

20

m/s7.34

)50)(8.9(2)15(

2

−=

−−+−=

+= asvv

A youngster throws a rock from a bridge into the river 50m below. The rock has a speed of 15 m/s when it leaves the youngster. Calculated the velocity of the rock when it strikes the water (a) if it is thrown straight upward, (b) if it is thrown straight downward