8/29/2013PHY 113 C Fall 2012 -- Lecture 21 PHY 113 C General Physics I 11 AM – 12:15 PM TR Olin...
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8/29/2013 PHY 113 C Fall 2012 -- Lecture 2 1 PHY 113 C General Physics I 11 AM – 12:15 PM TR Olin 101 Plan for Lecture 2: Chapter 2 – Motion in one dimension 1.Time and Position 2.Time and Velocity 3.Time and Acceleration 4.Special relationships for constant
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Transcript of 8/29/2013PHY 113 C Fall 2012 -- Lecture 21 PHY 113 C General Physics I 11 AM – 12:15 PM TR Olin...
PHY 113 C Fall 2012 -- Lecture 2 18/29/2013
PHY 113 C General Physics I11 AM – 12:15 PM TR Olin 101
Plan for Lecture 2:
Chapter 2 – Motion in one dimension
1. Time and Position
2. Time and Velocity
3. Time and Acceleration
4. Special relationships for constant acceleration
PHY 113 C Fall 2012 -- Lecture 2 28/29/2013
Some updates/announcements
PHY 113 C Fall 2012 -- Lecture 2 38/29/2013
iclicker exercises:Webassign Experiences so far
A. Have not tried itB. Cannot loginC. Can loginD. Have logged in and have completed one or more
example problems.
Textbook Experiences
E. Have no textbook (yet)F. Have complete physical textbookG. Have electronic version of textbookH. Textbook is on orderI. Other
PHY 113 C Fall 2012 -- Lecture 2 48/29/2013
Comment on Webassign #1 problem:
7. A pet lamb grows rapidly, with its mass proportional to the cube of its length. When the lamb's length changes by 16%, its mass increases by 16.8 kg. Find the lamb's mass at the end of this process.
Problem solving steps1. Visualize problem – labeling
variables2. Determine which basic physical
principle(s) apply3. Write down the appropriate
equations using the variables defined in step 1.
4. Check whether you have the correct amount of information to solve the problem (same number of knowns and unknowns).
5. Solve the equations.6. Check whether your answer makes
sense (units, order of magnitude, etc.).
PHY 113 C Fall 2012 -- Lecture 2 58/29/2013
7. A pet lamb grows rapidly, with its mass proportional to the cube of its length. When the lamb's length changes by 16%, its mass increases by 16.8 kg. Find the lamb's mass at the end of this process.
L
p L
LL
M MM
kLMkLM
i
if
ff
ffii
16.0
kg 8.16
33
kg 8.4611.16
1.16 kg 8.16
1/
/
1/
/ :clearsdust When the
3
3
3
3
3
3
33
3
if
iff
if
if
if
ff
LL
LLMM
LL
LL
kLkL
kL
M
M
PHY 113 C Fall 2012 -- Lecture 2 68/29/2013
i>clicker exercise:What did you learn from this problem?
A. It is a bad idea to have a pet lambB. This was a very hard questionC. I hope this problem will not be on an examD. My physics instructor is VERY meanE. All of the above.
PHY 113 C Fall 2012 -- Lecture 2 78/29/2013
Motion in one-dimension
PHY 113 C Fall 2012 -- Lecture 2 88/29/2013
Graphical representation of position (displacement) x(t)
t (s)
PHY 113 C Fall 2012 -- Lecture 2 98/29/2013
Comment: Your text mentions the notion of a “scalar quantity” in contrast to a “vector quantity” which will be introduced in Chapter 3. In most contexts, a scalar quantity – like one-dimensional distance or displacement can be positive or negative.
Another comment: Your text describes the time rate of change of displacement as “velocity” which, in one-dimension is a signed scalar quantity. In general “speed” is the magnitude of velocity – a positive scalar quantity.
PHY 113 C Fall 2012 -- Lecture 2 108/29/2013
Graphical representation of position (displacement): x(t) time rate of change of displacement = velocity: v(t)
t (s)
dt
dx
t
txttxtv
Δt
)()()(
:velocity
lim0
4 20
4040
2040
)20()40()(
m/s-s
mm--ss
xxtv
PHY 113 C Fall 2012 -- Lecture 2 118/29/2013
Instantaneous velocity
t (s)
vA
vB
vC
vE
vF
vD
PHY 113 C Fall 2012 -- Lecture 2 128/29/2013
Demonstration of tangent line limit
dt
dx
t
txttxtv
Δt
)()()(
: velocityousInstantane
lim0
PHY 113 C Fall 2012 -- Lecture 2 138/29/2013
Average velocity versus instantaneous velocity
dt
dx
t
txttxtv
Δt
)()()(
: velocityousInstantane
lim0
AB
AB
AB
t
tB
A
tt
txtx
tt
dttv
v
B
A
)()(
)(
: velocityAverage
PHY 113 C Fall 2012 -- Lecture 2 148/29/2013
Average velocity
AB
ABB
A tt
txtxv
)()(
:result stated Previouslyiclicker exercise:
This results is:
A. ExactB. Approximate
AB
AB
AB
t
t
AB
t
t
AB
t
t
N
ii
B
A tt
txtx
tt
tdx
tt
dtdt
tdx
tt
dttv
N
tvv
B
A
B
A
B
A
)()()(
)()()(
:Why
1
PHY 113 C Fall 2012 -- Lecture 2 158/29/2013
Webassign Example
PHY 113 C Fall 2012 -- Lecture 2 168/29/2013
Instantaneous velocity using calculus
2765)(
:Suppose42 ttttx
38146)( ttdt
dxtv
x
v
PHY 113 C Fall 2012 -- Lecture 2 178/29/2013
Anti-derivative relationship
:m/s 3.0 and 0 suppose -- Example
)( :Then
:Suppose
motionelocity Constant v
00
00
0
vx
tvxtx
vdt
dx
t
x
PHY 113 C Fall 2012 -- Lecture 2 188/29/2013
Velocity
AB
ABB
A tt
txtxv
dt
dxtv
)()(
: velocityAverage
)(
: velocitysInstanteou
PHY 113 C Fall 2012 -- Lecture 2 198/29/2013
Acceleration
AB
ABB
A tt
tvtva
dt
xd
dt
dx
dt
d
dt
dvta
)()(
:onaccelerati Average
)(
:onaccelerati sInstanteou
2
2
PHY 113 C Fall 2012 -- Lecture 2 208/29/2013
Rate of acceleration
4
4
3
3
3
3
2
2
2
2
3
3
2
2
:onaccelerati of rate of rate sInstanteou
:onaccelerati of rate sInstanteou
dt
xd
dt
dx
dt
d
dt
vd
dt
dv
dt
d
dt
ad
dt
da
dt
d
dt
xd
dt
dx
dt
d
dt
d
dt
vd
dt
dv
dt
d
dt
da
iclicker exerciseHow many derivatives of position are useful for describing motion:
A. 1 (dx/dt)B. 2 (d2x/dt2)C. 3 (dx3/dt3)D. 4 (dx4/dt4).E
PHY 113 C Fall 2012 -- Lecture 2 218/29/2013
Anti-derivative relationship
202
100
00
000
)(
)( :Then
0 ,0 and :Suppose
motionon acceleratiConstant
tatvxtx
tavtv
x)x(v)v(adt
dv
PHY 113 C Fall 2012 -- Lecture 2 228/29/2013
Examples
2000
202
10000
m/s 8.9 5m/s 0
)(
avx
tatvxx(t)tavtv
v(t)
x(t)
PHY 113 C Fall 2012 -- Lecture 2 238/29/2013
Examples
2000
202
10000
m/s 8.9 5m/s 0
)(
avx
tatvxx(t)tavtv
v(t)
x(t)
PHY 113 C Fall 2012 -- Lecture 2 248/29/2013
Summary
t
t
t
t
dttatvdt
dvta
dttvtxdt
dxtv
0
0
')'()( )(
')'()( )(
PHY 113 C Fall 2012 -- Lecture 2 258/29/2013
x(t)
v(t)
Another example
PHY 113 C Fall 2012 -- Lecture 2 268/29/2013
From webassign:
mssmsxst
smssmsvst
255/22
1)5( :5at position
/105/2)5( :5at velocity
22
2
PHY 113 C Fall 2012 -- Lecture 2 278/29/2013
Another example
t
t
dttvtx0
')'()(
m
sx
453302
1
)3(
m
sx
sx
105
230
)3(
)5(
(m/s)
(s)
PHY 113 C Fall 2012 -- Lecture 2 288/29/2013
Another example -- continued
dt
tdvta
)()(
(m/s)
(s)
2m/s 1003
030
03
)0()3()(
vv
ta
2m/s 035
00
35
)3()5()(
vv
ta
PHY 113 C Fall 2012 -- Lecture 2 298/29/2013
Special relationships between t,x,v,a for constant a:
t
t
t
t
dttatvdt
dvta
dttvtxdt
dxtv
0
0
')'()( )(
')'()( )(
:iprelationsh General
200
2
0
2
1
2
100)(
0)(
:iprelationsh Special
attvxattvxtx
atvatvtv
PHY 113 C Fall 2012 -- Lecture 2 308/29/2013
Special relationships between t,x,v,a for constant a:
20
20
0
200
0
2
1)(
2
1)(
)(
:iprelationsh Special
vtva
xtxa
vtvt
attvxtx
atvtv
PHY 113 C Fall 2012 -- Lecture 2 318/29/2013
Special case: constant velocity due to earth’s gravity
In this case, the “one” dimension is the vertical direction with “up” chosen as positive:
a = -g = -9.8 m/s2
y(t) = y0 + v0t – ½ gt2
y0 = 0 v0 = 20 m/s
At what time tm will the ball hit the ground ym = -50m ?
Solve: ym = y0 + v0tm – ½ gtm2
quadratic equation physical solution: tm = 5.83 s
PHY 113 C Fall 2012 -- Lecture 2 328/29/2013
Helpful mathematical relationships(see Appendix B of your text)
g
yygvvt
yytvgtt
gttvyy
mm
mmmm
mmm
02
00
002
200
2
02
1:for equation Quadratic
2
1
