chapter 2ww2.odu.edu/~agodunov/teaching/notes231/chapter_02.pdf · motion along a straight line 2...
TRANSCRIPT
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Chapter 2
Motion along a straight line
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Kinematics & Dynamics
Kinematics:
Description of Motion without regard to its cause.
Dynamics:
Study of principles that relate motion to its cause.
Basic physical variables in kinematics and dynamics:
Time
Position
Displacement
Velocity
Acceleration
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Part 1
Time, Position, Velocity, Acceleration
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Time
Time: that part of existence which is measured in seconds, minutes, hours, days, weeks, months, years, etc., or this process considered as a whole
Cambridge Dictionary
Time: something that is measured in minutes, hours, years etc using clocks
Clocks: an instrument in a room or in a public building that shows what time it is
Longman Dictionary of Contemporary English
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Time
”Time is like a river flowing.” – Newton
”Zeit ist wass enie Uhr mass.” – Einstein
(”Time is what a clock measures.”)
“Time is space between events” – Feynman
"Time is that quality of nature which keeps events from happening all at once” – Anonymous
Time is neither young nor old
a beginning or an end,
now or forever;
time is the empty distance in between.
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Position and Displacement
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Trajectory
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Velocity
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Velocity and Speed
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Instantaneous velocity
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Acceleration
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Analytic representation
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Velocity and acceleration
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One-Dimensional (1-D) Motion
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Part 2
Motion with constant Acceleration
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Constant Acceleration
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Calculating position
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Summary table of results
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1D motion with constant acceleration a
⎪⎩
⎪⎨⎧
++=
+=2
00
0
21 attvxx
atvv the system of equations with six variables
max: two unknowns
case 1: we may eliminate time from the system
)(2 020
2 xxavv −+=
case 2: we may eliminate acceleration from the system
tvvxx )(21
00 ++=
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example ⎪⎩
⎪⎨⎧
++=
+=2
00
0
21 attvxx
atvv
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example ⎪⎩
⎪⎨⎧
++=
+=2
00
0
21 attvxx
atvv
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example ⎪⎩
⎪⎨⎧
++=
+=2
00
0
21 attvxx
atvv
)(2 020
2 xxavv −+=
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problem 2.22 ⎪⎩
⎪⎨⎧
++=
+=2
00
0
21 attvxx
atvv
The catapult of the aircraft carrier USS Abraham Lincoln accelerates an F/A-18 Hornet jet fighter from rest to a takeoff speed of 173 mph in a distance of 307 ft. Assume constant acceleration.
1. Calculate the acceleration of the fighter in m/s.
2. Calculate the time required for the fighter to accelerate to takeoff speed.
Given: calculations
)(2 020
2 xxavv −+=
smmphvmftx
mxsmv
/33.770.17357.93307
0.0/0.0
0
0
====
== sm
xxvva /0.32
)(2 0
20
2
=−−
=
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problem 2.36 ⎪⎩
⎪⎨⎧
++=
+=2
00
0
21 attvxx
atvv
At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.2 m/s2 . At the same instant a truck, traveling with a constant speed of 20.0 m/s , overtakes and passes the car.How far beyond its starting point does the car overtake the truck?
Given:
x0 = 0.0 mv0t= 20.0 m/sv0c= 0.0 m/sxc = xtat = 0.0 m/s2
ac= 3.2 m/s2
)(2 020
2 xxavv −+=
c
tt
c
t
cctt
avx
avt
taxtvx
20
0
20
2
25.0
=
=
===
2 objects = 4 equationstime is the same
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Part 3
Free Fall
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Free fall ⎪⎩
⎪⎨⎧
−+=
−=2
00
0
21 gttvyy
gtvv
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example
my 0.300 =
smv /0.100 =
⎪⎩
⎪⎨⎧
−+=
−=2
00
0
21 gttvyy
gtvv
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Solution (cont)
my 0.300 =
smv /0.100 =
⎪⎩
⎪⎨⎧
−+=
−=2
00
0
21 gttvyy
gtvv
)(2 020
2 xxavv −+=
![Page 15: Chapter 2ww2.odu.edu/~agodunov/teaching/notes231/Chapter_02.pdf · Motion along a straight line 2 Kinematics & Dynamics ... accelerates an F/A-18 Hornet jet fighter from rest to a](https://reader033.vdocuments.us/reader033/viewer/2022052407/5b5c134b7f8b9ac8618b8925/html5/thumbnails/15.jpg)
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The second question
my 0.300 =
smv /0.100 =
⎪⎩
⎪⎨⎧
−+=
−=2
00
0
21 gttvyy
gtvv
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Solution for the second question
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Discussion of result for question 2
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The third question
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Being practical: 1D motion with const. a
⎪⎩
⎪⎨⎧
++=
+=2
00
0
21 attvxx
atvv1. initial value problem:knowing initial conditions (x0,v0) and acceleration a, one may find (x,v) at any moment in time.
2. “who cares about time”then the system of equation can be reduced to a simple equation
3. two object problemin this case we need co consider a system of 4 equation with the same time variables. In most problems we need only a couple equations from the system
)(2 020
2 xxavv −+=
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
++=
+=
++=
+=
2202022
2022
2101011
1011
21
21
tatvxx
tavv
tatvxx
tavv
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example
The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. The freight train is traveling at 15.0 m/s in the same direction as the passenger train.
The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of -0.10 m/s2, while the freight train continues with constant speed.
1 .Will the cows nearby witness a collision?
2. If so, where will it take place?
21
01
01
/1.0
/0.250.0
sma
smvmx
−=
==
22
02
02
/0.0
/0.150.200
sma
smvmx
=
==
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⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
++=
+=
++=
+=
2202022
2022
2101011
1011
21
21
tatvxx
tavv
tatvxx
tavv
21
01
01
/1.0
/0.250.0
sma
smvmx
−=
==
22
02
02
/0.0
/0.150.200
sma
smvmx
=
==
⎪⎪
⎩
⎪⎪
⎨
⎧
+==
++=
+=
tvxxvv
tatvx
tavv
02022
022
21011
1011
210.0
tvxtatv
xxmeanscollision
02022
101
21
21
+=+
=
solutions:
t1 = 22.5 s
t2 = 177.0 s
negative solutions?