Ch 2. Motion in a Straight Line Definitions 1. Kinematics - Motion Kinetic Energy - Energy...
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Transcript of Ch 2. Motion in a Straight Line Definitions 1. Kinematics - Motion Kinetic Energy - Energy...
Ch 2. Motion in a Straight Line
Definitions1. Kinematics - Motion
Kinetic Energy - Energy associated with motion2. Motion in physics is broken down into categories
a.) Translational Motion - motion such that an object moves from one position to another along a straight line.
b.) Rotational Motion - motion such that an object moves from one position to another along a circular path.
c.) Vibrational Motion - motion such that an object moves back and forth in some type of periodicity.
One Dimensional (x-axis only)
“Dinophysics : Velocity-Raptor”
Straight Line
Spinning
Motion in 3-D space can be complicated
Up and Back
Example: Diatomic Molecule Moving Through Space.
Rotational
Vibrational
Translational
i
f
Net Translation
X - Dir
Note: In this chapter all objects are going to be considered POINT PARTICLES – No Spatial Extent – No Rotations – No Vibrations
Speed
1. Speed - How fast an object is moving regardless of what direction it is moving.
Distance TraveledSpeed = Change in time
One way travel = 130 mi.Total Distance Traveled = 260 mi.Total time elapsed = 5.2 hrs. or (5 hrs 12 min)
Example 1 Traveling from your parking space at Conestoga to New York City and back to Conestoga. Find your average speed for the round trip.
x(mi)
y(mi)
up
back
NY
Conestoga 970
≡
Equality by Definition
Speed Calculations are EASY Always distance / time
Round Trip Average Speed
26050
5.2miles
hr
milesspeed
hr
v
Displacement - Change in position (straight line distance with direction) Must specify a coordinate system.
x = Change in x = xf
xi
m- 2 m = + 4 m x 6
∆ “Delta”
Delta x is the displacement or change in the x position
Example: Cartesian coordinate system
up
back
Mathematical Notation for Direction
x(m)
y(m)
xi= 2m
x1
xf= 6m
x2
xi = xinitial= Initial Position
xf = xfinal = Final Position
Average Velocity
Average Velocity = Change in positionChange in time
Avg. Velocity - How fast an object is moving and in what direction it is moving.
f i
avgf i
x xxv
t t t
≡
Equality by Definition
Notation for Displacement & Velocity = x “hat”, and has a value of one. The sole purpose of is to indicate the
directionExample Problem: A particle initially at position x = 5 m at time t= 2 s moves to position x = -2 m and
arrives at time t = 4 s.a.) Find the displacement of the particle.b.) Find the average speed and velocity of the particle.
x̂ x̂
x(m)
y(m)Given:
5 @ 2
2 @ 4i m i s
f m f s
x t
x t
a.)
ˆ ˆ ˆ[ 2 ( 5 )] 7
f i
m m
x x x
x x x x
ix
fx x
ˆ7ˆ3.5
2ms
avg
mavg
s
xv
tx
v x
distanceb.)
t7
3.52
ms
speed
mspeed
s
v
v
Example Problem 1 revisitedExample 1. Traveling from your parking space at Conestoga to New York City
and back to Conestoga. The straight line distance from Conestoga to Y is 97 mi.
One way travel = 130 mi.
Total Distance Traveled = 260 mi.
Travel time Con. to NY = 2.6 hrs.
Travel time NY to Con. = 2.6 hrs. x(mi)
y(mi)
up
back
NY
Conestoga 970
a.) What was the avg speed from Conestoga to NY?
b.) What was the avg velocity from Conestoga to NY?
c.) What was the avg speed for the round trip?
d.) What was the avg velocity for the round trip?
Speed in PATH DEPENDENT. Velocity is PATH INDEPENDENT. It only depends on the initial and final positions.
1302.6 50miles
mileshr hr
speedv
ˆ972.6
ˆ37.3mimi
hr hr
xavgv x
2605.2 50miles
mileshr hr
speedv
ˆ05.2 0mi
mihr hr
xavgv
Scalar vs. Vector QuantitiesScalar - Quantity that has magnitude only.
- Mass - Speed- Length - Energy
Vector - A quantity that has both magnitude and direction.- Position - Acceleration- Velocity - Forces
A number (with units) that describes how big or small
x(m)21 3 5 640-2 -1-3-4-5-6
Pt. APt. B
Example: Length vs. Position
= x “hat”, and is called a unit vector in the x-direction. It has a magnitude of one (hence the name unit) and is used solely to specify direction.
x̂
Scalar Vector
ˆ3mA x
ˆ4
4m
m
B x
B
3mA
Concepts Check – The Negatives
Q. Can velocity be negative?
A . YES! Negative displacement means an object moved backwards.
E.g. An object with a displacement ∆x of moved backwards 10m.ˆm -10 x
A. NO! – The least speed an object can have is zero – it is at rest
Q. Can speed be negative?
Q. Can displacement be negative?
A. NO! – The least distance an object can move is zero – it is at rest
Q. Can distance be negative?
A . YES! Negative velocity means an object is moving backwards.
E.g. An object moving is moving backwards with a speed of 10 m/sˆms-10 x
Position vs. Time Graph
Position vs. Time
0
5
10
15
20
25
30
0 1 2 3 4 5 6
Time (sec)
X -
Po
sit
ion
(m
)
x Y1 Y2
Time (s) Position (m)
Position (m)
0 0 0
1 1 5
2 4 10
3 9 15
4 16 20
5 25 25
Both of these movements describe an object moving in one dimension along the x-axis! NOT up and to the right!
x=riset=run
For any time interval
SLOPE is Avg. Velocity
x riseavg t runv slope
0 25m20m15m10m5m
Movement 1
Movement 2
● ● ● ● ●
● ● ● ● ●
1s 2s 3s 4s 5s
1s 2s 3s 4s 5s
Movement 1Movement 2
Position vs. Time Graph for a Complete Trip
A
B C
D
E
F
-150
-100
-50
0
50
100
150
200
250
300
0 10 20 30 40 50 60
Time (s)
Po
sit
ion
(m
)
x y
Time (s) Position (m)
0 0
10 200
20 200
25 150
45 -100
60 0
Find the average velocity as the object moves from:a.) A to B b.) B to Cc.) C to D d.) A to E
200 0
10 0a.) 20m
mss
xA B tv Slope
0 0
20 10b.) 0, Stoppedm
s
xB C tv
150 200
25 20c.) 10m
mss
xD E tv
100 0
45 0d.) 2.22m
mss
xA E tv
Slope of the secant line is vavg
Velocity vs. Time (Constant Velocity)
Velocity Function
0
1
2
3
4
0 1 2 3 4 5
t(sec)
v (m
/s)
Position Function
0
1
2
3
4
5
6
7
8
0 1 2 3 4 5
t(sec)
x(m
)
t
avg
xv
t
rise Sloperun
Area
x
x
Slope
rise
run
v = height
∆t = base( ) ( )x v t height base area
A
B C
D
E
F
-150
-100
-50
0
50
100
150
200
250
300
0 10 20 30 40 50 60
Time (s)
Po
sit
ion
(m
)
A B
B C
CD
D E
E F
-25
-20
-15
-10
-5
0
5
10
15
20
25
0 10 20 30 40 50 60
Time (s)
Ve
loc
ity
(m
/s)
Velocity vs. Time Graph for a Complete Trip
+X
− X
Velocity vs. Time Graph for a Complete Trip
Area = 200m
Area =
-50m
Area = -250m
∆t = 1.5 sec
0
10
20
30
40
50
60
0 1 2 3 4
t(sec)
x(m
)
●
●●
Instantaneous Velocity recall: f iavg
f i
x t x txˆv x
t t t
(Average velocity)
x t m + 10 m
s t - 0.5
m
s t
22
44
3 avg
50.5 - 35.0ˆv 10.3 x
3.5 - 2.0ms
m m
s s
mavg
39.7 - 35.0ˆv 23.5 x
2.2 - 2.0ms
m
s s
Consider the function x(t): A.
B. ∆t = 0.2 sec
The instantaneous velocity at the time t = ti is the limiting value we get by letting the upper value of the tf approach ti.
Mathematically this is expressed as:
The velocity function is the time derivative of the position function . Differentiation (Calculus)
v t
dX t
dtlim
X t X t
t tt f ti
f i
f i
v t X t
f iavg
f i
v vv ˆa xt tt
avg 2
m/s ma s s
t 3(t)v 23
s
m
6 t 12 s s
AccelerationWhen the instantaneous velocity of a particle is changing with time, the particle is accelerating
Units:
Example: If a particle is moving with a velocity in the x-direction given by
a.) What is the average acceleration over the time interval
(Average Acceleration)
2 2
2
3243(12) 3(6)(12 6) 6 54
f i
f i
ms
ms
v vvavg t t t
avg s
a
a
Example: Instantaneous Acceleration
a.) Find aavg. over the time interval 5 t 8
b.) What is the acceleration at time t = 6 s ?
c.) What is the acceleration when the velocity of the particle is zero?
Velocity vs. Time
-15
-10
-5
0
5
10
15
20
25
30
35
0 2 4 6 8 10
time (s)
ve
loc
ity
(m
/s)
time (s) vel. (m/s)0 -101 -22 -53 54 125 146 127 218 30
2
30 148 5 5.3 m
s
vavg ta
Stopped
Slope of tangent – pick 2 points on the tangent line.Answer will we smaller than the answer to part a, 2(5.3 ).m
s
Objects with zero velocity can be accelerating!0Slope
Positive and Negative Accelerations
v(m/s)
t (s)
A
C
E
D
B
FA→B:
B→C:
C→D:
D→E:
E→F:
Moving Forward
Stopped 0
Moving Backward
Slowing Down, Moving Backward, Pt. B=Stop( ) ( )v negative a slope positive
Slowing Down, Moving Forward, Pt. E=Stop( ) ( )v Positive a slope negative
( ) ( )v Positive a slope positive Speeding Up, Moving Forward
( ) v positive a slope ZERO Constant Speed, Moving Forward
( ) ( )v negative a slope negative Speeding Up, Moving Backward
Special Case: Constant Acceleration
a(m/s2)
t (s)ti = 0 tf = t
a
0
We make the assumption that the acceleration does not change. Near the surface of the earth, (where most of us spend most of our time) the acceleration due to gravity is approximately constant ag =
9.8 m/s2
Area! Slope!
v(m/s)
t (s)ti = 0 tf = t
vf
0
x(m)
t (s)ti = 0 tf = t
xf
xi
Area!Slope!
vi
v = v + a tf i
x x + v tf i i2= +
1
2a t
1.
2.
v( )
(
vt
f i f i
a v a t height base
v v a t t
0
)
y b mx
1x2x
1 2
12 ( )i f i
x x x
x v t t v v
12 ( )ix v t t at
xavv if 2223.
Solving for the 3rd constant acceleration equationSolve equation 1 for t and substitute t into equation 2 to get the following equation.
f iv vt
a
212ix v t at
212
f i f iv v v v
i a ax v a
22 21
2 2f i iv v v
f f i ia ax v v v v
2 2 f ia x v v 2 22 2i f f iv v v v 2iv
2 2
2 2
2
2
i f
f i
a x v v
v v a x
tf t/2
y(t)
tf t/2
FREE-FALL ACCELERATION (9.8 m/s2 = 32 ft/s2)Consider a ball is thrown straight up.It is in “Free Fall” the moment it leaves you hand.Plot y(t) vs. t for the example above.
Plot v(t) vs. t
2Why? Because 9.8
from the moment it leaves
your hand.
ms
a
iv
20 ftv t
ivGround Level
212yf i iy y v t at
top
Moving up
Stopped 0
Moving down
f iv v at
x(t)
Area Under Curve
Slopev(t)
a(t)
FINAL NOTES ON CH 2.Remember , when going between the following graphs
Problem Solving with the constant acceleration equations1.Write down all three equations in the margin2.a = 9.8 m/s2 for free fall problems3.Analyze the problem in terms of initial and final sections.