Download Ch 2.1 and 2.2 PPT Chap 2.1 and 2.2

Chapter 2: One Dimension
Motion
2.1 – 2.4, 2.5, 2.6
Position
Distance and Displacement
Speed and Velocity
Acceleration
Motion equations
Problem-Solving
Objectives:
Student is able
1. 3.A.1.1: to express the motion od an object
using narrative, mathematical, and graphical
representations.
2. 3.A.1.3: To analyze experimental data
describing the motion of an object and to
express the result using above representation.
Learning Objective
• Define position, displacement, distance in a
particular frame reference.
• Distinguish between displacement and distance
• Distinguish between speed and velocity
• Define acceleration and uniform or non-uniform
motion.
• Solve problems involving initial and final velocity,
acceleration, displacement, and time.
How to present a motion
Motion can be described by
• Words
• Diagram, a graph, a picture
• equations
One dimension motion:
• Motion is along a straight line (horizontal,
vertical or slanted).
• The moving object is treated as though it
were a point particle.
• Particle model – representing object
For Example: long distance runner, an
airplane, and throwing a ball, etc
Picturing a Motion
You are free to choose the origin and positive
direction as you like, but once your choice is made,
stick with it.
Picturing Motion
The locations of your house, your friend’s house,
and the grocery store in terms of a onedimensional coordinate system.
Figure 2-3
One-dimensional motion along the x axis
What is a position?
• Location of an object in a particular time.
A teacher paces left and
right while lecturing.
Her position relative to
Earth is given by x .
The +2.0 m
displacement of the
teacher relative to Earth
is represented by an
arrow pointing to the
right.
Distance and Displacement
Distance is the length of the actual path
taken by an object. Consider travel from
point A to point B in diagram below:
s = 20 m
A
B
Distance s is a scalar
quantity (no direction):
Contains magnitude only
and consists of a
number and a unit.
(20 m, 40 mi/h, 10 gal)
Distance and Displacement
Displacement is the straight-line separation
of two points in a specified direction.
D = 12 m, 20o
A
q
B
A vector quantity:
Contains magnitude
AND direction, a
number, unit & angle.
(12 m, 300; 8 km/h, N)
Distance and Displacement
• For motion along x or y axis, the displacement is
determined by the x or y coordinate of its final
position. Example: Consider a car that travels 8 m, E
then 12 m, W.
Net displacement D is
from the origin to the
final position:
D = 4 m, W
What is the distance
traveled? 20 m !!
D
8 m,E
x = -4
x
x = +8
12 m,W
What is a displacement equation?
The Signs of Displacement
• Displacement is positive (+) or
negative (-) based on LOCATION.
Examples:
The displacement is
the y-coordinate.
Whether motion is
up or down, + or - is
based on LOCATION.
2m
-1 m
-2 m
The direction of motion does not matter!
It is usually convenient
to consider motion
upward or to the right
as positive ( + ) and
motion downward or
to the left as negative
(−).
Speed and velocity
• The motion of these racing snails can be described by
their speeds and their velocities.
Definition of Speed
• Speed is the distance traveled per unit of
time (a scalar quantity).
s = 20 m
A
Time t = 4 s
B
v=
s
t
=
20 m
4s
v = 5 m/s
Not direction dependent!
Definition of Velocity
• Velocity is the displacement per
unit of time. (A vector quantity.)
s = 20 m
A
D=12 m
20o
Time t = 4 s
B
D 12 m
v 
t
4s
v = 3 m/s at 200 N of E
Direction required!
Example 1. A runner runs 200 m, east, then
changes direction and runs 300 m, west. If
the entire trip takes 60 s, what is the average
speed and what is the average velocity?
Recall that average
s2 = 300 m
speed is a function
only of total distance
start
and total time:
s1 = 200 m
Total distance: s = 200 m + 300 m = 500 m
total path 500 m
Average speed 

time
60 s
Direction does not matter!
Avg. speed
8.33 m/s
Example 1 (Cont.) Now we find the average
velocity, which is the net displacement divided
by time. In this case, the direction matters.
v
x f  x0
t = 60 s
xf = -100 m
x1= +200 m
t
x0 = 0 m; xf = -100 m
100 m  0
v
 1.67 m/s
60 s
xo = 0
Direction of final
displacement is to
the left as shown.
Average velocity: v  1.67 m/s, West
Note: Average velocity is directed to the west.
Example 2. A sky diver jumps and falls for
600 m in 14 s. After chute opens, he falls
another 400 m in 150 s. What is average
speed for entire fall?
Total distance/ total time:
xA  xB 600 m + 400 m
v

t A  tB
14 s + 150 s
1000 m
v
164 s
v  6.10 m/s
Average speed is a function
only of total distance traveled
and the total time required.
14 s
A
600 m
B
400 m
142 s
The Signs of Velocity
 Velocity is positive (+) or negative (-)
based on direction of motion.
+
-
+
-
+
First choose + direction;
then v is positive if motion
is with that direction, and
negative if it is against that
direction.
Examples of Speed
Orbit
2 x 104 m/s
Light = 3 x 108 m/s
Jets = 300 m/s
Car = 25 m/s
Speed Examples (Cont.)
Runner = 10 m/s
Glacier = 1 x 10-5 m/s
Snail = 0.001 m/s
Average Speed and
Instantaneous Velocity
 The average speed depends ONLY on the
distance traveled and the time required.
A
s = 20 m
C
Time t = 4 s
B
The instantaneous
velocity is the magnitude and direction of
the speed at a particular instant. (v at
point C)
Average and Instantaneous v
x2
Dx
x1
Dt
t1
t2
Instantaneous Velocity:
Dx
vinst 
(Dt  0)
Dt
Displacement, x
Average Velocity:
Dx x2  x1
vavg 

Dt t2  t1
slope
Dx
Dt
Time
Quiz
You and your dog go to for a walk to
the park. On the way, your dog takes
many side trips to chase squirrels or
examine fire hydrants. When you arrive
at the park, do you and your dog have
the same displacement?
1. Yes
2. No
Quiz (con)
If the position of a car is zero, does its
speed nave to be zero?
1. Yes
2. No
3. It depends on the position
Warm up problem:
Average Round-Trip Speed
• A person travels from city A to city B with a
speed of 40 mph and returns with a speed of 60
mph. What is his average round-trip speed?
(A) 100 mph
(B) 50 mph
(C) 48 mph
(D) 10 mph
(E) None of these
Critical thinking question
• And The Winner Is... Two marbles roll
along two horizontal tracks. One track
has a dip, and the other has a bump of
the same shape. Which marble wins?
Definition of Acceleration
 An acceleration is the rate at which
velocity changes (A vector quantity.)
 A change in velocity requires the
application of a push or pull (force).
A formal treatment of force and acceleration will
be given later. For now, you should know that:
• The direction of acceleration is same as
direction of force.
• The acceleration is
proportional to the
magnitude of the force.
Acceleration and Force
F
a
2F
2a
Pulling the wagon with twice the force
produces twice the acceleration and
acceleration is in direction of force.
Why acceleration opposite
direction in this situation?
• A subway train in Sao Paulo, Brazil, decelerates as it comes into a
station. It is accelerating in a direction opposite to its direction of
motion. (credit: Yusuke Kawasaki, Flickr)
Example of Acceleration
+
Force
t=3s
v0 = +2 m/s
vf = +8 m/s
The wind changes the speed of a boat
from 2 m/s to 8 m/s in 3 s. Each
second the speed changes by 2 m/s.
Wind force is constant, thus acceleration is constant.
Constant Acceleration
Acceleration:
a avg
Dv v f  v0


Dt t f  t0
Setting to = 0 and solving for v, we have:
v f  v0  at
Final velocity = initial velocity + change in velocity
The Signs of Acceleration
• Acceleration is positive (+) or negative
(-) based on the direction of force.
+
F
a (-)
F
a(+)
Choose + direction first.
Then acceleration a will
have the same sign as
that of the force F —
regardless of the
direction of velocity.
Figure 2.14
a) This
car is
is slowing
speedingdown
up
b.
This car
it moves
toward
as as
it moves
toward
thethe
right.
right. It therefore
has
Therefore,
it has negative
positive acceleration
in
acceleration
in our
our coordinate
system.
coordinate
system,
because
its acceleration is toward the
left. The car is also
decelerating: the direction of
its acceleration is opposite to
its direction of motion.
Can you describe c and d?
Example 3 (No change in direction): A constant
force changes the speed of a car from 8 m/s to
20 m/s in 4 s. What is average acceleration?
+
Force
t=4s
v1 = +8 m/s
Step
Step
Step
Step
1.
2.
3.
4.
v2 = +20 m/s
Draw a rough sketch.
Choose a positive direction (right).
Label given info with + and - signs.
Indicate direction of force F.
Example 3 (Continued): What is average
acceleration of car?
+
Force
t=4s
v1 = +8 m/s
v2 = +20 m/s
20 m/s - 8 m/s
Step 5. Recall definition
a
 3 m/s
of average acceleration.
4s
aavg
Dv v2  v1


Dt t2  t1
a  3 m/s, rightward
Average and Instantaneous a
aavg
Dv v2  v1


Dt t2  t1
ainst
Dv

(Dt  0)
Dt
slope
v2
Dv
Dv
v1
Dt
Dt
t1
t2
time
Example 4: A wagon moving east at 20 m/s
encounters a very strong head-wind, causing it
to change directions. After 5 s, it is traveling
west at 5 m/s. What is the average
acceleration? (Be careful of signs.)
+
vf = -5 m/s
E
Force
vo = +20 m/s
Step 1. Draw a rough sketch.
Step 2. Choose the eastward direction as positive.
Step 3. Label given info with + and - signs.
Quiz: Acceleration
•
If the velocity of a car is non-zero, can
the acceleration of the car be zero?
1. Yes
2. No
3. Depends on the velocity
Review of Symbols and Units
• Displacement (x, xo); meters (m)
• Velocity (v, vo); meters per second (m/s)
• Acceleration (a); meters per s2 (m/s2)
• Time (t); seconds (s)
Review sign convention for each symbol
Velocity for constant a
Average velocity:
vavg
Average velocity:
Dx x f  x0


Dt t f  t 0
vavg 
v0  v f
2
Setting to = 0 and combining we have:
x  x0 
v0  v f
2
t
Formulas based on definitions:
x  x0 
v0  v f
2
t
v f  v0  at
Derived formulas:
x  x0  v0t  at
1
2
2
x  x0  v f t  at
1
2
2
2a( x  x0 )  v  v
2
f
2
0
For constant acceleration only
Use of Initial Position x0 in Problems.
0
x  x0 
v0  v f
2
0
t
x  x0  v0t  at
1
2
0
2
x  x0  v f t  at
1
2
0
If you choose the
origin of your x,y
axes at the point of
the initial position,
you can set x0 = 0,
simplifying these
equations.
2
2a( x  x0 )  v  v
v f  v0  at
2
f
2
0
The xo term is very
useful for studying
problems involving
motion of two bodies.
Example 5: A ball 5 m from the bottom of an
incline is traveling initially at 8 m/s. Four seconds
later, it is traveling down the incline at 2 m/s. How
far is it from the bottom at that instant?
+
x
vo
5m
8 m/s
x = xo +
vo + vf
2
F
vf
-2 m/s
Careful
t=4s
t =5m+
8 m/s + (-2 m/s)
2
(4 s)
(Continued)
+
F
x
vo
vf
-2 m/s
5m
t=4s
8 m/s
x=5m+
x=5m+
8 m/s + (-2 m/s)
2
8 m/s - 2 m/s
2
(4 s)
(4 s)
x = 17 m
Acceleration in our Example
v f  v0  at
a
v f  v0
t
+
5m
x
vo
v
F
-2 m/s
t=4s
8 m/s
(2 m/s)  (8 m/s)
2
a
 2 m/s
4s
a = -2.50 m/s2
The force
What
is the
changing
meaning
of negative
signplane!
for a?
speed
is down
Example 6: A airplane flying initially at 400
ft/s lands on a carrier deck and stops in a
distance of 300 ft. What is the acceleration?
+400 ft/s
v=0
300 ft
+
F
vo
X0 = 0
Step 1. Draw and label sketch.
Step 2. Indicate + direction and F direction.
Example: (Cont.)
v=0
+400 ft/s
300 ft
+
Step 3. List given; find
information with signs.
List t = ?, even though
time was not asked for.
F
vo
X0 = 0
Given: vo = +400 ft/s
v=0
x = +300 ft
Find: a = ?; t = ?
Problem Solving Strategy:
 Draw and label sketch of problem.
 Indicate + direction
 List givens and state what is to be found.
Given: ____, _____, _____ (x,v,vo,t)
Find: ____, _____
 Select equation containing one and not
the other of the unknown quantities, and
solve for the unknown.