Transcript Motion in One Dimension

Kinematics:
Motion in One Dimension
2.1 Displacement & Velocity
Learning Objectives
• Describe motion in terms of displacement,
time, and velocity
• Calculate the displacement of an object
traveling at a known velocity for a specific
time interval
• Construct and interpret graphs of position
versus time
Essential Concepts
•
•
•
•
Frames of reference
Vector vs. scalar quantities
Displacement
Velocity
– Average velocity
– Instantaneous velocity
• Acceleration
• Graphical representation of motion
Reference Frames
• Motion is relative
• When we say an object is moving, we
mean it is moving relative to something
else (reference frame)
Scalar Quantities & Vector
Quantities
• Scalar quantities have magnitude
• Example: speed
15 m/s
• Vector quantities have magnitude and
direction
• Example: velocity
15 m/s North
Displacement
• Displacement is a vector quantity
• Indicates change in location (position) of
a body
∆x = xf - xi
• It is specified by a magnitude and a
direction.
• Is independent of the path traveled by an
object.
Displacement is change in position
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Displacement vs. Distance
• Distance is the length of the path that an
object travels
• Displacement is the change in position of
an object
Describing Motion
Describing motion requires a frame of reference
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Determining Displacement
In these examples, position is determined with
respect to the origin, displacement wrt x1
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Indicating Direction of Displacement
Direction can be indicated by sign, degrees, or
geographical directions.
When sign is used, it follows the conventions
of a standard graph
Positive
Right
Up
Negative
Left
Down
Displacement
• Linear change in position of an object
• Is not the same as distance
Displacement
•
•
•
•
Distance = length (blue)
How many units did the object move?
Displacement = change in position (red)
How could you calculate the magnitude of line
AB?
• ≈ 5.1 units, NE
Reference Frames &
Displacement
• Direction is relative to the initial position, x1
• x1 is the reference point
Average Velocity
Speed: how far an object travels in a given time interval
Velocity includes directional information:
Average Velocity
displacement
average velocity =
time interval
Dx x f - xi
v=
=
Dt
t
Velocity
• Example
• A squirrel runs in a straight line, westerly
direction from one tree to another,
covering 55 meters in 32 seconds.
Calculate the squirrel’s average velocity
• vavg = ∆x / ∆t
• vavg = 55 m / 32 s
• vavg = 1.7 m/s west
Velocity can be represented graphically:
Position Time Graphs
Velocity can be interpreted
graphically: Position Time Graphs
Find the average velocity between t = 3 min to t = 8 min
Calculate the average velocity for
the entire trip
Formative Assessment:
Position-Time Graphs
Object at rest?
Traveling slowly in a
positive direction?
Traveling in a
negative direction?
Traveling quickly in a
positive direction?
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Average vs. Instantaneous Velocity
• Velocity at any given moment in time or
at a specific point in the object’s path
Position-time when velocity is not
constant
Average velocity compared to
instantaneous velocity
Instantaneous velocity is the slope of the tangent
line at any particular point in time.
Instantaneous Velocity
• The instantaneous velocity is the average
velocity, in the limit as the time interval becomes
infinitesimally short.
2.2 Acceleration
2.2 Acceleration
Learning Objectives
• Describe motion in terms of changing
velocity
• Compare graphical representations of
accelerated and non-accelerated motions
• Apply kinematic equations to calculate
distance, time, or velocity under conditions
of constant acceleration
X-t graph when velocity is changing
Acceleration
Acceleration is the rate of change of velocity.
Acceleration: Change in Velocity
• Acceleration is the rate of change of
velocity
• a = ∆v/∆t
• a = (vf – vi) / (tf – ti)
• Since velocity is a vector quantity, velocity
can change in magnitude or direction
• Acceleration occurs whenever there is a
change in magnitude or direction of
movement.
Acceleration
Because acceleration is a vector, it must have direction
Here is an example of negative acceleration:
Customary Dimensions of
Acceleration
• a = ∆v/∆t
•
= m/s/s
•
= m/s2
• Sample problems 2B
A bus traveling at 9.0 m/s slows down with an
average acceleration of -1.8 m/s. How long
does it take to come to a stop?
Negative Acceleration
• Both velocity & acceleration can have (+)
and (-) values
• Negative acceleration does not always
mean an object is slowing down
Is an object speeding up or
slowing down?
• Depends upon the signs of both velocity and
acceleration
Velocity
+
+
-
Accel
+
+
Motion
Speeding up in + dir
Speeding up in - dir
Slowing down in + dir
Slowing down in - dir
• Construct statement summarizing this table.
Velocity-Time Graphs
• Is this object accelerating?
• How do you know?
• What can you say about its motion?
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Velocity-Time Graph
•
•
•
•
Is this object accelerating?
How do you know?
What can you say about its motion?
What feature of the graph represents
acceleration?
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Velocity-Time Graph
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Displacement with Constant
Acceleration (C)
x
vavg 
t
Since
Then
Thus
Or
and
vavg 
vi  v f
vavg  vavg
x vi  v f

t
2
 vi  v f
x  
 2

 t

1
x  vi  v f t
2
2
Displacement on v-t Graphs
How can you find displacement on the v-t graph?
x
v
, so x  vt
t
Displacement on v-t Graphs
x  vt
Displacement is
the area under the
line!
Graphical Representation of
Displacement during Constant
Acceleration
Displacement on a Non-linear v-t graph
• If displacement is the area under the v-t
graph, how would you determine this area?
Final velocity of an
accelerating object
v
Since a 
t
v f  vi
and
a
t
Then v f  vi  at
Displacement During Constant
Acceleration (D)
1
We know : Δx  vi  v f Δt
2
We know : v f  vi  aΔt
Substituti ng v f into the above equation :
1
Δx  vi  vi  aΔt Δt
2
1
Δx  2vi  aΔt Δt
2
1
Δx  vi Δt  aΔt 2
2
Graphical Representation
Derivation of the Equation
Final velocity after any
displacement (E)
v f  vi  2ax
2
2
A baby sitter pushes a stroller from rest,
accelerating at 0.500 m/s2. Find the velocity after
the stroller travels 4.75m. (p. 57)
Identify the variables.
Solve for the unknown.
Substitute and solve.
Kinematic Equations
x  x f  xi
vavg
1
x  (vi  v f )t
2
1
x  vi t  at 2
2
x

t
v
a
t
v f  vi  at
v f  vi  2ax
2
2
2.3 Falling Objects
Objectives
1. Relate the motion of a freely falling body
to motion with constant acceleration.
2. Calculate displacement, velocity, and
time at various points in the motion of a
freely falling object.
3. Compare the motions of different objects
in free fall.
Motion Graphs of Free Fall
What do motion graphs of an object in
free fall look like?
Motion Graphs of Free Fall
What do motion graphs of an object in
free fall look like?
x-t graph
v-t graph
Do you think a heavier object falls
faster than a lighter one?
Why or why not?
Yes because ….
No, because ….
Free Fall
• In the absence of air resistance, all objects
fall to earth with a constant acceleration
• The rate of fall is independent of mass
• In a vacuum, heavy objects and light
objects fall at the same rate.
• The acceleration of a free-falling object is
the acceleration of gravity, g
• g = 9.81m/s2
memorize this value!
Free Fall
• Free fall is the motion of a body when
only the force due to gravity is acting on
the body.
• The acceleration on an object in free fall
is called the acceleration due to
gravity, or free-fall acceleration.
• Free-fall acceleration is denoted with by
ag (generally) or g (on Earth’s surface).
Free Fall Acceleration
• Free-fall acceleration is the same for all
objects, regardless of mass.
• This book will use the value g = 9.81 m/s2.
• Free-fall acceleration on Earth’s surface is –
9.81 m/s2 at all points in the object’s motion.
• Consider a ball thrown up into the air.
– Moving upward: velocity is decreasing,
acceleration is –9.81 m/s2
– Top of path: velocity is zero, acceleration is –9.81
m/s2
– Moving downward: velocity is increasing,
acceleration is –9.81 m/s2
Sample Problem
• Falling Object
• A player hits a volleyball so that it
moves with an initial velocity of
6.0 m/s straight upward.
• If the volleyball starts from 2.0 m
above the floor,
• how long will it be in the air
before it strikes the floor?
Sample Problem, continued
1. Define
Given:
vi = +6.0 m/s
a = –g = –9.81 m/s2
Δ y = –2.0 m
Diagram:
Place the origin at the
Starting point of the ball
(yi = 0 at ti = 0).
Unknown:
Δt = ?
2. Plan
Choose an equation or situation:
Both ∆t and vf are unknown.
v f  vi  2ay
2
2
v f  vi  at
We can determine ∆t if we know vf
Solve for vf then substitute & solve for ∆t
3. Calculate
Rearrange the equation to isolate the unknowns:
v f   vi  2ay
2
vf = - 8.7 m/s
t 
v f  vi
a
Δt = 1.50 s
Summary of Graphical Analysis of Linear
Motion
This is a graph of x vs. t for
an object moving with
constant velocity. The
velocity is the slope of the
x-t curve.
Comparison of v-t and x-t Curves
On the left we have a graph of velocity vs. time for
an object with varying velocity; on the right we have
the resulting x vs. t curve. The instantaneous
velocity is tangent to the curve at each point.
Displacement an v-t Curves
The displacement, x, is
the area beneath the v
vs. t curve.