motion in one dimension chapter 2
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Motion in One Dimension Chapter 2. 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 - PowerPoint PPT PresentationTRANSCRIPT
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Motion in One DimensionChapter 2
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2.1 Displacement & VelocityLearning 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
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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
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Displacement
• Net change in position of an object• Straight line distance from beginning to end• Is not the same as distance
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Displacement
• Displacement is a vector quantity • Indicates change in location of a body
∆x = xf - xi
• It is specified by a magnitude and a direction.
• Displacement does not take into account the path followed between xi and xf .
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Displacement is change in position
www.cnx.org
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Describing MotionDescribing motion requires a frame of reference
http://www.sfu.ca/phys/100/lectures/lecture5/lecture5.html
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Determining DisplacementIn these examples, position is determined with respect to the origin, displacement wrt x1
http://www.sfu.ca/phys/100/lectures/lecture5/lecture5.html
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Indicating Direction of Displacement
When sign is used, it follows the conventions of a standard graph
Positive Right Up
Negative Left Down
Direction can be indicated by sign, degrees, or geographical directions.
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Displacement• In this example, the body moved 10 units to get from A to B.• Distance = length traveled = 10 units• Displacement = net change in position ≈ 5.1 units, NE
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Velocity
t
xx
t
xv if
in time change
ntdisplaceme velocity average
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Velocity• Example
• A squirrel runs in a 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
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Velocity is not Speed
Velocity is a vector quantity
Speed is a scalar quantity
Magnitude Magnitude
Direction No direction
= ∆ displacement /∆ time
vavg = ∆x /∆t
= distance traveled/ time
= d /∆t
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Position-time graphs(x-t graphs)
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Velocity can be interpreted graphically: Position Time Graphs
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Velocity can be interpreted graphically: Position Time Graphs
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Calculate the average velocity for the entire trip
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Position-Time Graphs
dev.physicslab.org
Object at rest?
Traveling slowly in a positive direction?
Traveling in a negative direction?
Traveling quickly in a positive direction?
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Instantaneous Velocity
• Is not average velocity
• Is velocity of an object at any given moment in time or at a specific point in the object’s path
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Position-time when velocity is not constant
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Instantaneous Velocity
• Velocity at any point on an x-t graph• Velocity at a given point in time• Is the slope of a line tangent to the x-t curve
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Average velocity compared to instantaneous velocity
Instantaneous velocity is the slope of the tangent line at any particular point in time.
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2.2 Acceleration
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2.2 AccelerationLearning 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
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X-t graph when velocity is changing
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Changes 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 speed or direction of movement.
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Dimensions of acceleration
• a = ∆v/∆t
• = meters per second per second
• = m/s/s
• = m/s2
• Sample problems 2B
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Acceleration is a vector quantity
• Has magnitude and direction
• Both velocity & acceleration can have (+) and (-) values
• Whether –a indicates slowing down or speeding up depends upon the sign of velocity
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www.gcsescience.com
Velocity-Time Graphs
What would a position-time graph look like?
Acceleration-time graph?
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Free-fall motion graphs
x-t graph v-t graph
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www.gcsescience.com
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dev.physicslab.org
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Velocity & Acceleration
Velocity acceleration_ __ Motion
+ + speeding up
(-) (-) speeding up
+ (-) slowing down
(-) + slowing down
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Displacement with Constant Acceleration
tvvx
tvv
x
vv
t
x
vv
vvv
t
xv
fi
fi
fi
avgavg
fiavgavg
2
1 Or
2
Thus
2 Then
Since2
and
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tavvThent
vvaSince
if
if
Final velocity of an accelerating object
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2Δ2
1ΔΔ tatvx i
Displacement during constant acceleration
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Final velocity after any displacement
xavv if 222
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.
Practice 2E, p. 58
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Kinematic Equations
xavvtatvx
tavvtvvx
t
va
t
xvxxx
ifi
iffi
if
2 2
1
)(2
1
222
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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.
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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!
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• 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 the symbols ag (generally) or g (on Earth’s surface).
Free Fall
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• 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
Free Fall Acceleration
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Free-Fall Problem• A tennis ball is tossed vertically with an
initial velocity 0f 6.0 m/s. How high will the ball go? How long will the ball be in the air until it returns to its starting point?
• Known: vi = 6.0 m/s; a = -g = -9.81 m/s2
• How high? vf2 = vi
2 + 2aΔy solve for Δy• vf = vi + aΔt solve for Δt• Since the ball goes up Δt and comes down
Δt, time in the air is 2Δt
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Sample Problem
• Falling Object• Jason 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?
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Sample Problem, continued
1. DefineGiven: Unknown:
vi = +6.0 m/s Δt = ?
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).
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2. Plan Choose an equation or situation:
Both ∆t and vf are unknown.
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:
yavv if 222 tavv if
yavv if 22a
vvt if
vf = - 8.7 m/s Δt = 1.50 s