Chabot College
Physics 4A
Spring 2015
Chapter 2
•Motion Along a Straight Line!
Goals for Chapter 2
• Describe straight-line motion in
terms of velocity and
acceleration
• Distinguish between average
and instantaneous velocity and
acceleration
Position
x(t)
• Interpret graphs
•
position versus time, x(t) versus t
(slope = velocity!)
time
Goals for Chapter 2
• Describe straight-line motion in
terms of velocity and
acceleration
• Distinguish between average
and instantaneous velocity and
acceleration
Velocity
v(t)
• Interpret graphs
•
velocity versus time, v(t) versus t
Slope = acceleration!
time
Goals for Chapter 2
• Understand straight-line motion
with constant acceleration
• Examine “freely-falling” bodies
• Analyze straight-line motion when
the acceleration is not constant
Introduction
Kinematics is the study of motion.
Displacement, velocity and acceleration are
important physical quantities.
A bungee jumper speeds up during the first part of
his fall and then slows to a halt.
Introduction
•
A bungee jumper speeds up during the first
part of his fall and then slows to a halt.
•
•
•
Model his position in time with function x(t)
Model his velocity in time v(t)
Model his acceleration in time a(t)
2-1 Position, Displacement, and Average Velocity

For this chapter, we restrict motion in three
ways:
1.We consider motion along a straight line only
2.We discuss only the motion itself, not the forces
that cause it
3.We consider the moving object to be a particle
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-1 Position, Displacement, and Average Velocity

A particle is either:


A point-like object (such as an electron)
Or an object that moves such that each part
travels in the same direction at the same rate
(no rotation or stretching)
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-1 Position, Displacement, and Average Velocity

Position is measured relative to a reference
point:
o
The origin, or zero point, of an axis
x at time
t1
Origin
x=0
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-1 Position, Displacement, and Average Velocity

Position is still measured relative to the same
reference point:
o
The origin, or zero point, of an axis
x at time
t2
Origin
x=0
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-1 Position, Displacement, and Average Velocity

Position has a sign:
o
o
Positive direction is in the direction of increasing
numbers
Negative direction is opposite the positive
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-1 Position, Displacement, and Average Velocity

A change in position is called displacement
o
∆x is the change in x
o
Always (final position) – (initial position)
© 2014 John Wiley & Sons, Inc. All
rights reserved.
Displacement
Displacement is written:
•SIGN matters! Direction matters!
• It is a VECTOR!!
Vectors have THREE things…
Magnitude
Direction
Units
Displacement could be:
3 m North, or
– 5 cm on the x axis, or
+83 miles NW
Displacement
Displacement is written:
•SIGN matters! Direction matters!
• It is a VECTOR!!
Displacement is negative =>
<= Positive displacement
2-1 Position, Displacement, and Average Velocity
Examples A particle moves . . .
o
From x = 5 m to x = 12 m?
o
∆x = 7 m (positive direction)
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-1 Position, Displacement, and Average Velocity
Examples A particle moves . . .
o
From x = 5 m to x = 12 m: ∆x = 7 m (positive direction)
o
From x = 5 m to x = 1 m?
o
∆x = -4 m (negative direction)
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-1 Position, Displacement, and Average Velocity
Examples A particle moves . . .
o
From x = 5 m to x = 12 m: ∆x = 7 m (positive direction)
o
From x = 5 m to x = 1 m: ∆x = -4 m (negative direction)
o
From x = 5 m to x = 200 m to x = 5 m:
o

∆x = 0 m !!
The actual distance covered is irrelevant
© 2014 John Wiley & Sons, Inc. All
rights reserved.
Displacement vs. Distance
• Displacement (blue line) = how far the object is
from its starting point, regardless of path
• Distance traveled (dashed line) is measured
along the actual path.
Displacement vs. Distance
Q: You make a round trip to the store 1 mile away.
•What distance do you travel?
•What is your displacement?
Displacement vs. Distance
Q: You walk 70 meters across the campus, hear a
friend call from behind, and walk 30 meters back
the way you came to meet her.
•What distance do you travel?
•What is your displacement?
2-1 Position, Displacement, and Average Velocity
Answer: pairs (b) and (c)
(b) -7 m – -3 m = -4 m
(c) -3 m – 7 m = -10 m
© 2014 John Wiley & Sons, Inc. All
rights reserved.
Speed vs. Velocity
Speed is the ratio of how far an object travels in a
given time interval (in any direction)
Ex: Go 10 miles to Chabot in 30 minutes
Average speed = 10 mi / 0.5 hr = 20 mph
Speed vs. Velocity
Velocity includes directional information:
VECTOR!
Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes
Average velocity = 20 mi / 0.333 hr = 60 mph NORTH
Always think about your answer …. Is it reasonable??
Speed vs. Velocity
Not with traffic is 60 mph even possible! 
Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes
Average velocity = 20 mi / 0.333 hr = 60 mph NORTH
2-1 Position, Displacement, and Average Velocity

Average velocity is the ratio of:
o
o

A displacement, ∆x
To the time interval in which the displacement
occurred, ∆t
Average velocity has units of (distance) / (time)
o
Meters per second, m/s
© 2014 John Wiley & Sons, Inc. All
rights reserved.
Speed vs. Velocity
Velocity includes directional information:
VECTOR!
Ex: Go 10 miles on 880 Northbound to Chabot in 30 minutes
Average velocity = 10 mi / 0.5 hr = 20 mph NORTH
Speed vs. Velocity
Speed is a SCALAR
• 60 miles/hour, 88 ft/sec, 27 meters/sec
Velocity is a VECTOR
• 60 mph North
• 88 ft/sec East
• 27 m/s @ azimuth of 173 degrees
Example of Average Velocity
• Position of runner as a function of time is
plotted as moving along the x axis of a
coordinate system.
• During a 3.00-s time interval, a runner’s
position changes from x1 = 50.0 m to
x2 = 30.5 m
• What was the runner’s
average velocity?
Example of Average Velocity
During a 3.00-sec interval, runner’s position
changes from x1 = 50.0 m to x2 = 30.5 m
What was the runner’s average velocity?
Vavg = (30.5 - 50.0) meters/3.00 sec
= -6.5 m/s
in the x direction.
Note!
Dx = FINAL – INITIAL
position
The answer must have
value1,
units2, &
DIRECTION3
Example of Average SPEED
During a 3.00-s time interval, the runner’s
position changes from x1 = 50.0 m to x2 = 30.5 m.
What was the runner’s average speed?
Savg = |30.5-50.0| meters/3.00 sec = 6.5 m/s
The answer must have
value & units but it is a
scalar! No direction
needed
Negative velocity???
• Average x-velocity is negative during a time
interval if particle moves in negative xdirection for that time interval.
Displacement, time, and average velocity
A racing car starts from rest, and after 1 second is
19 meters from the starting line.
After the next 3 seconds, the car is 277 meters
from the starting line.
What was its average velocity in those 3 seconds?
Displacement, time, and average velocity
Q
A racing car starts from rest, and after 1 second is 19
meters from the starting line. After the next 3 seconds,
the car is 277 meters from the starting line. What was its
average velocity in those 3 seconds?
Solution Method:
1.
What do you know? What do you need to find? What
are the units? What might be a reasonable estimate?|
2.
DRAW it! Visualize what is happening. Create a
coordinate system, label the drawing with everything.
3.
Find what you need from what you know
Displacement, time, and average velocity
A racing car starts from rest, and after 1 second is
19 meters from the starting line. After the next 3
seconds, the car is 277 meters from the starting line.
What was its average velocity in those 3 seconds?
“starts from rest” = initial velocity = 0
car moves along straight (say along an x-axis)
has coordinate x = 0 at t=0 seconds
has coordinate x=+19 meters at t =1 second
Has coordinate x=+277 meters at t = 1+3 = 4 seconds.
Displacement, time, and average velocity
Q A racing car starts from rest, and after 1 second is 19 meters
from the starting line. After the next 3 seconds, the car is 277
meters from the starting line. What was its average velocity in
those 3 seconds?
A position-time graph
x(t) vs t
A position-time graph (an “x-t” graph) shows
the particle’s position x as a function of time
t.
Average x-velocity is related to the slope of
an x-t graph.
A position-time graph
x(t) vs t
2-2 Instantaneous Velocity and Speed
Example



The graph shows the
position and velocity of an
elevator cab over time.
The slope of x(t), and so also
the velocity v, is zero from 0
to 1 s, and from 9s on.
During the interval bc, the
slope is constant and
nonzero, so the cab moves
with constant velocity
(4 m/s).
Instantaneous Speed
Instantaneous speed is the average
speed in the limit as the time interval
becomes infinitesimally short.
Ideally, a speedometer would measure
instantaneous speed; in fact, it
measures average speed, but over a
very short time interval.
Note: It doesn’t measure direction!
Instantaneous Speed
Instantaneous velocity is the average
velocity in the limit as the time interval
becomes infinitesimally short.
Velocity is a vector; you must
include direction!
V = 27 m/s west…
Instantaneous Velocity Example
A jet engine moves along an
experimental track (the x axis) as
shown.
Its position as a function of time is
given by the equation
X(t)
x = At2 + B
where A = 2.10 m/s2 and B = 2.80 m.
time
Instantaneous Velocity Example
A jet engine’s position as a function of
time is x = At2 + B, where A = 2.10
m/s2 and B = 2.80 m.
(a)Determine the displacement of the
engine during the time interval from
t1 = 3.00 s to t2 = 5.00 s.
(b)Determine the average velocity
during this time interval.
(c)Determine the magnitude of the
instantaneous velocity at t = 5.00 s.
Instantaneous Velocity Example
A jet engine’s position as a function of
time is x = At2 + B, where A = 2.10
m/s2 and B = 2.80 m.
(a)Determine the displacement of the
engine during the time interval from
t1 = 3.00 s to t2 = 5.00 s.
@ t = 3.00 s x1 = 21.7
@ t = 5.00 s x2 = 55.3
x2 – x1 = 33.6 meters in +x direction
Instantaneous Velocity Example
A jet engine’s position as a function of
time is x = At2 + B, where A = 2.10
m/s2 and B = 2.80 m.
b) Determine the average velocity
during this time interval.
Vavg = 33.6 m/ 2.00 sec
= 16.8 m/s
in the + x direction
Instantaneous Velocity Example
A jet engine’s position as a function of
time is x = At2 + B, where A = 2.10
m/s2 and B = 2.80 m.
(c) Determine the magnitude of the
instantaneous velocity at t = 5.00 s
|v| = dx/dt @ t = 5.00 seconds
= 21.0 m/s
2-2 Instantaneous Velocity and Speed
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-2 Instantaneous Velocity and Speed
Answers:
(a) Situations 1 and 4 (zero)
(b) Situations 2 and 3
© 2014 John Wiley & Sons, Inc. All
rights reserved.
Acceleration
Acceleration = the rate of change of velocity.
Units: meters/sec/sec or m/s^2 or m/s2 or ft/s2
Since velocity is a vector, acceleration is ALSO a vector,
so direction is crucial…
acceleration = 2.10 m/s2 in the +x direction
Acceleration Example
A car accelerates along a straight road from rest
to 90 km/h in 5.0 s.
What is the magnitude of its average
acceleration?
KEY WORDS:
“straight road” = assume constant acceleration
“from rest” = starts at 0 km/h
Acceleration Example
A car accelerates along a straight road from rest to 90
km/h in 5.0 s
What is the magnitude of its average acceleration?
|a|
= (90 km/hr – 0 km/hr)/5.0 sec
= 18 km/h/sec along road
better – convert to more reasonable units
90 km/hr = 90 x 103 m/hr x 1hr/3600 s = 25 m/s
So
|a| = 5.0 m/s2
(note – magnitude only is requested)
Acceleration
Acceleration = the rate of change of velocity.
Acceleration vs. Velocity?
(a) If the velocity of an object is zero, does it
mean that the acceleration is zero?
(b) If the acceleration is zero, does it mean that
the velocity is zero?
Think of some examples.
Acceleration Example
• An automobile is moving to the right along
a straight highway. Then the driver puts on
the brakes.
• If the initial velocity (when the driver hits
the brakes) is v1 = 15.0 m/s, and it takes 5.0
s to slow down to v2 = 5.0 m/s, what was the
car’s average acceleration?
Acceleration Example
An automobile is moving to the right along a straight highway.
Then the driver puts on the brakes. If the initial velocity (when
the driver hits the brakes) is v1 = 15.0 m/s, and it takes 5.0 s to
slow down to v2 = 5.0 m/s, what was the car’s average
acceleration?
Acceleration Example
A semantic difference between negative acceleration
and deceleration:
“Negative” acceleration is acceleration in the
negative direction (defined by coordinate system).
“Deceleration” occurs when the acceleration is
opposite in direction to the velocity.
Finding velocity on an x-t graph
• At any point on an x-t graph,
instantaneous x-velocity is equal to
slope of tangent to curve at that point.
Motion diagrams
A motion diagram shows position of a
particle at various instants, and arrows
represent its velocity at each instant.
Motion diagrams
A motion diagram shows position of a
particle at various instants, and arrows
represent its velocity at each instant.
Average acceleration
• Acceleration describes the rate of change of
velocity with time.
• The average x-acceleration is aav-x = Dvx/Dt.
Use x vs. t graph to find instantaneous
acceleration and average acceleration.
Finding acceleration on a vx-t graph
Instantaneous acceleration
• The instantaneous acceleration is
ax = dvx/dt.
• Q A racing car starts from rest, and after 1 second is 19
meters from the starting line.
• After the next 3 seconds, the car is 277 meters from the
starting line.
• What was its average acceleration in the first second?
• What was its average acceleration in the first 4
seconds?
An x-t graph and a motion diagram
An x-t graph and a motion diagram
A vx-t graph and motion diagram
A vx-t graph and motion diagram
Motion with constant acceleration
For a particle with constant
acceleration, the velocity
changes at the same rate
throughout the motion.
Acceleration given x(t)
• A particle is moving in a straight line with
position given by x = (2.10 m/s2)t2 + (2.80 m)
• Calculate (a) its average acceleration during the
interval from t1 = 3.00 s to t2 = 5.00 s, &
• (b) instantaneous acceleration as function of time.
Acceleration given x(t)
A particle is moving in a straight line with its
position is given by x = (2.10 m/s2)t2 + (2.80 m)
Calculate (a) its average acceleration during the
interval from t1 = 3.00 s to t2 = 5.00 s
V = dx/dt = (4.2 m/s) t
V1 = 12.6 m/s
V2 = 21 m/s
Dv/Dt = 8.4 m/s/2.0 s = 4.2 m/s2
Acceleration given x(t)
A particle is moving in a straight line with its
position is given by x = (2.10 m/s2)t2 + (2.80 m).
Calculate (b) its instantaneous acceleration as a
function of time.
Analyzing acceleration
Graph shows Velocity as a function of time for
two cars accelerating from 0 to 100 km/h in a
time of 10.0 s
Compare (a) the average acceleration; (b)
instantaneous acceleration; and (c) total
distance traveled for the two cars.
Analyzing acceleration
Velocity as a function of time for two cars
accelerating from 0 to 100 km/h in a time of 10.0 s
Compare (a) the average acceleration; (b)
instantaneous acceleration; and (c) total distance
traveled for the two cars.
Same final speed in time =>
Same average acceleration
But Car A accelerates faster…
2-4 Constant Acceleration



In many cases acceleration is
constant, or nearly so.
For these cases, 5 special
equations can be used.
Note that constant acceleration
means a velocity with a constant
slope, and a position with varying
slope (unless a = 0).
Constant Acceleration Equations
FIVE key variables:
Dxdisplacement vinitial ,
FIVE key equations:
Dx = ½ (vi+vf)t
Dx = vit + ½ at2
Dx = vft – ½ at2
vf = vi + at
vf2 = vi2 + 2aDx
vfinal , acceleration
time
2-4 Constant Acceleration
First basic equation

o
o
When the acceleration is constant, the average and
instantaneous accelerations are equal
Rewrite Eq. 2-7 and rearrange
Eq. (2-11)

This equation reduces to v = v0 for t = 0

Its derivative yields the definition of a, dv/dt
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-4 Constant Acceleration
Second basic equation

Eq. (2-12)
o
Average = ((initial) + (final)) / 2:
Eq. (2-13)
Eq. (2-14)
o
Substitute 2-14 into 2-12
Eq. (2-15)
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-4 Constant Acceleration

These are enough to solve any constant
acceleration problem
o

Solve as simultaneous equations
Additional useful forms:
Eq. (2-16)
Eq. (2-17)
Eq. (2-18)
© 2014 John Wiley & Sons, Inc. All
rights reserved.
The equations of motion with constant acceleration
Equation of Motion
Variables Present
vx  v0x  axt
•
Initial velocity, final velocity, acceleration, time
x  x0  v0xt  12 axt 2
•
Displacement (x – x0), initial velocity, time,
acceleration
2  2a  x  x 
vx2  v0x
x
0
•
Initial velocity, final velocity, acceleration,
displacement
v0x  vx 
x  x0 
t
2 
•
Displacement, initial velocity, final velocity, time
•
Displacement (x – x0), final velocity, time,
acceleration






x = x0 + vxt – 1/2ax
t2
The equations of motion with constant acceleration
vx  v0x  axt
•
Initial velocity, final velocity, acceleration, time
x  x0  v0xt  12 axt 2
•
Displacement (x – x0), initial velocity, time,
acceleration
2  2a  x  x 
vx2  v0x
x
0
•
Initial velocity, final velocity, acceleration,
displacement
v0x  vx 
x  x0 
t
2 
•
Displacement, initial velocity, final velocity, time






x = x0 + vxt – 1/2axt2
Equation of Motion
Find 3 of 4, solve for 4th!
2-4 Constant Acceleration

Table 2-1 shows
5 key equations
& the quantities
missing from
each.
2-4 Constant Acceleration
Answer: Situations 1 (a = 0) and 4.
© 2014 John Wiley & Sons, Inc. All
rights reserved.
A motorcycle with constant
acceleration
Follow Example 2.4 for an accelerating
motorcycle.
Two bodies with different
(constant) accelerations
• A motorist is moving at a constant speed of 15m/s (in
a 25 mph zone, through a school crosswalk) passes a
police officer on a motorcycle, who is stopped.
• The officer immediately accelerates at a constant 3.0
m/s2 in pursuit, and overtakes the motorist.
• How far are they from the point where the motorist
first passed the officer?
• How much time elapsed?
Two bodies with different
accelerations
• A motorist is moving at a constant speed of 15m/s (in
a 25 mph zone, through a school crosswalk) passes a
police officer on a motorcycle, who is stopped.
• The officer immediately accelerates at a constant 3.0
m/s2 in pursuit, and overtakes the motorist.
Two bodies with different
accelerations
Two different initial/final velocities and
accelerations
TIME and displacement are linked!
Two bodies with different
accelerations
Two different initial/final
velocities and accelerations
TIME and displacement are
linked!
Two bodies with different
accelerations (part 2!)
tp
•
•
tm
A motorist is moving at a constant speed of 15m/s (in a 25
mph zone, through a school crosswalk) passes a police
officer on a motorcycle, who is stopped.
The officer waits 3 seconds, and then accelerates at a
constant 3.0 m/s2 in pursuit, and overtakes the motorist.
Two bodies with different
accelerations
Displacement
(m)
Two different initial/final velocities
and accelerations & times?
TIME and displacement are STILL
linked, but times aren’t equal…
tm
Motorist
Officer
tp
Time
(s)
Freely falling bodies
• Free fall is the motion of an
object under the influence of
only gravity.
• In the figure, a strobe light
flashes with equal time
intervals between flashes.
• The velocity change is the
same in each time interval, so
the acceleration is constant.
Freely Falling Objects
• In the absence of air
resistance,
• all objects fall with the
same acceleration,
• although this may be
tricky to tell by testing
in an environment
where there is air
resistance.
Freely Falling Objects
• The acceleration due to
gravity at the Earth’s
surface is approximately
9.80 m/s2.
• At a given location on
the Earth and in the
absence of air
resistance, all objects
fall with the same
constant acceleration.
Example: Falling from a tower.
Suppose that a ball is
dropped (v0 = 0) from a
tower 70.0 m high.
How far will it have fallen
after a time t1 = 1.00 s,
t2 = 2.00 s, and
t3 = 3.00 s?
Ignore air resistance.
Example: Falling from a tower.
Suppose that a ball is
dropped (v0 = 0) from a
tower 70.0 m high.
How far will it have fallen
after a time t1 = 1.00 s,
t2 = 2.00 s, and
t3 = 3.00 s?
Ignore air resistance.
Example: Thrown down from a tower
• Suppose a ball is thrown downward with an
initial velocity of 3.00 m/s, instead of being
dropped.
• (a) What then would be its position after t =
1.00 s and 2.00 s?
• (b) What is its speed after 1.00 s and 2.00 s?
Compare with the speeds of a dropped ball.
Freely Falling Objects
Example: Ball thrown up!
A person throws a ball upward
into the air with an initial
velocity of 15.0 m/s
Calculate (a) how high it goes,
& (b) how long the ball is in the
air before it comes back to the
hand. Ignore air resistance.
Freely Falling Objects
Example: Ball thrown up!
A person standing on top of a
building throws a ball upward into
the air with an initial velocity of
15.0 m/s
Calculate (a) how high it goes, &
(b) how long the ball is in the air
before it comes back to the hand.
Ignore air resistance.
Up-and-down
motion in free
fall
• An object is in
free fall even
when it is moving
upward.
Is the acceleration zero at the highest
point?
The vertical
velocity, but
not the
acceleration,
is zero at the
highest point.
Is the acceleration zero at the highest
point?
The vertical
velocity, but
not the
acceleration,
is zero at the
highest point.
Ball thrown upward (cont.)
• Consider again a ball thrown
upward, & calculate…
• (a) how much time it takes for
the ball to reach the maximum
height,
• (b) the velocity of the ball when
it returns to the thrower’s hand
(point C).
Freely Falling Objects
• Give examples to show the error in these
two common misconceptions:
• (1) that acceleration and velocity are always
in the same direction
• (2) that an object thrown upward has zero
acceleration at the highest point.
The quadratic formula
For a ball thrown upward at an initial speed of
15.0 m/s, calculate at what time t the ball
passes a point 8.00 m above the person’s
hand.
Ball thrown at edge of cliff
A ball is thrown upward at a speed of
15.0 m/s by a person on the edge of a
cliff, so that the ball can fall to the base
of the cliff 50.0 m below.
(a) How long does it take the ball to
reach the base of the cliff?
(b) What is the total distance traveled by
the ball?
Ignore air resistance (likely to be
significant, so our result is an
approximation).
Variable Acceleration; Integral
Calculus
Deriving the kinematic equations through
integration:
For constant acceleration,
Variable Acceleration; Integral
Calculus
Then:
For constant acceleration,
Variable Acceleration; Integral
Calculus
Example: Integrating a time-varying
acceleration.
An experimental vehicle starts from rest
(v0 = 0) at t = 0 and accelerates at a rate
given by a = (7.00 m/s3)t. What is
(a) its velocity and
(b) its displacement 2.00 s later?
Graphical Analysis and Numerical
Integration
The total displacement of an
object can be described as the
area under the v-t curve:
Graphical Analysis and Numerical
Integration
Similarly, the velocity may be written as the
area under the a-t curve.
However, if the velocity or acceleration is not
integrable, or is known only graphically,
numerical integration may be used instead.
Example: Numerical integration
An object starts from rest at t = 0 and accelerates
at a rate a(t) = (8.00 m/s4)t2. Determine its velocity
after 2.00 s using numerical methods.
Velocity and position by integration
t
t
0
0
vx  vox   axdt and x  x0   vxdt.
The acceleration of a car is not
always constant.
The motion may be integrated over
many small time intervals to give
Motion with changing
acceleration
2-3 Acceleration

Acceleration is a vector quantity:
o
Positive sign means in the positive coordinate direction
o
Negative sign means the opposite
o
Units of (distance) / (time squared)
2-3 Acceleration

Note: accelerations can be expressed in units of g
Eq. (2-10)
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2-3 Acceleration
Example If a car with velocity v = -25 m/s is braked to a
stop in 5.0 s, then a = + 5.0 m/s2.
Acceleration is positive, but speed has decreased.
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rights reserved.
2-3 Acceleration
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-3 Acceleration
Answers:
(a) + (b) - (c) -
(d) +
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2-3 Acceleration
Example






The graph shows the velocity and
acceleration of an elevator cab
over time.
When acceleration is 0 (e.g.
interval bc) velocity is constant.
When acceleration is positive
(ab) upward velocity increases.
When acceleration is negative
(cd) upward velocity decreases.
Steeper slope of the velocitytime graph indicates a larger
magnitude of acceleration: the
cab stops in half the time it takes
to get up to speed.
2-5 Free-Fall Acceleration
Learning Objectives


2.16 Identify that if a particle
is in free flight (whether
upward or downward) and if
we can neglect the effects of
air on its motion, the particle
has a constant downward
acceleration with a magnitude
g that we take to be 9.8m/s2.
2.17 Apply the constant
acceleration equations (Table
2-1) to free-fall motion.
© 2014 John Wiley & Sons, Inc. All
rights reserved.
Figure 2-12
2-5 Free-Fall Acceleration

Free-fall acceleration is the rate at which an object
accelerates downward in the absence of air
resistance
o
o
o

Varies with latitude and elevation
Written as g, standard value of 9.8 m/s2
Independent of the properties of the object (mass,
density, shape, see Figure 2-12)
The equations of motion in Table 2-1 apply to
objects in free-fall near Earth's surface
o
o
In vertical flight (along the y axis)
Where air resistance can be neglected
© 2014 John Wiley & Sons, Inc. All
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2-5 Free-Fall Acceleration

The free-fall acceleration is downward (-y
direction)
o
Value -g in the constant acceleration equations
Answers:
(a) The sign is positive (the ball moves upward); (b) The sign is negative (the ball
moves downward); (c) The ball's acceleration is always -9.8 m/s2 at all points
along its trajectory
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rights reserved.
2-6 Graphical Integration in Motion Analysis
•
•
2.18 Determine a particle's
change in velocity by graphical
integration on a graph of
acceleration versus time.
2.19 Determine a particle's
change in position by
graphical integration on a
graph of velocity versus time.
© 2014 John Wiley & Sons, Inc. All
rights reserved.
2-6 Graphical Integration in Motion Analysis
Integrating acceleration:

o

Given a graph of an object's acceleration a versus
time t, we can integrate to find velocity
The Fundamental Theorem of Calculus gives:
Eq. (2-27)
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2-6 Graphical Integration in Motion Analysis

The definite integral can be evaluated from a graph:
Eq. (2-28)
© 2014 John Wiley & Sons, Inc. All
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2-6 Graphical Integration in Motion Analysis
Example


•
•
•
•
The graph shows the
acceleration of a person's head
and torso in a whiplash
incident.
To calculate the torso speed at
t = 0.110 s (assuming an initial
speed of 0), find the area
under the pink curve:
area A = 0
area B = 0.5 (0.060 s) (50 m/s2) = 1.5 m/s
area C = (0.010 s) (50 m/s2) = 0.50 m/s
total area = 2.0 m/s
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rights reserved.
2
Summary
Position


Displacement
Relative to origin

Positive and negative
directions
Average Velocity

Change in position (vector)
Eq. (2-1)
Average Speed
Displacement / time (vector)

Distance traveled / time
Eq. (2-2)
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Eq. (2-3)
2
Summary
Instantaneous Velocity

At a moment in time

Speed is its magnitude
Average Acceleration

Ratio of change in velocity to
change in time
Eq. (2-7)
Eq. (2-4)
Instantaneous Acceleration

First derivative of velocity

Second derivative of position
Constant Acceleration

Includes free-fall, where
a = -g along the vertical axis
Eq. (2-8)
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rights reserved.
Tab. (2-1)