Chapter 6A. Acceleration
AA PowerPoint
PowerPoint Presentation
Presentation by
by
Paul
Paul E.
E. Tippens,
Tippens, Professor
Professor of
of Physics
Physics
Southern
Southern Polytechnic
Polytechnic State
State University
University
©
2007
The Cheetah: A cat that is built for speed. Its strength and
agility allow it to sustain a top speed of over 100 km/h. Such
speeds can only be maintained for about ten seconds.
Photo © Vol. 44 Photo Disk/Getty
Objectives: After completing this
module, you should be able to:
• Define and apply concepts of average and
instantaneous velocity and acceleration.
• Solve problems involving initial and final
velocity, acceleration, displacement, and time.
• Demonstrate your understanding of directions
and signs for velocity, displacement, and
acceleration.
• Solve problems involving a free-falling body in
a gravitational field.
Uniform Acceleration
in One Dimension:
• Motion is along a straight line (horizontal,
vertical or slanted).
• Changes in motion result from a CONSTANT
force producing uniform acceleration.
• The cause of motion will be discussed later.
Here we only treat the changes.
• The moving object is treated as though it
were a point particle.
Distance and Displacement
Distance
Distance isis the
the length
length of
of the
the actual
actual path
path
taken
taken by
by an
an object.
object. Consider
Consider travel
travel from
from
point
point AA to
to point
point BB in
in diagram
diagram below:
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
-line separation
Displacement isis the
the straight
straight-line
separation
of
of two
two points
points in
in aa specified
specified direction.
direction.
D = 12 m, 20o
A

B
A vector quantity:
Contains magnitude
AND direction, a
number, unit & angle.
(12 m, 300; 8 km/h, N)
Distance and Displacement
•• For
For motion
motion along
along xx or
or yy axis,
axis, the
the displacement
displacement isis
determined
determined by
by the
the xx or
or yy coordinate
coordinate of
of its
its final
final
position.
position. Example:
Example: Consider
Consider aa car
car that
that travels
travels 88 m,
m, EE
then
then 12
12 m,
m, W.
W.
Net displacement D is
from the origin to the
final position:
D
D=
= 44 m,
m, W
W
What is the distance
traveled? 20 m !!
D
8 m,E
x = -4
x
x = +8
12 m,W
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
The direction
direction of
of motion
motion does
does not
not matter!
matter!
Definition of Speed
•• Speed
Speed isis the
the distance
distance traveled
traveled per
per unit
unit
of
of time
time (a
(a scalar
scalar quantity).
quantity).
s = 20 m
A
Time t = 4 s
B
v=
s
t
=
20 m
4s
vv == 55 m/s
m/s
Not direction dependent!
Definition of Velocity
•• Velocity
Velocity is
is the
the displacement
displacement per
per
unit
unit of
of time.
time. (A
(A vector
vector quantity.)
quantity.)
s = 20 m
A
D=12 m
20o
Time t = 4 s
B
D 12 m
v 
4s
t
00 N of E
vv =
3
m/s
at
20
= 3 m/s at 20 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
Avg. speed
8.33 m/s
Direction does not matter!
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
xf = -100 m
t = 60 s
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:
x A  xB 600 m + 400 m
v

t A  tB
14 s + 150 s
1000 m
v
164 s
v  6.10 m/s
Average
Average speed
speed isis aa function
function
only
only of
of total
total distance
distance traveled
traveled
and
and the
the total
total time
time required.
required.
14 s
A
625 m
B
356 m
142 s
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
The average
average speed
speed depends
depends ONLY
ONLY on
on the
the
distance
distance traveled
traveled and
and the
the time
time required.
required.
A
s = 20 m
C
Time t = 4 s
B
The
The instantaneous
instantaneous
velocity
velocity isis the
the magnmagnitude
itude and
and direction
direction of
of
the
the speed
speed at
at aa parparticular
ticular instant.
instant. (v
(v at
at
point
point C)
C)
The Signs of Velocity
 Velocity
-)
Velocity isis positive
positive (+)
(+) or
or negative
negative ((-)
based
based on
on direction
direction of
of motion.
motion.
+
-
+
-
+
First
First choose
choose +
+ direction;
direction;
then
then vv isis positive
positive ifif motion
motion
isis with
with that
that direction,
direction, and
and
negative
negative ifif itit isis against
against that
that
direction.
direction.
Average and Instantaneous v
x2
x
x1
t
t1
t2
Instantaneous Velocity:
x
(t  0)
vinst 
t
Displacement, x
Average Velocity:
x x2  x1
vavg 

t t2  t1
slope
x
t
Time
Definition of Acceleration
 An acceleration is the change in velocity
per unit of time. (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
Pulling the
the wagon
wagon with
with twice
twice the
the force
force
produces
produces twice
twice the
the acceleration
acceleration and
and
acceleration
acceleration isis in
in direction
direction of
of force.
force.
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
Wind force
force isis constant,
constant, thus
thus acceleration
acceleration isis constant.
constant.
The Signs of Acceleration
•• Acceleration
+) or
Acceleration isis positive
positive ((+)
or negative
negative
((-)
-) based
based on
on the
the direction
direction of
of force
force..
+
F
a (-)
F
a(+)
Choose
Choose +
+ direction
direction first.
first.
Then
Then acceleration
acceleration aa will
will
have
have the
the same
same sign
sign as
as
that
that of
of the
the force
force FF —
—
regardless
regardless of
of the
the
direction
direction of
of velocity.
velocity.
Average and Instantaneous a
aavg
v v2  v1


t t2  t1
ainst
v

(t  0)
t
slope
v2
v
v
v1
t
t
t1
t2
time
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
v2 = +20 m/s
Step 1. Draw a rough sketch.
Step 2. Choose a positive direction (right).
Step 3. Label given info with + and - signs.
Step 4. 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
v v2  v1


t t2  t1
a  3 m/s, rightward
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.
Example 4 (Cont.): Wagon moving east at 20 m/s
encounters a head-wind, causing it to change
directions. Five seconds later, it is traveling west at
5 m/s. What is the average acceleration?
Choose
Choose the
the eastward
eastward direction
direction as
as positive.
positive.
Initial
Initial velocity,
velocity, vvoo =
= +20
+20 m/s,
m/s, east
east (+)
(+)
Final
5 m/s,
-)
Final velocity,
velocity, vvff =
= --5
m/s, west
west ((-)
The
The change
change in
in velocity,
velocity, vv =
= vvff -- vv00
vv =
-5 m/s)
25 m/s
= ((-5
m/s) -- (+20
(+20 m/s)
m/s) =
= --25
m/s
Example 4: (Continued)
+
vf = -5 m/s
aavg =
v
t
E
vo = +20 m/s
Force
v = (-5 m/s) - (+20 m/s) = -25 m/s
=
vf - vo
tf - to
aa =
= -- 55 m/s
m/s22
a=
-25 m/s
5s
Acceleration is directed to
left, west (same as F).
Signs for Displacement
+
D
vf = -5 m/s
C
A
vo = +20 m/s
E
Force
B
a = - 5 m/s2
Time t = 0 at point A. What are the signs
(+ or -) of displacement at B, C, and D?
At B, x is positive, right of origin
At C, x is positive, right of origin
At D, x is negative, left of origin
+
D
vf = -5 m/s
Signs for Velocity
x=0
A
vo = +20 m/s
C
E
Force
a = - 5 m/s2
What are the signs (+ or -) of velocity at
points B, C, and D?
 At B, v is zero - no sign needed.
 At C, v is positive on way out and
negative on the way back.
 At D, v is negative, moving to left.
B
Signs for Acceleration
+
D
vf = -5 m/s
C
A
vo = +20 m/s
E
Force
B
a = - 5 m/s2
What are the signs (+ or -) of acceleration at
points B, C, and D?
 At B, C, and D, a = -5 m/s, negative
at all points.
 The force is constant and always directed
to left, so acceleration does not change.
Definitions
Average velocity:
vavg
x x2  x1


t t2  t1
Average acceleration:
aavg
v v2  v1


t t2  t1
Velocity for constant a
Average velocity:
vavg
Average velocity:
x x f  x0


t t f  t0
vavg 
v0  v f
2
Setting to = 0 and combining we have:
x  x0 
v0  v f
2
t
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 + v f
2
F
vf
-2 m/s
Careful
t=4s
t =5m+
8 m/s + (-2 m/s)
2
(4 s)
(Continued)
+
F
x
vo
5m
vf
-2 m/s
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)
xx == 17
17 m
m
Constant Acceleration
Acceleration:
a avg
v v f  v0


t t f  t0
Setting to = 0 and solving for v, we have:
v f  v0  at
Final velocity = initial velocity + change in velocity
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
22
aa =
-2.50
m/s
= -2.50 m/s
The force
What
is the
changing
meaning
of negative
signplane!
for a?
speed
is down
Formulas based on definitions:
x  x0 
v0  v f
2
t
v f  v0  at
Derived formulas:
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
IfIf you
you choose
choose the
the
origin
origin of
of your
your x,y
x,y
axes
axes at
at the
the point
point of
of
the
the initial
initial position,
position,
you
you can
can set
set xx00 =
= 0,
0,
simplifying
simplifying these
these
equations.
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.
Review of Symbols and Units
•• Displacement
Displacement ((x,
x, xxoo);
); meters
meters ((m
m))
•• Velocity
v, vvoo);
Velocity ((v,
); meters
meters per
per second
second ((m/s
m/s))
22 (m/s22)
•• Acceleration
(
a
);
meters
per
s
Acceleration (a); meters per s (m/s )
•• Time
t); seconds
Time ((t);
seconds ((ss))
Review sign convention for each symbol
The Signs of Displacement
•• Displacement
Displacement isis positive
positive (+)
(+) or
or
negative
-) based
negative ((-)
based on
on LOCATION
LOCATION..
2m
-1 m
-2 m
The displacement is
the y-coordinate.
Whether motion is
up or down, + or - is
based on LOCATION.
The Signs of Velocity
•• Velocity
-)
Velocity isis positive
positive (+)
(+) or
or negative
negative ((-)
based
based on
on direction
direction of
of motion
motion..
+
-
+
-
+
First choose + direction;
then velocity v is positive
if motion is with that +
direction, and negative if
it is against that positive
direction.
Acceleration Produced by Force
•• Acceleration
+) or
-) based
Acceleration isis ((+)
or ((-)
based on
on
NOT based
direction
based on
on vv).
).
direction of
of force
force ((NOT
F
F
a(-)
a(+)
A push or pull (force) is
necessary to change
velocity, thus the sign of
a is same as sign of F.
More will be said later
on the relationship
between F and a.
Problem Solving Strategy:
 Draw and label sketch of problem.
 Indicate + direction and force direction.
 List givens and state what is to be found.
Given: ____, _____, _____ (x,v,vo,a,t)
Find: ____, _____
 Select equation containing one and not
the other of the unknown quantities, and
solve for the unknown.
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 = ?
Continued . . . x
v=0
+400 ft/s
300 ft
+
F
vo
X0 = 0
Step 4. Select equation
that contains a and not t.
a=
-vo2
2x
=
-(400 ft/s)2
2(300 ft)
0
0
2a(x -xo) = v2 - vo2
Initial position and
final velocity are 2zero.
aa =
= -- 267
267 ft/s
ft/s2
Why isForce
the acceleration
negative?
Because
is in a negative
direction!
Acceleration Due to Gravity
•• Every
Every object
object on
on the
the earth
earth
experiences
experiences aa common
common force:
force:
the
the force
force due
due to
to gravity.
gravity.
•• This
This force
force isis always
always directed
directed
toward
toward the
the center
center of
of the
the earth
earth
(downward).
(downward).
•• The
The acceleration
acceleration due
due to
to gravity
gravity
isis relatively
relatively constant
constant near
near the
the
Earth
’s surface.
Earth’s
surface.
g
W
Earth
Gravitational Acceleration
•• In
In aa vacuum,
vacuum, all
all objects
objects fall
fall
with
with same
same acceleration.
acceleration.
•• Equations
Equations for
for constant
constant
acceleration
acceleration apply
apply as
as usual.
usual.
•• Near
’s surface:
Near the
the Earth
Earth’s
surface:
a = g = 9.80 m/s2 or 32 ft/s2
Directed downward (usually negative).
Experimental Determination
of Gravitational Acceleration.
The apparatus consists of a
device which measures the time
required for a ball to fall a given
distance.
Suppose the height is 1.20 m
and the drop time is recorded
as 0.650 s. What is the
acceleration due to gravity?
t
y
Experimental Determination of
Gravity (y0 = 0; y = -1.20 m)
t
y = -1.20 m; t = 0.495 s
y  v0t  12 at 2 ; v0  0
2 y 2(1.20 m)
a 2 
2
t
(0.495 s)
Acceleration
2
of Gravity: a  9.79 m/s
Acceleration a is negative
because force W is negative.
y
+
W
Sign Convention:
A Ball Thrown
Vertically Upward
avy==
=-0
+
avy=
==-++
yva==
=+--
UP = +
Release Point
vya==
=-0-
•• Displacement
Displacementisispositive
positive
(+)
-) based
(+)or
ornegative
negative((-)
based
on
.
onLOCATION
LOCATION.
•• Velocity
Velocityisispositive
positive(+)
(+)or
or
negative
-) based
negative((-)
basedon
on
direction
.
directionof
ofmotion
motion.
yv=
=a=Negative
Negative
Tippens
• Acceleration is (+) or (-)
based on direction of force
(weight).
Same Problem Solving
Strategy Except a = g:
 Draw and label sketch of problem.
 Indicate + direction and force direction.
 List givens and state what is to be found.
Given: ____, _____, a = - 9.8 m/s2
Find: ____, _____
 Select equation containing one and not
the other of the unknown quantities, and
solve for the unknown.
Example 7: A ball is thrown vertically upward with
an initial velocity of 30 m/s. What are its position
and velocity after 2 s, 4 s, and 7 s?
Step 1. Draw and
label a sketch.
Step 2. Indicate + direction
and force direction.
+
a=g
Step 3. Given/find info.
a = -9.8 ft/s2
t = 2, 4, 7 s
vo = + 30 m/s y = ? v = ?
vo = +30 m/s
Finding Displacement:
Step 4. Select equation
that contains y and not v.
0
y  y0  v0t  at
1
2
+
a=g
2
y = (30 m/s)t + ½(-9.8 m/s2)t2
Substitution of t = 2, 4, and 7 s
will give the following values:
vo = 30 m/s
yy == 40.4
40.4 m;
m; yy =
= 41.6
41.6 m;
m; yy =
= -30.1
-30.1 m
m
Finding Velocity:
Step
Step 5.
5. Find
Find vv from
from equation
equation
that
that contains
contains vv and
and not
not xx::
v f  v0  at
v f  30 m/s  (9.8 m/s )t
+
a=g
2
Substitute t = 2, 4, and 7 s:
vo = 30 m/s
vv == +10.4
+10.4 m/s;
m/s; vv =
= -9.20
-9.20 m/s;
m/s; vv =
= -38.6
-38.6 m/s
m/s
Example 7: (Cont.) Now find
the maximum height attained:
Displacement is a maximum
when the velocity vf is zero.
v f  30 m/s  (9.8 m/s )t  0
2
30 m/s
t
; t  3.06 s
2
9.8 m/s
To find ymax we substitute
t = 3.06 s into the general
equation for displacement.
+
a=g
vo = +96 ft/s
y = (30 m/s)t + ½(-9.8 m/s2)t2
Example 7: (Cont.) Finding the maximum height:
y = (30 m/s)t + ½(-9.8 m/s2)t2
t = 3.06 s
Omitting units, we obtain:
+
a=g
y  (30)(3.06)  (9.8)(3.06)
1
2
y = 91.8 m - 45.9 m
vo =+30 m/s
ymax = 45.9 m
2
Summary of Formulas
x  x0 
v0  v f
2
t
v f  v0  at
Derived Formulas:
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
Summary: Procedure
 Draw and label sketch of problem.
 Indicate + direction and force direction.
 List givens and state what is to be found.
Given: ____, _____, ______
Find: ____, _____
 Select equation containing one and not
the other of the unknown quantities, and
solve for the unknown.
CONCLUSION OF
Chapter 6 - Acceleration