Chapter 6A. Acceleration
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State 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 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
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!
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:
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 speed is a function
only of total distance traveled
and the total time 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 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)
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.
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
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 the wagon with twice the force
produces twice the acceleration and
acceleration is in direction of 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 force is constant, thus acceleration is constant.
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.
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 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
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 the eastward direction as positive.
Initial velocity, vo = +20 m/s, east (+)
Final velocity, vf = -5 m/s, west (-)
The change in velocity, Dv = vf - v0
Dv = (-5 m/s) - (+20 m/s) = -25 m/s
Example 4: (Continued)
+
vf = -5 m/s
aavg =
Dv
Dt
E
vo = +20 m/s
Force
Dv = (-5 m/s) - (+20 m/s) = -25 m/s
=
vf - vo
tf - to
a = - 5 m/s2
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
Dx x2  x1


Dt t2  t1
Average acceleration:
aavg
Dv v2  v1


Dt t2  t1
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
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
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
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
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.
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
The Signs of Displacement
• Displacement is positive (+) or
negative (-) based on 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 is positive (+) or negative (-)
based on direction of 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 is (+) or (-) based on
direction of force (NOT based on v).
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 . . .
v=0
x
+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.
a = - 267 ft/s
Why isForce
the acceleration
negative?
Because
is in a negative
direction!
Acceleration Due to Gravity
• Every object on the earth
experiences a common force: the
force due to gravity.
• This force is always directed
toward the center of the earth
(downward).
• The acceleration due to gravity is
relatively constant near the
Earth’s surface.
g
W
Earth
Gravitational Acceleration
• In a vacuum, all objects fall
with same acceleration.
• Equations for constant
acceleration apply as usual.
• Near the 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?
Dt
y
Experimental Determination of
Gravity (y0 = 0; y = -1.20 m)
Dt
y = -1.20 m; t = 0.495 s
y  v0t  at ; v0  0
1
2
2
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=
==-++
ya
=+-v ==
•
UP = +
Release Point
Displacement is positive
(+) or negative (-) based
on LOCATION.
vya==
=-0-
•
yv=
=a=Negative
Negative
Tippens
Velocity is positive (+) or
negative (-) based on
direction of motion.
• 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
y = 40.4 m; y = 41.6 m; y = -30.1 m
Finding Velocity:
Step 5. Find v from equation
that contains v and not x:
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
v = +10.4 m/s; v = -9.20 m/s; v = -38.6 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:
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