The document discusses acceleration and related concepts: - Acceleration is the change in velocity per unit of time and is a vector quantity. It results from an applied force and is proportional to the force's magnitude. - Velocity is speed in a given direction, while speed is the distance traveled per time and does not consider direction. - Average acceleration is calculated as the change in velocity divided by the time interval. Instantaneous acceleration is the slope of the velocity-time graph at an instant. - Examples demonstrate calculating average speed and acceleration from initial and final velocities and time intervals. Direction and signs of displacement, velocity, and acceleration must be considered carefully.

1. Chapter 6A. AccelerationChapter 6A. Acceleration
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
A PowerPoint Presentation byA PowerPoint Presentation by
Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics
Southern Polytechnic State UniversitySouthern Polytechnic State University
© 2007

2. The CheetahThe Cheetah: A cat that is built for speed. Its strength and: A cat that is built for speed. Its strength and
agility allow it to sustain a top speed of over 100 km/h. Suchagility allow it to sustain a top speed of over 100 km/h. Such
speeds can only be maintained for about ten seconds.speeds can only be maintained for about ten seconds.
Photo © Vol. 44 Photo Disk/Getty

3. Objectives: After completing thisObjectives: After completing this
module, you should be able to:module, you should be able to:
•• Define and apply concepts ofDefine and apply concepts of averageaverage andand
instantaneousinstantaneous velocity and acceleration.velocity and acceleration.
•• Solve problems involving initial and finalSolve problems involving initial and final
velocityvelocity,, accelerationacceleration,, displacementdisplacement, and, and timetime..
•• Demonstrate your understanding of directionsDemonstrate your understanding of directions
and signs for velocity, displacement, andand signs for velocity, displacement, and
acceleration.acceleration.
•• Solve problems involving a freeSolve problems involving a free--falling body infalling body in
aa gravitational fieldgravitational field..

4. Uniform AccelerationUniform Acceleration
in One Dimension:in One Dimension:
•• Motion is along a straight line (horizontal,Motion is along a straight line (horizontal,
vertical or slanted).vertical or slanted).
•• Changes in motion result from a CONSTANTChanges in motion result from a CONSTANT
force producing uniform acceleration.force producing uniform acceleration.
•• The cause of motion will be discussed later.The cause of motion will be discussed later.
Here we only treat the changes.Here we only treat the changes.
•• The moving object is treated as though itThe moving object is treated as though it
were a point particle.were a point particle.

5. Distance and DisplacementDistance 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:
Distance is the length of the actual path
taken by an object. Consider travel from
point A to point B in diagram below:
A
B
s = 20 m
DistanceDistance ss is ais a scalarscalar
quantity (no direction):quantity (no direction):
ContainsContains magnitudemagnitude onlyonly
and consists of aand consists of a
numbernumber and aand a unit.unit.
(20 m, 40 mi/h, 10 gal)(20 m, 40 mi/h, 10 gal)

6. Distance and DisplacementDistance and Displacement
Displacement is the straight-line separation
of two points in a specified direction.
DisplacementDisplacement is the straightis the straight--line separationline separation
of two points in a specified direction.of two points in a specified direction.
A vector quantity:
Contains magnitude
AND direction, a
number, unit & angle.
(12 m, 300; 8 km/h, N)
A
BD = 12 m, 20o


7. Distance and DisplacementDistance 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.
•• For motion along x or y axis, theFor motion along x or y axis, the displacementdisplacement isis
determined by the x or y coordinate of its finaldetermined by the x or y coordinate of its final
position. Example: Consider a car that travels 8 m, Eposition. Example: Consider a car that travels 8 m, E
then 12 m, W.then 12 m, W.
Net displacementNet displacement DD isis
from the origin to thefrom the origin to the
final position:final position:
What is theWhat is the distancedistance
traveled?traveled? 20 m !!
12 m,W
D
D = 4 m, WD = 4 m, W
x
8 m,E
x = +8x = -4

8. The Signs of DisplacementThe Signs of Displacement
•• Displacement is positive (+) orDisplacement is positive (+) or
negative (negative (--) based on) based on LOCATIONLOCATION..
2 m
-1 m
-2 m
The displacement is
the y-coordinate.
Whether motion is
up or down, + or - is
based on LOCATION.
Examples:
The direction of motion does not matter!The direction of motion does not matter!

9. Definition of SpeedDefinition of Speed
• Speed is the distance traveled per unit
of time (a scalar quantity).
•• SpeedSpeed is the distance traveled per unitis the distance traveled per unit
of time (a scalar quantity).of time (a scalar quantity).
v = =
s
t
20 m
4 s
v = 5 m/sv = 5 m/s
Not direction dependent!
A
Bs = 20 m
Time t = 4 s

10. Definition of VelocityDefinition of Velocity
• Velocity is the displacement per
unit of time. (A vector quantity.)
•• VelocityVelocity is the displacement peris the displacement per
unit of time. (A vector quantity.)unit of time. (A vector quantity.)
v = 3 m/s at 200 N of Ev = 3 m/s at 200 N of E
Direction required!
A
Bs = 20 m
Time t = 4 s
12 m
4 s
D
v
t
 
D=12 m
20o

11. Example 1.Example 1. A runner runsA runner runs 200 m, east,200 m, east, thenthen
changes direction and runschanges direction and runs 300 m, west300 m, west. If. If
the entire trip takesthe entire trip takes 60 s60 s, what is the average, what is the average
speed and what is the average velocity?speed and what is the average velocity?
Recall thatRecall that averageaverage
speedspeed is a functionis a function
onlyonly ofof total distancetotal distance
andand total timetotal time::
Total distance:Total distance: ss = 200 m + 300 m = 500 m= 200 m + 300 m = 500 m
500 m
60 s
total path
Average speed
time
  Avg. speed
8.33 m/s
Direction does not matter!Direction does not matter!
startstart
ss11 = 200 m= 200 mss22 = 300 m= 300 m

12. Example 1 (Cont.)Example 1 (Cont.) Now we find the averageNow we find the average
velocity, which is thevelocity, which is the net displacementnet displacement divideddivided
byby timetime. In this case, the direction matters.. In this case, the direction matters.
xxoo = 0= 0
tt = 60 s= 60 s
xx11= +200 m= +200 mxxff == --100 m100 m0fx x
v
t


xx00 = 0 m;= 0 m; xxff == --100 m100 m
100 m 0
1.67 m/s
60 s
v
 
  
Direction of finalDirection of final
displacement is todisplacement is to
the left as shown.the left as shown.
Average velocity: 1.67 m/s, Westv 
Note: Average velocity is directed to the west.Note: Average velocity is directed to the west.

13. Example 2.Example 2. A sky diver jumps and falls forA sky diver jumps and falls for
600 m in 14 s. After chute opens, he falls600 m in 14 s. After chute opens, he falls
another 400 m in 150 s. What is averageanother 400 m in 150 s. What is average
speed for entire fall?speed for entire fall?
625 m
356 m
14 s
142 s
A
B
600 m + 400 m
14 s + 150 s
A B
A B
x x
v
t t

 

1000 m
164 s
v  6.10 m/sv 
Average speed is a function
only of total distance traveled
and the total time required.
Average speed is a functionAverage speed is a function
onlyonly of total distance traveledof total distance traveled
and the total time required.and the total time required.
Total distance/ total time:Total distance/ total time:

14. Examples of SpeedExamples of Speed
Light = 3 x 108 m/s
Orbit
2 x 104 m/s
Jets = 300 m/s Car = 25 m/s

15. Speed Examples (Cont.)Speed Examples (Cont.)
Runner = 10 m/s
Snail = 0.001 m/s
Glacier = 1 x 10-5 m/s

16. Average Speed andAverage Speed and
Instantaneous VelocityInstantaneous Velocity
The instantaneous
velocity is the magn-
itude and direction of
the speed at a par-
ticular instant. (v at
point C)
The instantaneous
velocity is the magn-
itude and direction of
the speed at a par-
ticular instant. (v at
point C)
 The average speed depends ONLY on the
distance traveled and the time required.
 TheThe averageaverage speedspeed dependsdepends ONLYONLY on theon the
distance traveled and the time required.distance traveled and the time required.
A
Bs = 20 m
Time t = 4 s
C

17. The Signs of VelocityThe Signs of Velocity
First choose + direction;
then v is positive if motion
is with that direction, and
negative if it is against that
direction.
First choose + direction;
then v is positive if motion
is with that direction, and
negative if it is against that
direction.
 Velocity is positive (+) or negative (-)
based on direction of motion.
 Velocity is positive (+) or negative (Velocity is positive (+) or negative (--))
based onbased on direction of motion.direction of motion.
-+
-+
+

18. Average and InstantaneousAverage and Instantaneous vv
x
tt
xx22
xx11
tt22tt11
2 1
2 1
avg
x x x
v
t t t
 
 
 
( 0)inst
x
v t
t

  

x
t
Time
slope
Displacement,x
Average Velocity:Average Velocity: Instantaneous Velocity:Instantaneous Velocity:

19. Definition of AccelerationDefinition of Acceleration
 AnAn accelerationacceleration is the change in velocityis the change in velocity
per unit of time. (Aper unit of time. (A vectorvector quantity.)quantity.)
 AA changechange inin velocityvelocity requires therequires the
application of a push or pull (application of a push or pull (forceforce).).
A formal treatment of force and acceleration willA formal treatment of force and acceleration will
be given later. For now, you should know that:be given later. For now, you should know that:
• The direction of accel-
eration is same as
direction of force.
• The acceleration is
proportional to the
magnitude of the force.

20. Pulling the wagon with twice the force
produces twice the acceleration and
acceleration is in direction of force.
Pulling the wagon with twice the forcePulling the wagon with twice the force
produces twice the acceleration andproduces twice the acceleration and
acceleration is in direction of force.acceleration is in direction of force.
Acceleration and ForceAcceleration and Force
F
a
2F 2a

21. Example of AccelerationExample of Acceleration
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.Wind force is constant, thus acceleration is constant.
+
vf = +8 m/sv0 = +2 m/s
t = 3 s
Force

22. The Signs of AccelerationThe Signs of Acceleration
• Acceleration is positive (+) or negative
(-) based on the direction of force.
•• Acceleration is positive (Acceleration is positive (++) or negative) or negative
((--) based on the) based on the directiondirection ofof forceforce..
Choose + direction first.
Then acceleration a will
have the same sign as
that of the force F —
regardless of the
direction of velocity.
Choose + direction first.Choose + direction first.
ThenThen accelerationacceleration aa willwill
have thehave the same signsame sign asas
that of thethat of the force Fforce F ——
regardless of theregardless of the
direction of velocity.direction of velocity.
F
F
+
a (-)
a(+)

23. Average and InstantaneousAverage and Instantaneous aa
v
t
v2
v1
t2t1
v
t
time
slope
2 1
2 1
avg
v v v
a
t t t
 
 
 
( 0)inst
v
a t
t

  


24. Example 3 (No change in direction):Example 3 (No change in direction): A constantA constant
force changes the speed of a car fromforce changes the speed of a car from 8 m/s8 m/s toto
20 m/s20 m/s inin 4 s4 s. What is average acceleration?. What is average acceleration?
Step 1. Draw a rough sketch.Step 1. Draw a rough sketch.
Step 2. Choose a positive direction (right).Step 2. Choose a positive direction (right).
Step 3. Label given info with + andStep 3. Label given info with + and -- signs.signs.
Step 4. Indicate direction of force F.Step 4. Indicate direction of force F.
+
v1 = +8 m/s
t = 4 s
v2 = +20 m/s
Force

25. Example 3 (Continued):Example 3 (Continued): What is averageWhat is average
acceleration of car?acceleration of car?
Step 5. Recall definitionStep 5. Recall definition
of average acceleration.of average acceleration.
2 1
2 1
avg
v v v
a
t t t
 
 
 
20 m/s - 8 m/s
3 m/s
4 s
a   
3 m/s, rightwarda  
+
v1 = +8 m/s
t = 4 s
v2 = +20 m/s
Force

26. Example 4:Example 4: A wagon moving east atA wagon moving east at 20 m/s20 m/s
encounters a very strong headencounters a very strong head--wind, causing itwind, causing it
to change directions. Afterto change directions. After 5 s5 s, it is traveling, it is traveling
west atwest at 5 m/s5 m/s. What is the average. What is the average
acceleration?acceleration? (Be careful of signs.)(Be careful of signs.)
Step 1. Draw a rough sketch.Step 1. Draw a rough sketch.
+ Force
Step 2. Choose the eastward direction as positive.Step 2. Choose the eastward direction as positive.
vo = +20 m/svf = -5 m/s
Step 3. Label given info with + andStep 3. Label given info with + and -- signs.signs.
E

27. Example 4 (Cont.):Example 4 (Cont.): Wagon moving east atWagon moving east at 20 m/s20 m/s
encounters a headencounters a head--wind, causing it to changewind, causing it to change
directions. Five seconds later, it is traveling west atdirections. Five seconds later, it is traveling west at
5 m/s5 m/s. What is the average acceleration?. 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, v = vf - v0
v = (-5 m/s) - (+20 m/s) = -25 m/s
Choose the eastward direction as positive.Choose the eastward direction as positive.
Initial velocity,Initial velocity, vvoo == +20 m/s, east (+)+20 m/s, east (+)
Final velocity,Final velocity, vvff == --5 m/s, west (5 m/s, west (--))
The change in velocity,The change in velocity, vv == vvff -- vv00
vv = (= (--5 m/s)5 m/s) -- (+20 m/s) =(+20 m/s) = --25 m/s25 m/s

28. Example 4: (Continued)Example 4: (Continued)
aaavgavg == =
vv
tt
vvff -- vvoo
ttff -- ttoo
aa ==
--25 m/s25 m/s
5 s5 s
a = - 5 m/s2a = - 5 m/s2 Acceleration is directed to
left, west (same as F).
+ Force
vo = +20 m/s
vf = -5 m/s
E
v = (-5 m/s) - (+20 m/s) = -25 m/s

29. Signs for DisplacementSigns for Displacement
Time t = 0 at pointTime t = 0 at point AA. What are the signs. What are the signs
(+ or(+ or --) of) of displacementdisplacement atat BB,, CC, and, and DD??
At B, x is positive, right of origin
At C, x is positive, right of origin
At D, x is negative, left of origin
+ Force
vo = +20 m/svf = -5 m/s
E
a = - 5 m/s2
A B
C
D

30. Signs for VelocitySigns for Velocity
What are the signs (+ orWhat are the signs (+ or --) of velocity at) of velocity at
points B, C, and D?points B, C, and D?
 AtAt B,B, vv isis zerozero -- no sign needed.no sign needed.
 AtAt CC,, vv isis positivepositive on way out andon way out and
negativenegative on the way back.on the way back.
 AtAt DD,, vv isis negativenegative, moving to left., moving to left.
+ Force
vo = +20 m/svf = -5 m/s
E
a = - 5 m/s2
A B
C
D
x = 0

31. What are the signs (+ or -) of acceleration at
points B, C, and D?
 The force is constant and always directed
to left, so acceleration does not change.
 AtAt B, C, and DB, C, and D,, aa == --5 m/s,5 m/s, negativenegative
at all points.at all points.
Signs for AccelerationSigns for Acceleration
+ Force
vo = +20 m/svf = -5 m/s
E
a = - 5 m/s2
A B
C
D

32. DefinitionsDefinitions
Average velocity:
Average acceleration:
2 1
2 1
avg
x x x
v
t t t
 
 
 
2 1
2 1
avg
v v v
a
t t t
 
 
 

33. Velocity for constantVelocity for constant aa
Average velocity: Average velocity:
Setting to = 0 and combining we have:
0
0
f
avg
f
x xx
v
t t t

 
 
0
2
f
avg
v v
v


0
0
2
fv v
x x t

 

34. Example 5:Example 5: A ballA ball 5 m5 m from the bottom of anfrom the bottom of an
incline is traveling initially atincline is traveling initially at 8 m/s8 m/s.. FourFour secondsseconds
later, it is traveling down the incline atlater, it is traveling down the incline at 2 m/s2 m/s. How. How
far is it from the bottom at that instant?far is it from the bottom at that instant?
x = xo + t
vo + vf
2
= 5 m + (4 s)
8 m/s + (-2 m/s)
2
5 m
x
8 m/s
-2 m/s
t = 4 s
vo
vf
+ F
CarefulCareful

35. x = 5 m + (4 s)
8 m/s - 2 m/s
2
x = 17 mx = 17 m
x = 5 m + (4 s)
8 m/s + (-2 m/s)
2
5 m
x
8 m/s
-2 m/s
t = 4 s
vo
vf
+ F
(Continued)(Continued)

36. Constant AccelerationConstant Acceleration
Acceleration:Acceleration:
Setting tSetting too = 0 and solving for= 0 and solving for v,v, we have:we have:
Final velocity = initial velocity + change in velocityFinal velocity = initial velocity + change in velocity
0fv v at 
0
0
f
avg
f
v vv
a
t t t

 
 

37. Acceleration in our ExampleAcceleration in our Example
a = -2.50 m/s2a = -2.50 m/s2
What is the meaningWhat is the meaning
of negative sign forof negative sign for aa??
0fv v
a
t


0fv v at 
5 m
x
8 m/s8 m/s
-2 m/s
t = 4 s
vo
v
+ F
2( 2 m/s) ( 8 m/s)
2 m/s
4 s
a
  
  
The force changingThe force changing
speed is down plane!speed is down plane!

38. Formulas based on definitions:Formulas based on definitions:
DerivedDerived formulasformulas:
For constant acceleration onlyFor constant acceleration only
21
0 0 2x x v t at   21
0 2fx x v t at  
0
0
2
fv v
x x t

  0fv v at 
2 2
0 02 ( ) fa x x v v  

39. Use of Initial PositionUse of Initial Position xx00 in Problems.in Problems.
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.
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.
21
0 0 2x x v t at  
21
0 2fx x v t at  
0
0
2
fv v
x x t

 
2 2
0 02 ( ) fa x x v v  
0fv v at 
TheThe xxoo term is veryterm is very
useful for studyinguseful for studying
problems involvingproblems involving
motion of two bodies.motion of two bodies.
00
00
00
00

40. Review of Symbols and UnitsReview 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)
• Displacement ((x, xx, xoo); meters (); meters (mm))
• Velocity ((v,v, vvoo); meters per second (); meters per second (m/sm/s))
• Acceleration ((aa); meters per s); meters per s22 ((m/sm/s22
))
•• TimeTime ((tt); seconds (); seconds (ss))
Review sign convention for each symbolReview sign convention for each symbol

41. The Signs of DisplacementThe Signs of Displacement
• Displacement is positive (+) or
negative (-) based on LOCATION.
•• Displacement is positive (+) orDisplacement is positive (+) or
negative (negative (--) based on) based on LOCATIONLOCATION..
The displacement is
the y-coordinate.
Whether motion is
up or down, + or - is
based on LOCATION.
2 m2 m
--1 m1 m
--2 m2 m

42. The Signs of VelocityThe Signs of Velocity
• Velocity is positive (+) or negative (-)
based on direction of motion.
•• Velocity is positive (+) or negative (Velocity is positive (+) or negative (--))
based onbased on direction of motiondirection 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.
+
-
-
+
+

43. Acceleration Produced by ForceAcceleration Produced by Force
• Acceleration is (+) or (-) based on
direction of force (NOT based on v).
•• Acceleration is (Acceleration is (++) or () or (--) based on) based on
direction ofdirection of forceforce ((NOTNOT based onbased on vv).).
A push or pull (force) is
necessary to change
velocity, thus the sign of
a is same as sign of F.
FF aa((--))
FF aa(+)(+) More will be said later
on the relationship
between F and a.

44. Problem Solving Strategy:Problem Solving Strategy:
 Draw and label sketch of problem.Draw and label sketch of problem.
 IndicateIndicate ++ direction anddirection and forceforce direction.direction.
 List givens and state what is to be found.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.

45. Example 6:Example 6: A airplane flying initially atA airplane flying initially at 400400
ft/sft/s lands on a carrier deck and stops in alands on a carrier deck and stops in a
distance ofdistance of 300 ft.300 ft. What is the acceleration?What is the acceleration?
300 ft
+400 ft/s
vo
v = 0
+ F
Step 1. Draw and label sketch.
Step 2. Indicate + direction and FF direction.
XX00 = 0= 0

46. Example: (Cont.)Example: (Cont.)
300 ft
+400 ft/s
vo
v = 0
+ F
Step 3.Step 3. List given; find
information with signs.
Given:Given: vvoo = +400 ft/s= +400 ft/s
vv = 0= 0
xx = +300 ft= +300 ft
Find:Find: aa = ?; t= ?; t = ?= ?
List t = ?, even thoughList t = ?, even though
time was not asked for.time was not asked for.
XX00 = 0= 0

47. Step 4.Step 4. Select equation
that contains aa and not tt.
300 ft
+400 ft/s
vo
v = 0
+ F
x
2a(x -xo) = v2 - vo
2
0 0
a = =
-vo
2
2x
-(400 ft/s)2
2(300 ft) a = - 267 ft/s2aa == -- 267 ft/s267 ft/s22
Why is the acceleration negative?Why is the acceleration negative?
Continued . . .Continued . . .
Initial position andInitial position and
final velocity are zero.final velocity are zero.
XX00 = 0= 0
Because Force is in a negative direction!Because Force is in a negative direction!

48. Acceleration Due to GravityAcceleration 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.
•• Every object on the earthEvery object on the earth
experiences a common force:experiences a common force:
the force due to gravity.the force due to gravity.
•• This force is always directedThis force is always directed
toward the center of the earthtoward the center of the earth
(downward).(downward).
•• The acceleration due to gravityThe acceleration due to gravity
is relatively constant near theis relatively constant near the
EarthEarth’’s surface.s surface.
Earth
Wg

49. Gravitational AccelerationGravitational Acceleration
• In a vacuum, all objects fall
with same acceleration.
• Equations for constant
acceleration apply as usual.
• Near the Earth’s surface:
•• In a vacuum, all objects fallIn a vacuum, all objects fall
with same acceleration.with same acceleration.
•• Equations for constantEquations for constant
acceleration apply as usual.acceleration apply as usual.
•• Near the EarthNear the Earth’’s surface:s surface:
aa = g == g = 9.80 m/s9.80 m/s22
or 32 ft/sor 32 ft/s22
Directed downward (usually negative).Directed downward (usually negative).

50. Experimental DeterminationExperimental Determination
of Gravitational Acceleration.of Gravitational Acceleration.
The apparatus consists of aThe apparatus consists of a
device which measures the timedevice which measures the time
required for a ball to fall a givenrequired for a ball to fall a given
distance.distance.
Suppose the height is 1.20 mSuppose the height is 1.20 m
and the drop time is recordedand the drop time is recorded
as 0.650 s. What is theas 0.650 s. What is the
acceleration due to gravity?acceleration due to gravity?
yy
tt

51. Experimental Determination ofExperimental Determination of
GravityGravity (y(y00 = 0; y == 0; y = --1.20 m)1.20 m)
yy
tt
y =y = --1.20 m; t = 0.495 s1.20 m; t = 0.495 s
21
0 02 ; 0y v t at v  
2 2
2 2( 1.20 m)
(0.495 s)
y
a
t

 
2
9.79 m/sa  
Acceleration
of Gravity:
Acceleration a is negative
because force W is negative.
AccelerationAcceleration aa isis negativenegative
because forcebecause force WW isis negative.negative.
++
WW

52. Sign Convention:Sign Convention:
A Ball ThrownA Ball Thrown
Vertically UpwardVertically Upward
• Velocity is positive (+) or
negative (-) based on
direction of motion.
•• Velocity is positive (+) orVelocity is positive (+) or
negative (negative (--) based on) based on
direction of motiondirection of motion..
• Displacement is positive
(+) or negative (-) based
on LOCATION.
•• Displacement is positiveDisplacement is positive
(+) or negative ((+) or negative (--) based) based
onon LOCATIONLOCATION..
Release Point
UP = +
TippensTippens
•• Acceleration is (+) or (Acceleration is (+) or (--))
based on direction ofbased on direction of forceforce
(weight).(weight).
y = 0
y = +
y = +
y = +
y = 0
y = -
NegativeNegative
v = +
v = 0
v = -
v = -
v= -
NegativeNegative
a = -
a = -
a = -
a = -
a =a = --

53. Same Problem SolvingSame Problem Solving
Strategy ExceptStrategy Except aa = g= g::
 Draw and label sketch of problem.Draw and label sketch of problem.
 IndicateIndicate ++ direction anddirection and forceforce direction.direction.
 List givens and state what is to be found.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.

54. Example 7:Example 7: A ball is thrown vertically upward withA ball is thrown vertically upward with
an initial velocity ofan initial velocity of 30 m/s30 m/s. What are its position. What are its position
and velocity afterand velocity after 2 s2 s,, 4 s4 s, and, and 7 s7 s??
Step 1. Draw and
label a sketch.
a = g
+
vo = +30 m/s
Step 2. Indicate + direction
and force direction.
Step 3. Given/find info.
a = -9.8 ft/s2 t = 2, 4, 7 s
vo = + 30 m/s y = ? v = ?

55. Finding Displacement:Finding Displacement:
a = g
+
vo = 30 m/s
0
y =y = (30 m/s)(30 m/s)tt ++ ½½((--9.8 m/s9.8 m/s22
))tt22
Substitution of t = 2, 4, and 7 sSubstitution of t = 2, 4, and 7 s
will give the following values:will give the following values:
y = 40.4 m; y = 41.6 m; y = -30.1 my = 40.4 m; y = 41.6 m; y = -30.1 m
21
0 0 2y y v t at  
Step 4. Select equation
that contains y and not v.

56. Finding Velocity:Finding Velocity:
Step 5. Find v from equation
that contains v and not x:
Step 5. Find v from equation
that contains v and not x:
Substitute t = 2, 4, and 7 s:Substitute t = 2, 4, and 7 s:
v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/sv = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s
a = g
+
vo = 30 m/s
0fv v at 
2
30 m/s ( 9.8 m/s )fv t  

57. Example 7: (Cont.)Example 7: (Cont.) Now findNow find
thethe maximum heightmaximum height attained:attained:
Displacement is a maximumDisplacement is a maximum
when the velocitywhen the velocity vvff is zero.is zero.
a = g
+
vo = +96 ft/s
2
30 m/s ( 9.8 m/s ) 0fv t   
2
30 m/s
; 3.06 s
9.8 m/s
t t 
To findTo find yymaxmax we substitutewe substitute
tt = 3.06 s= 3.06 s into the generalinto the general
equation for displacement.equation for displacement.
y =y = (30 m/s)(30 m/s)tt ++ ½½((--9.8 m/s9.8 m/s22
))tt22

58. Example 7: (Cont.)Example 7: (Cont.) Finding theFinding the maximum height:maximum height:
y =y = (30 m/s)(30 m/s)tt ++ ½½((--9.8 m/s9.8 m/s22
))tt22
a = g
+
vo =+30 m/s
tt = 3.06 s= 3.06 s
21
2(30)(3.06) ( 9.8)(3.06)y   
yy = 91.8 m= 91.8 m -- 45.9 m45.9 m
Omitting units, we obtain:Omitting units, we obtain:
ymax = 45.9 m

59. Summary of FormulasSummary of Formulas
DerivedDerived FormulasFormulas:
For Constant Acceleration Only
21
0 0 2x x v t at   21
0 2fx x v t at  
0
0
2
fv v
x x t

  0fv v at 
2 2
0 02 ( ) fa x x v v  

60. Summary: ProcedureSummary: Procedure
 Draw and label sketch of problem.Draw and label sketch of problem.
 IndicateIndicate ++ direction anddirection and forceforce direction.direction.
 List givens and state what is to be found.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.

61. CONCLUSION OFCONCLUSION OF
Chapter 6Chapter 6 -- AccelerationAcceleration