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1. Chapter 2
Kinematics in One Dimension
Giancoli, PHYSICS,6/E © 2004. Electronically
reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey

2. Module 2
Displacement, Velocity and Acceleration
Giancoli, Sec 2-1, 2, 3, 4, 8
AP Physics C Lesson 1
Giancoli, PHYSICS,6/E © 2004. Electronically
reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey
The following is an excellent lecture on this material.

3. Reference Frames
• Any measurement of position, displacement, velocity
or acceleration must be made with respect to a defined
reference frame—this is first step in problem solution.
Possible reference frames:
•Window with up = + or –
•Un-stretched net with up = + or -
•Stretched net with up = + or –
•Ground = not sufficient
information
Module 2-1

4. Coordinate Axis
• We will use a set of coordinate axis where x is
horizontal and y is vertical
+y
-y
+x-x
x1 x2
•Many problems will be motion in one dimension
so we will plot x vs. time.
Module 2-2

5. Displacement
Displacement: change in position
12 xxx −=∆
+y
-y
+x-x
x1 x2
Displacement is a vector, so it has magnitude and
direction. In one dimension we use + or minus sign to
indicate direction.
Module 2-3

6. Don’t Confuse Displacement and Distance
A person walks 70 m East and 30 m West.
Distance traveled =
Displacement =
100 m
40 m East or + 40 m
Module 2-4

7. Negative Displacement
In the figure below the displacement is negative.
A negative displacement may indicate motion toward
the West or something else depending on the
situation and the coordinate system chosen.
12 xxx −=∆ mmm 203010 −=−=
Module 2-5

8. Average Speed and Velocity
elapsedtime
traveleddistance
speedaverage =
elapsedtime
ntdisplaceme
(velocityaverage =)v
t
x
tt
xx
v
∆
∆
=
−
−
=
12
12
Average velocity is a vector, so it has magnitude
and direction. In one dimension we use + or minus
sign to indicate direction.
Module 2-6

9. t
x
t
x
v =
∆
∆
=
21
ttt ∆+∆=
h
km
km
h
km
km
t
990
2800
790
3100
+=
Example 1. An airplane travels east 3100 km at a velocity of 790 km/h,
and then encounters a tailwind that boosts its velocity to 990 km/h for
the next 2800 km. What was the total time for the trip? (Assume three
significant figures)
hhht 75.683.292.3 =+=
2
2
1
1
v
x
v
x
t
∆
+
∆
=
Module 2-7

10. Example 1 (continued) An airplane travels east 3100 km at a velocity of 790
km/h, and then encounters a tailwind that boosts its velocity to 990 km/h for
the next 2800 km. What was the total time for the trip? (Assume three
significant figures)
What was the average velocity of the plane for this trip?
ht 75.6=
t
x
v
∆
∆
= h
kmkm
75.6
28003100 +
=
h
km874=
Note: A simple average of v1 and v2 gives 890 km/h and is not correct
Module 2-8

11. Instantaneous Velocity
instantaneous velocity is defined as the average velocity
over an infinitesimally short time interval.
t
x
v
∆
∆
→∆
=
0t
lim
Module 2-9

12. Graphical Analysis of Linear Motion
graphtimevs.ofslopeisVelocity x
t
x
v =
∆
∆
=
Module 2-10

13. Acceleration
Average Acceleration: change in velocity
divided by the time taken to make this change.
t
v
tt
vv
a
∆
∆
=
−
−
=
12
12
2
UNITS
s
m
s
s
m
==
Module 2-11

14. s
mvst
s
mvt
0.50.5
0.150
22
11
+==
+==
12
12
tt
vv
a
−
−
=
ss
s
m
s
m
00.5
0.150.5
−
−
=
Example 2. A car traveling at 15.0 m/s slows down to 5.0 m/s in
5.0 seconds. Calculate the car’s acceleration.
Coordinate System: + is to the right
20.2
s
m−= )lefttheto(
Module 2-12

15. Acceleration
Instantaneous Acceleration: same definition as
before but over a very short ∆ t.
t
v
a
∆
∆
→∆
=
0t
lim
Module 2-13

16. Module 3
Motion with Constant Acceleration
Giancoli, Sec 2-1, 2, 3, 4, 8
Giancoli, PHYSICS,6/E © 2004. Electronically
reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey

17. Motion with Constant Acceleration
t (s) v ( m/s) a ( m/ s2
)
0 0 15
1 15 15
2 30 15
3 45 15
4 60 15
5 75 15
Consider the case of a car that accelerates from rest with a constant
acceleration of 15 m / s2
.
t
v
tt
vv
a
∆
∆
=
−
−
=
12
12
)( 1212 ttavv −+= ( ) )0(150 22 −+= t
s
m
We can make a table
Module 3-1

18. Derivations
•In the next 4 slides we will combine several known
equations under the assumption that the acceleration is
constant.
•This process is called a derivation.
•In general you will need to know the initial
assumptions, the resultant equations and how to apply
them.
•You do not need to memorize derivations
•But, I could ask you to derive an equation for a
specific problem. This is very similar to an ordinary
problem without a numeric answer.
Module 3-2

19. Motion at Constant Acceleration - Derivation
•Consider the special case acceleration equals a
constant: a = constant
• Use the subscript “0” to refer to the initial
conditions
• Thus t0 refers to the initial time and we will set t0 =
0.
• At this time v0 is the initial velocity and x0 is the
initial displacement.
•At a later time t, v is the velocity and x is the
displacement
•In the equations t1t0 and t2  t Module 3-3

20. Motion at Constant Acceleration - Derivation
1)(Eqn.0
0
0
t
xx
tt
xx
v
−
=
−
−
=
)2Eqn.(Constant0
=
−
=
t
vv
a
The average velocity during this time is:
The acceleration is assumed to be constant
From this we can write
)3Eqn.(0 tavv +=
t
v
ov
Module 3-4

21. Motion at Constant Acceleration - Derivation
Because the velocity increases at a uniform rate, the average
velocity is the average of the initial and final velocities
)4Eqn.(
2
0 vv
v
+
=
From the definition of average velocity
t
atvv
xt
vv
xtvxx )
2
()
2
( 00
0
0
00
++
+=
+
+=+=
And thus
)5Eqn.(
2
1 2
00 tatvxx ++=
t
v
ov
Module 3-5

22. Motion at Constant Acceleration
• The book derives one more equation by eliminating
time
• The 4 equations listed below only apply when
a = constant
tavv += 0
2
00
2
1
tatvxx ++=
)(2 0
2
0
2
xxavv −+=
2
0vv
v
+
=
Module 3-6

23. mx
s
mv
xv
0.155.11
00 00
==
==
( )0
2
0
2
2 xxavv −+=
)(2 0
2
0
2
xx
vv
a
−
−
=
Example 3. A world-class sprinter can burst out of the blocks to
essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race.
What is the average acceleration of this sprinter, and how long does it
take her to reach that speed? (Note: we have to assume a=constant)
)00.15(2
0)5.11( 22
−
−
=
m
s
m
a
241.4
s
ma =
atvv += 0
a
vv
t 0
−
=
241.4
05.11
s
m
s
m
t
−
=
st 61.2=
Module 3-7

24. 2
00
2
1
tatvxx tttt
++=
t
s
mx t )25(=
2
00
2
1
tatvxx cccc ++=
Example 4. A truck going at a constant speed of 25 m/s passes a car at rest.
The instant the truck passes the car, the car begins to accelerate at a constant
1.00 m / s2
. How long does it take for the car to catch up with the truck.
How far has the car traveled when it catches the truck?
2
2 )0.1(
2
1
t
s
mx c =
When the car catches the truck:
tc
xx =
t
s
mt
s
m )25()0.1(
2
1 2
2 =
s
mt
s
m 25)0.1(
2
1
2 =
st 50=
2
2 )50)(0.1(
2
1
s
s
mxc
= m1250=
Module 3-8

25. Ch 2 25
Quiz
Calculate the average acceleration from t = 0 s to t = 30s.
A 0.17 m/s2
B 0.25 m/s2
C 15 m/s2
D none of the above

26. Ch 2 26
Quiz
Calculate the acceleration from t = 0 s to t = 15 s.
A 0 m/s2
B 1.0 m/s2
C 15 m/s2
D none of the above

27. ConcepTest 2.8aConcepTest 2.8a Acceleration IAcceleration I
If the velocity of a car is non-zeroIf the velocity of a car is non-zero
((vv ≠≠ 00), can the acceleration of), can the acceleration of
the car be zero?the car be zero?
A) yes
B) no
C) depends on the velocity

28. ConcepTest 2.8aConcepTest 2.8a Acceleration IAcceleration I
Sure it can! An object moving with constantconstant velocityvelocity
has a non-zero velocity, but it has zerozero accelerationacceleration
since the velocity is not changing.
If the velocity of a car is non-zeroIf the velocity of a car is non-zero
((vv ≠≠ 00), can the acceleration of), can the acceleration of
the car be zero?the car be zero?
A) yes
B) no
C) depends on the velocity

29. Example 6: Calculate the acceleration between points A and B and B
and C.
t
v
a
∆
∆
→∆
=
0t
lim
C)B&BAforlinestraight( →→
∆
∆
=
t
v
a
20.0
0.150.20
0.15
s
m15.0
s
m
ss
s
m
=
−
−
=
12
12
tt
vv
a BA
−
−
=
20.2
0.200.25
0.15
s
m5.0
s
m
ss
s
m
−=
−
−
=
12
12
tt
vv
aBC
−
−
=
Module 3-10

30. Graphical Analysis of Linear Motion
t
x
v
∆
∆
→∆
=
0t
lim
t
v
a
∆
∆
→∆
=
0t
lim
v is slope of position
vs. time graph.
a is slope of velocity
vs. time graph.
Module 3-9

31. Module 4
Falling Objects
Giancoli, Sec 2-1, 2, 3, 4, 8
Giancoli, PHYSICS,6/E © 2004. Electronically
reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey

32. Falling Objects
•Galileo showed that for object falling from rest
with no air resistance
y ∝ t2
•Note that this is true when acceleration is
constant
Module 4-1

33. Falling Objects
Galilei showed that
•near the surface of the earth
•at the same location
•in the absence of air resistance
all objects fall with the same constant
acceleration g, the acceleration
due to gravity
g = 9.8 m/s2
Note: g is a positive number. When you define your
coordinate system, if up is positive then a = - g.
smv /8.91 =
smv /6.192 =
smv /4.293 =
Module 4-2

34. Up and Down Motion
For object that is thrown upward
and returns to starting position:
• assumes up is positive
•velocity changes sign (direction)
but acceleration does not
•Velocity at top is zero
•time up = time down
•Velocity returning to starting
position = velocity when it was
released but opposite sign
2
00
2
1
gttvyy −+=
gtvv −= 0
Module 4-3

35. Acceleration due to Gravity
Module 4-4

36. Example 5
myy
s
ma
s
mv
700
8.90.12
0
20
−==
−=+=
2
00
2
1
gttvyy −+=
2
9.412070 tt −+=−
(2-47) A stone is thrown vertically upward with a speed of 12.0
m/s from the edge of a cliff. The distance from the bottom to the release point
is 70.0 m.(a) How much later does it reach the bottom of the cliff? (b) What is
its speed just before hitting? (c) What total distance did it travel?
(a) UP = POSITIVE
070129.4 2
=−− tt
)9.4)(2(
)70)(9.4)(4()12()12( 2
−−−±−−
=t
sst 20.5,75.2−=
Use quadratic equation:
Answer = 5.20 s
Module 4-5

37. Example 5
atvv += 0
)2.5)(8.9(12 2 s
s
m
s
m −+=
2-47 A stone is thrown vertically upward with a speed of 12.0 m/s
from the edge of a cliff. The distance from the bottom to the release point is
70.0 m.(a) How much later does it reach the bottom of the cliff? (b) What is its
speed just before hitting? (c) What total distance did it travel?
(b)
s
m9.38−=
(c) Find maximum height, where 0=v
)(2 0
2
0
2
yyavv −+=
))(2(
2
0
2
0
a
vv
yy
−
+=
)8.9)(2(
)12(0
0
2
2
s
m
s
m
y
−
−
+= m35.7=
Total distance = mmm 0.7035.735.7 ++ m7.84=
)(2 0
2
0
2
yyavv −=−
Module 4-6
timetotal
distancetotal
SpeedAverage = sm
s
m
3.16
20.5
7.84
==

38. Example 5 (2-47) continued A stone is thrown vertically upward with a speed of
12.0 m/s from the edge of a cliff. The distance from the bottom to the release
point is 70.0 m. Excel Calculation—use the equation for displacement and
velocity to calculate y and vy vs time.
Time Y (m) v y (m/s)
(s)
0.00 0.00 12.00
0.25 2.69 9.55
0.50 4.78 7.10
0.75 6.24 4.65
1.00 7.10 2.20
1.25 7.34 -0.25
1.50 6.98 -2.70
1.75 5.99 -5.15
2.00 4.40 -7.60
2.25 2.19 -10.05
2.50 -0.63 -12.50
2.75 -4.06 -14.95
3.00 -8.10 -17.40
3.25 -12.76 -19.85
3.50 -18.03 -22.30
3.75 -23.91 -24.75
4.00 -30.40 -27.20
4.25 -37.51 -29.65
4.50 -45.23 -32.10
4.75 -53.56 -34.55
5.00 -62.50 -37.00
5.25 -72.06 -39.45Module 4-7

39. Problem Solving Tips
1. Read the whole problem and make sure you
understand it. Then read it again.
2. Decide on the objects under study and what the
time interval is.
3. Draw a diagram and choose coordinate axes.
4. Write down the known (given) quantities, and
then the unknown ones that you need to find.
5. What physics applies here? Plan an approach to a
solution.
Module 4-8

40. Problem Solving Tips
6. Which equations relate the known and
unknown quantities? Are they valid in this
situation? Solve algebraically for the unknown
quantities, and check that your result is
sensible (correct dimensions).
7. Calculate the solution and round it to the
appropriate number of significant figures.
8. Look at the result – is it reasonable? Does it
agree with a rough estimate?
9. Check the units again.
Module 4-9