K
K
i n e
i n e
m
m
a
a
t i
t i
c s
c s
i n O
i n O
n
n
e
e
D i
D i
m
m
e
e
n
n
s i o n
s i o n
Chapter 2
Chapter 2
Outline
Outline
•
•
Displacement
Displacement
••
Speed & velocity
Speed & velocity
••
Acceleration
Acceleration
••
Application of equati
Application
of equations
ons
••
Free falling bodies
Free falling bodies
•Outline
Outline
•
•
Displacement
Displacement
••
Speed & velocity
Speed & velocity
••
Acceleration
Acceleration
••
Application of equati
Application
of equations
ons
••
Free falling bodies
Free falling bodies
•K i n e m a t i c s
K i n e m a t i c s
deals with the concepts that
deals with the concepts that
are needed to describe motion.
are needed to describe motion.
D y n a m i c s
D y n a m i c s
deals with the effect that forces
deals with the effect that forces
have on motion.
have on motion.
Together, kinematics and dynamics form
Together, kinematics and dynamics form
the branch of physics known as
2.1 2.1 D i s p l a c em e n tD i s p l a c em e n t
pos
posiiti
tion
on
initial
initial
o o x x pos
posiiti
tion
on
final
final
x x nt
nt
displaceme
displaceme
o o x x x x x x 2.1 D i s p l a c em e n t
m
0
.
2
o x m
0
.
7
x m
0
.
5
x m
0
.
5
m
2.0
m
7.0
o x x x 2.1 D i s p l a c em e n t
m
0
.
2
x m
0
.
7
o x m
0
.
5
x m
0
.
5
m
7.0
m
2.0
o x x x 2.1 D i s p l a c em e n t
m
0
.
2
o x m
0
.
7
x m
0
.
5
x
2
.0
m
7
.
0
m
m
5.0
o x x x 2.2 S p eed a n d V el o c i t y
A verage sp eed is the distance traveled divided by the time
required to cover the distance.
time
Elapsed
Distance
speed
Average
2.2 S p eed a n d V el o c i t y
Exam ple 1
Distance Run by a Jogger
How far does a jogger run in 1.5 hours (5400 s) if his average speed is 2.22 m/s? time Elapsed Distance speed Average
2
.
22
m
s
5400
s
12000
m
time
Elapsed
speed
Average
Distance
2.2 S p eed a n d V el o c i t y
A v er ag e v e lo c i t y is the displacement divided by the elapsed
time. time Elapsed nt Displaceme velocity Average
t
t
t
o o
x x x v 2.2 S p eed a n d V el o c i t y
E x am p l e 2 The World’s Fastest Jet-Engine Car
Andy Green in the car ThrustSSC set a world record of 341.1 m/s in 1997. To establish such a record, the driver makes two runs through the course, one in each direction, to nullify wind effects. From the data, determine the average velocity for each run.
2.2 S p eed a n d V el o c i t y s m 5 . 339 s 4.740 m 1609 t x v s m 7 . 342 s 4.695 m 1609 t x v
2.2 S p eed a n d V el o c i t y
The
i n s t a n t an e o u s v e lo c i t yindicates how fast
the car moves and the direction of motion at each
instant of time.
t
t
x v 0lim
2.3 A c c e l e r at i o n
The notion of acceleration emerges when a change in velocity is combined with the time during which the
2.3 A c c e l e r at i o n t t t o o v v v a
2.3 A c c e l e r at i o n
E x am p l e 3 Acceleration and Increasing Velocity Determine the average acceleration of the plane.
s m 0 o v h km 260 v s 0 o t t
29
s
s h km 0 . 9 s 0 s 29 h km 0 h km 260 o o t t v v a 2.3 A c c e l e r at i o n
E x a m p l e 3 Acceleration and Decreasing Velocity 2 s m 0 . 5 s 9 s 12 s m 28 s m 13 o o t t v v a
2.4 E q u at i o n s o f K i n e m a t i c s fo r C o n s t a n t A c c e l er at i o n
It is customary to dispense with the use of boldface symbols overdrawn with arrows for the displacement, velocity, and acceleration vectors. We will, however, continue to convey the directions with a plus or minus sign.
o o t t v v a o o t t x x v o o t t x x v o o
t
t
v
v
a
2.4 E q u at i o n s o f K i n e m a t i c s fo r C o n s t a n t A c c e l er at i o n o o t t x x v
0
ox
t
o
0
v
v
t
t
v
x
21 o
Let the object be at the origin when the clock starts.
t x v
2.4 E q u at i o n s o f K i n e m a t i c s fo r C o n s t a n t A c c e l er at i o n o o
t
t
v
v
a
t
v
v
a
o
ov
v
at
at
v
v
o
2.4 E q u at i o n s o f K i n e m a t i c s fo r C o n s t a n t A c c e l er at i o n
Five kinematic variables:
1. displacement, x
2. acceleration (constant), a
3. final velocity (at time t ), v
4. initial velocity, v
o2.4 E q u at i o n s o f K i n e m a t i c s fo r C o n s t a n t A c c e l er at i o n
at
v
v
o
v
v
t
v
v
at
t
x
21 o
12 o
o
2 2 1at
t
v
x
o
2.4 E q u at i o n s o f K i n e m a t i c s fo r C o n s t a n t A c c e l er at i o n
m
110
s
0
.
8
s
m
0
.
2
s
0
.
8
s
m
0
.
6
12 2 2 2 2 1
v
t
at
x
o2.4 E q u at i o n s o f K i n e m a t i c s fo r C o n s t a n t A c c e l er at i o n
Ex amp le 6 Catapulting a Jet
Find its displacement.
s
m
0
o v??
x 2s
m
31
as
m
62
v2.4 E q u at i o n s o f K i n e m a t i c s fo r C o n s t a n t A c c e l er at i o n
a
v
v
v
v
t
v
v
x
o o o
2 1 2 1 t v v a o a v v t o a
v
v
x
o2
2 2
2.4 E q u at i o n s o f K i n e m a t i c s fo r C o n s t a n t A c c e l er at i o n
31
m
s
62
m
2
s
m
0
s
m
62
2
2 2 2 2 2
a
v
v
x
o2.4 E q u at i o n s o f K i n e m a t i c s fo r C o n s t a n t A c c e l er at i o n
Equations of Kinematics for Constant Acceleration
v
v
t
x
21 o
2 2 1at
t
v
x
o
at
v
v
o
ax
v
v
2
o2
2
2.5 A p p l i c at i o n s o f t h e Eq u a t i o n s o f K i n e m a ti c s
Reasoning Strategy
1. Make a drawing.
2. Decide which directions are to be called positive (+) and negative (-).
3. Write down the values that are given for any of the five kinematic variables.
4. Verify that the information contains values for at least three of the five kinematic variables. Select the appropriate equation. 5. When the motion is divided into segments, remember that
the final velocity of one segment is the initial velocity for the next. 6. Keep in mind that there may be two possible answers to a
2.5 A p p l i c at i o n s o f t h e Eq u a t i o n s o f K i n e m a ti c s
Ex amp le 8 An Accelerating Spacecraft
A spacecraft is traveling with a velocity of +3250 m/s. Suddenly the retrorockets are fired, and the spacecraft begins to slow down with an acceleration whose magnitude is 10.0 m/s2. What is
the velocity of the spacecraft when the displacement of the craft is +215 km, relative to the point where the retrorockets began firing?
x
a
v
v
o
t
2.5 A p p l i c at i o n s o f t h e Eq u a t i o n s o f K i n e m a ti c s
ax
v
v
2
o2
2
x
a
v
v
o
t
+215000 m
-10.0 m/s
2?
+3250 m/s
ax
v
v
o2
2
s
m
2500
m
215000
s
m
0
.
10
2
s
m
3250
2 2
v
2.6 F r ee l y F a ll i ng B o d i es
In the absence of air resistance, it is found that all bodies at the same location above the Earth fall vertically with the same acceleration. If the distance of the fall is small compared to the radius of the Earth, then the acceleration remains essentially constant throughout the descent.
This idealized motion is called free-fall and the acceleration of a freely falling body is called the acceleration due to
gravity . 2 2
s
ft
2
.
32
or
s
m
80
.
9
g
2.6 F r ee l y F a ll i ng B o d i es 2
s
m
80
.
9
g
2.6 F r ee l y F a ll i ng B o d i es
Exam ple 10 A Falling Stone
A stone is dropped from the top of a tall building. After 3.00s of free fall, what is the displacement y of the stone?
2.6 F r ee l y F a ll i ng B o d i es
y
a
v
v
o
t
2.6 F r ee l y F a ll i ng B o d i es
y
a
v
v
o
t
?
-9.80 m/s
20 m/s
3.00 s
m
1
.
44
s
00
.
3
s
m
80
.
9
s
00
.
3
s
m
0
12 2 2 2 2 1
v
t
at
y
o2.6 F r ee l y F a ll i ng B o d i es
Exam ple 12 How High Does it Go?
The referee tosses the coin up with an initial speed of 5.00m/s. In the absence if air resistance, how high does the coin go above its point of release?
2.6 F r ee l y F a ll i ng B o d i es
y
a
v
v
o
t
?
-9.80 m/s
20 m/s
+5.00
2.6 F r ee l y F a ll i ng B o d i es
y
a
v
v
o
t
?
-9.80 m/s
20 m/s
+5.00
m/s
ay
v
v
2
o2
2
a
v
v
y
o2
2 2
9
.
80
m
s
1
.
28
m
2
s
m
00
.
5
s
m
0
2
2 2 2 2 2 a
v
v
y
o2.6 F r ee l y F a ll i ng B o d i es
Conc eptual Exam pl e 14 Acceleration Versus Velocity
There are three parts to the motion of the coin. On the way up, the coin has a vector velocity that is directed upward and has decreasing magnitude. At the top of its path, the coin
momentarily has zero velocity. On the way down, the coin
has downward-pointing velocity with an increasing magnitude.
In the absence of air resistance, does the acceleration of the coin, like the velocity, change from one part to another?
2.6 F r ee l y F a ll i ng B o d i es
Conc eptual Exam pl e 15 Taking Advantage of Symmetry
Does the pellet in part b strike the ground beneath the cliff with a smaller, greater, or the same speed as the pellet
2.7 G r ap h i c a l A n a l y s i s o f V e lo c i t y a n d Ac c e l er a ti o n
