1
Chapter 3 Kinematics I: rectilinear motion
Kinematics is the theory about the description
of motion.
Physical theories are creations of the human
intellect; they must be invented rather than
discovered.
Motion implies change, and it is change
make life-and physics-visible and interesting.
2
§3.1 Position and displacement of a particle
1. Several concepts
Frame of reference —any object that is
chosen for reference of a motion.
Particle—a single mass point without shape.
An object whose part are all move in exactly
the same way can be treated as a particle; or
a complex object can be treated as a particle
if there are no internal motions or the
internal motions can be neglected for the
problem which you are discussing.
Coordinate system—the abstract of reference
2. The position vector and displacement vector
of a particle in rectilinear motion
x
O
ixr
ii
?
=
r
x
O
ixr
ff
?
=
r
Note:
ixixxrrr
ifif
??
)( ?? =?=?=
rrr
Displacement—the change of the position vector
§3.1 Position and displacement of a particle
1 The magnitude of the displacement is not
necessarily equal to the total distance
traveled by the particle during the time
interval ?t . .zeronotbemay,0when sr ?? =
r
s?
3
Note:
3 The position of a moving particle is a
function of time,
r
r
?2 is independent of the the specific
coordinate system we choose.
itxtr
?
)()( =
r
x
t
Agraphofx versus t
§3.1 Position and displacement of a particle
4the path of the
particle is a straight
line.
2. Instantaneous velocity
and instantaneous speed
§3.2 The speed and velocity of a moving particle
1. Average speed and average velocity
Define:
t
s
v
?
?
=
ave
t
r
v
?
?
r
r
=
ave
aveave
vvrs
rr
Q ≠∴≠ ??
t
x
1
t?
2
t?
)(tx
)(
1
tx
)(
2
tx
itxtrr
i
?
)()( ==
rr
ittxttrr
f
?
)()( ?? +=+=
rr
4
itxttxtrttrr
?
)]()([)()( ?+=?+= ???
rrr
t
itxttx
t
r
v
?
?
?
?
?
)]()([
ave
?+
==
r
r
i
t
x
t
r
t
trttr
vtv
tt
?
d
d
d
d)()(
limlim)(
s0
ave
s0
==
?+
==
→→
rrr
rr
?
?
??
Instantaneous velocity:
The component of the velocity:
t
tx
tv
x
d
)(d
)( =
§3.2 The speed and velocity of a moving particle
v
t
s
t
r
t
r
v
srr
tt
t
====∴
==
→→
→
d
d
d
d
limlim
ddlim
s0
ave
s0
s0
rr
r
rr
Q
?
?
?
??
?
Instantaneous speed:
t
ts
t
s
vtv
tt
d
)(d
limlim)(
s0
ave
s0
===
→→
?
?
??
The direction of the velocity:
0
d
)(d
>
t
tx
0
d
)(d
<
t
tx
same as ,
i
?
i
?
opposite of
§3.2 The speed and velocity of a moving particle
5
§3.3 The acceleration in rectilinear motion
Define:
t
vv
t
v
a
if
??
?
rr
r
r
?
==
ave
1. Average acceleration
i
t
tvtv
iaa
ixfx
x
?
)()(
?
aveave
?
?
==
r
2. Instantaneous acceleration
if
ixvttvvvv
ittvvitvv
xxif
xfxi
?
)]()([
?
)(,
?
)(
?+=?=
+==
??
?
rrr
rr
then
t
tv
t
itvttv
t
v
a
xx
tt
d
)(d
?
)]()([
limlim
s0s0
rr
r
=
?+
==
→→
?
?
?
?
??
or
i
t
tx
t
tr
i
t
tv
t
tv
a
x
?
d
)(d
d
)(d
d
)(d
d
)(d
2
2
2
2
====
r
r
r
r
The instantaneous acceleration of a particle is
the time rate of change of the velocity vector
or the first derivative of the instantaneous
velocity vector with respect to time;or the
second derivative of the instantaneous position
vector with respect to time.
§3.3 The acceleration in rectilinear motion
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2
2
d
)(d
d
)(d
)(
t
tx
t
tv
ta
x
x
==
The component of the acceleration:
§3.3 The acceleration in rectilinear motion
3. How to detect and measure the
instantaneous acceleration
Accelerometer—a plum bob
Direction of deviation of plum bob
from vertical is opposite to the
direction of the acceleration of the
particle in the horizontal plane.
a
r
θ
a
r
θ
§3.4 Rectilinear motion with a constant
acceleration
1. Some rules
§3.4 Rectilinear motion with a constant
acceleration
1ignore the effects of air resistance
2the origin of the coordinate could be chosen
discretionarily
3write the constant acceleration as
4choose the initial time instant to be
iaa
x
?
=
r
0=
i
t
tttt
if
=?=?
5let
00
)0()()0()(
xxixi
vvtvxxtx ====
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2. Rectilinear motion with a constant acceleration
from
x
x
a
t
tv
=
d
)(d
We have
tavtv
tatv
xxx
x
tv
v
x
x
x
+=
=
∫∫
0
t
0
)(
)(
d)(d
0
Likewise, from
)(
d
)(d
tv
t
tx
x
=
We have
2
00
0
0
t
0
)(
2
1
)(
d)(d)(d
0
tatvxtx
ttavtvtx
xx
t
xxx
tx
x
++=
+==
∫∫∫
§3.4 Rectilinear motion with a constant
acceleration
One –dimensional motion with constant
acceleration:
itvtvtavtv
xxxx
?
)()(where)(
0
=+=
r
itxtrtatvxtx
xx
?
)()(where
2
1
)(
2
00
=++=
r
iaaa
xx
?
whereconstant ==
r
§3.4 Rectilinear motion with a constant
acceleration
Eliminate t in equations about v
x
and x
)(2
0
2
0
2
xxavv
xxx
?=?
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§3.5 Geometric interpretations
1. The change in the position vector component
∫∫
==?=
f
i
f
i
t
t
x
tx
tx
x
ttvxxttvx d)(dd)(d
)(
)(
2. The change in the velocity component
∫∫
==?=
f
i
fx
ix
t
t
x
tv
tv
xxxx
ttavvtav d)(ddd
)(
)(
tt
)(tv
x )(tv
x
i
t
i
tf
t
f
t
Constant speed motionAccelerated motion
t
x
v
x
d
d
=
t
v
a
x
x
d
d
=
tt
)(ta
x )(ta
x
i
t
i
tf
t
f
t
Constant acceleration motionAccelerated motion
tt
)(ta
x
)(tv
x
Accelerated motion
§3.5 Geometric interpretations
3. What does the negative areas mean?
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x
a
x
v
x
§3.5 Geometric interpretations
§3.5 Geometric interpretations
A rebounding ball bearing
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Exercise 1
A geologist at the top of a 100-meter-deep
crevasse cannot resist the temptation to hurl
a gneiss rock down to the bottom. The rock
has an initial downward speed of 10.0m/s as
it leaves the geologist’s hand when t =0s.
Find
a. The time interval during which the particle
is in flight, called the time of flight; and
b. The speed of the particle at the instant just
before impact.
R. P96
A rocket is launched from rest from an
underwater base a distance of 125m below the
surface of a body of water. It moves vertically
upward with an unknown but assumed constant
acceleration , and it reaches the surface in a time
of 2.15s. When it breaks the surface its engines
automatically shut off and it continues to rise.
What maximum height does it reach?
Exercise 2
(H. P29)