Chapter 2 Motion in one
dimension
Kinematics
Dynamics
Section2-3 Position, velocity and
acceleration vectors
1. Position vector
At any particular time t, the
particle can be located by its x,
y and z coordinates, which are
the three components of the
position vector r :





r  x i  y j z k
 

where i , j and k are the
cartesian unit vectors.
z

r

k

O

j
x i
Fig 2-11
y
2. Displacement (位移)
y
We defined the displacement

vector r as the change in
position vector from t1 to t2.
t= t 2

r2
  
r  r2  r1
s

r

r1
O
x
t= t1
z
Fig 2-12
Note: 1) Displacement is not the same as the distance
traveled by the particle.
2) The displacement is determined only by the
starting and ending points of the interval.
  

if r1  x1i  y1 j  z1k




r2  x2i  y2 j  z2 k

Then the displacement is
  
r  r2  r1



 ( x2  x1 )i  ( y2  y1 ) j  ( z2  z1 )k
Direction: from start point to end point

Magnitude : r  x 2  y 2  z 2

The relationship between r and s :
In general,
z
r  s
Can

r  s
s
P1  r

r1 
?
Yes, for two cases:
1) 1D motion without
changing direction
2) When t  0,after

take limit: dr  ds
P2
O
x
r2
y


The difference between r and r (  r ):

(r   r )


z
r : magnitude of r

P1  r P2
r : the change of length

r1 
of position vectors
r  x2  y2  z 2
2
2
2
 x1  y1  z1
2
Note
2

r  r
2
O
x
r
r2
y
When t  0, after take limit:


r  dr


| r || dr |

s  ds | dr |
r  dr



 r  d r  dr  dr
3.velocity and speed
a.The average velocity in any interval is
defined to be displacement divided by the time

r
interval,

vav 
t
(2-7)
when we use the term velocity, we mean the
instantaneous velocity.
b. To find the instantaneous velocity, we
reduce the size of the time interval t , that is

t  0 and then  r  0 .




r (t  t )  r (t )
r
dr

v (t )  lim
 lim

t
dt
t 0
t 0 t
(2-9)
In cartesian coordinates:




dr  dx i  dy j  dz k

 dr dx  dy  dz 
v
 i
j k
dt dt
dt
dt

The vector v can also be written in terms of its




v  vx i  v y j  vz k
components as:
dx
v x  dt
dy
v y  dt
v
z
dz

dt
(2-11)
(2-12)
Discussion
The position vector of a moving particle at a

moment is r ( x, y ). The magnitude of the velocity of
the particle at the moment is:
dr
（A）
dt

dr
（C）
dt
√
√

| dr |
（B）
dt
（D）
dx 2
dy 2
( ) ( )
dt
dt
c.The terms average speed (平均速率)
and speed（速率）:
Average speed: vav
s

t
s is the total distance traveled.
r

|
Thus, vav | vav | |

t

| dr |
ds


| v |
Speed: v 
dt
dt

v | v |
d. Acceleration
We define the average acceleration as the
change in velocity per unit
time,
or


aav
v

t
(2-14)
And instantaneous acceleration


v d v
a  lim

(2-16)
dt
t 0 t
By analogy with Eq (2-12) ,we can write the
components acceleration vector as

dv x
ax 
dt
ay 
dv y
dt
dv z
az 
dt
(2-17)
Sample problem 2-4
A particle moves in the x-y plane
3
x(t )  At  Bt and y(t )  Ct 2  D , where
A  1.00m / s,3 B  32.0m / s, C  5.0m / s 3 and
D  12.0m . Find the position, velocity,
and acceleration of the particle when
t=3s.



Solution: r  x i  y j


 ( At  Bt ) i 
(Ct  D) j


r |t 3s  (69m) i  (57m) j
3



2


v  vx i  v y j  (3 At  B) i  2Ct j
2



v |t 3s  (5m / s) i  (30m / s) j





a  ax i  a y j  6 At i  2C j



a |t 3s  (18m / s ) i  (10m / s ) j
2
2
Sample problem
How do the velocity and
the acceleration vary with
time if position x(t) is known?
x(t)
dx
vx 
dt
dv x
ax 
dt
Colonel J. P. Stapp was in his braking rocket sled
Can you tell the direction of the acceleration from the figures?
His body is an accelerometer not a speedometer.
Out
in
Section 2-4 One-dimensional
kinematics
In one-dimensional kinematics, a particle
can move only along a straight line.
We can describe the motion of a particle
in two ways: with mathematical equations
and with graphs.
1.Motion at constant velocity
Suppose a puck (冰球) moves
along a straight line, which we
will use as the x-axis .

 vx  c
dx

 vx 
c
dt
x  x0  vt
x
A
t
0
(a)
vx
B
0
(b)
Fig 2-15
t
2. Accelerated motion (变速运动)
Two examples of accelerated motion are
x(t )  A  Bt  Ct
x(t )  D cos(t )
2
(2-20)
（2-21）
2-5 Motion with constant
acceleration
Let’s assume our motion is along the x axis, and
a x represents the x component of the acceleration.
Can we obtain
vx (t ) and x(t) from a x ?
dv x
or dv x  a x dt
 ax 
dt
 dv
x
(2-26)
  a x dt  a x  dt
vx  axt  v0
Note: the initial velocity
in the calculation.
v0
must be known
It is the similar way to find x(t) from v(t).
dx
 vx 
or dx  v dt
x
dt
1 2 '
 dx   (v0  axt )dt  v0t  2 axt  c
x  x0  v0t  axt
1
2
2
Note: the initial position x0 and velocity
must be known in the calculation.
v0
Discussion
 

Relationship between r (t ), v (t )and a (t )

r (t )
Derivative
Integral

( r (0) )

 dr
v
dt

 dv
a
dt
Derivative
v(t )
Integral

( v (0) )

a (t )
t 


r (t )  r0   v(t)dt
0
t 


v(t)  v0   a (t )dt
0
One example:
动画库\力学夹\1-02质点运动的
描述2.exe 4 (例3)
2-6 Freely falling bodies ……
Aristotle (384-322 B.C.) thought the heavier objects
would fall more rapidly because of their weight.
Galileo (1564-1642) made correct
assertion, that in the absence of
air resistance all objects fall with
the same speed.
In 1971, astronaut David Scott dropped a feather
and a hammer on the airless Moon, and he
observed that they reached the surface at about the
same time.
‘Feather and hammer’
experiment carried on the
airless moon
‘Feather and apple’ exp.
conducted in vacuum lab
Niagara Fall
Is it dangerous
to free fall from its
top?!
If the height of the
fall is 48 m, how
long will the person
reach the fall bottom
and what is his final
speed?