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Lecture 1
• Kinematics of a particle
Material covered
• Kinematics of a particle
- Introduction
- Rectilinear kinematics:
Continuous motion
- How to analyze problems
- Problems
Next lecture; Rectilinear
kinematics: Erratic motion
Kinematics of a particle: Introduction
Important contributors
Galileo Galilei, Newton, Euler
Mechanics
Statics Dynamics
Equilibrium of a
body that is at
rest/moves with
constant velocity
Accelerated
motion of a body
• Kinematics: geometric aspects of the motion
• Kinetics: Analysis of forces which cause the motion
Kinematics of a particle: Introduction
• Dynamics includes:
- Kinematics: study of the geometry of motion. Kinematics
is used to relate displacement, velocity, acceleration, and
time without reference to the cause of motion.
- Kinetics: study of the relations existing between the forces
acting on a body, the mass of the body, and the motion of the
body. Kinetics is used to predict the motion caused by given
forces or to determine the forces required to produce a given
motion.
• Rectilinear motion: position, velocity, and acceleration of a
particle as it moves along a straight line.
• Curvilinear motion: position, velocity, and acceleration of a
particle as it moves along a curved line in two or three
dimensions.
Today’s Objectives
Find the kinematic quantities (position, displacement,
velocity, and acceleration) of a particle traveling
along a straight path
Next lecture; Determine position, velocity, and
acceleration of a particle using graphs
Cartesian coordinates
Cylindrical coordinates
(Polar coordinates in 3D)
Choice of Coordinates
Rectilinear kinematics: Continuous Motion
A particle travels along a straight-line
path defined by the coordinate axis ‘s’
The POSITION of the particle at any
instant, relative to the origin, O, is
defined by the position vector r, or the
scalars. Scalars can be positive or
negative. Typical units for r and s are
meters (m) or feet (ft).
The displacement of the particle is defined
as its change in position.
Vector form: r = r’ - r Scalar form: s = s’ - s
The total distance traveled by the particle, sT, is a positive scalar that represents
the total length of the path over which the particle travels.
Velocity
Velocity is a measure of the rate of change in the position of a
particle. It is a vector quantity (it has both magnitude and
direction). The magnitude of the velocity is called speed, with
units of m/s or ft/s.
The average velocity of a particle during a
time interval t is
vavg = r/t
The instantaneous velocity is the time-derivative of position.
v = dr/dt
Speed is the magnitude of velocity: v = ds/dt
Average speed is the total distance traveled divided by elapsed
time: (vsp)avg = sT/ t
Acceleration
Acceleration is the rate of change in the velocity of a particle. It is a
vector quantity. Typical units are m/s2 or ft/s2.
The instantaneous acceleration is the time derivative of velocity.
Vector form: a = dv/dt
Scalar form: a = dv/dt = d2s/dt2
Acceleration can be positive (speed increasing)
or negative (speed decreasing).
As the book indicates, the derivative equations for velocity and
acceleration can be manipulated to get a ds = v dv
Constant acceleration
The three kinematic equations can be integrated for the special case
when acceleration is constant (a = ac) to obtain very useful equations.
A common example of constant acceleration is gravity; i.e., a body
freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2 ft/s2
downward. These equations are:
tavv co+=yields=
t
o
c
v
v
dtadv
o
2coo
s
t(1/2)at vss ++=yields= t
os
dtvds
o
)s-(s2a)(vv oc2
o2 +=yields=
s
s
c
v
v oo
dsadvv
1.Velocity as a Function of Time.
2. Position as a Function of Time.
3. Velocity as a Function of Position.
11
Uniform Rectilinear Motion
• For particle in uniform rectilinear motion, the
acceleration is zero and the velocity is constant.
0 0
0
0
x t
x
dx v dt
x x vt
x x vt
=
=
= +
constantdx
vdt
= =
Let the initial condition v = vo when to = 0.
ONLY CAN BE USED WHEN THE VELOCITY IS
CONSTANT.
12
Uniformly Accelerated Rectilinear
Motion
constantdv
adt
= =
0 0
v t
v
dv a dt= 0v v at =
For particle in uniformly accelerated rectilinear motion, the acceleration of
the particle is constant. Let initial condition v = vo when t = 0
0v v at= +
210 0 2
x x v t at= + +
2 2
0 02v v a x x= +
Let initial condition x = xo when t = 0
0
dxv v at
dt= = +
0
0
0
x t
x
dx v at dt= + 21
0 0 2x x v t at = +
Let initial condition v=vo at x = xo
constantdv
v adx
= =
0 0
v x
v x
v dv a dx= 2 210 02
v v a x x =
ONLY CAN BE USED WHEN THE ACCELARATION IS CONSTANT.
Rectilinear Motion: Position, Velocity &
Acceleration
• Particle moving along a straight line
is said to be in rectilinear motion.
• Position coordinate of a particle is
defined by positive or negative
distance of particle from a fixed
origin on the line.
• The motion of a particle is known if
the position coordinate for particle is
known for every value of time t.
Motion of the particle may be
expressed in the form of a function,
e.g., 326 ttx =
or in the form of a graph x vs. t.
Rectilinear Motion: Position, Velocity & Acceleration
• Instantaneous velocity may be positive or
negative. Magnitude of velocity is referred
to as particle speed.
• Consider particle which occupies position P
at time t and P’ at t+t,
t
xv
t
x
t
==
=
0lim
Average velocity
Instantaneous velocity
• From the definition of a derivative,
dt
dx
t
xv
t=
=
0lim
e.g.,
2
32
312
6
ttdt
dxv
ttx
==
=
Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle with velocity v at time t and
v’ at t+t,
Instantaneous accelerationt
va
t
==
0lim
tdt
dva
ttv
dt
xd
dt
dv
t
va
t
612
312e.g.
lim
2
2
2
0
==
=
==
=
• From the definition of a derivative,
• Instantaneous acceleration may be:
- positive: increasing positive velocity
or decreasing negative velocity
- negative: decreasing positive velocity
or increasing negative velocity.
Rectilinear Motion: Position, Velocity &
Acceleration
• Consider particle with motion given by
326 ttx =
2312 ttdt
dxv ==
tdt
xd
dt
dva 612
2
2
===
• at t = 0, x = 0, v = 0, a = 12 m/s2
• at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0
• at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2
• at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s2
Determination of the Motion of a Particle
• Recall, motion of a particle is known if position is known for all
time t.
• Typically, conditions of motion are specified by the type of
acceleration experienced by the particle. Determination of
velocity and position requires two successive integrations.
• Three classes of motion may be defined for:
- acceleration given as a function of time, a = f(t)
- acceleration given as a function of position, a = f(x)
- acceleration given as a function of velocity, a = f(v)
Important points
• Dynamics: Accelerated motion of
bodies
• Kinematics: Geometry of motion
• Average speed and average velocity
• Rectilinear kinematics or straight-
line motion
• Acceleration is negative when
particle is slowing down
• a ds = v dv; relation of acceleration,
velocity, displacement
Analyzing problems in dynamics
Coordinate system
• Establish a position coordinate S along the path and specify its fixed origin and positive direction
• Motion is along a straight line and therefore s, v and α can be represented as algebraic scalars
• Use an arrow alongside each kinematic equation in order to indicate positive sense of each scalar
Kinematic equations
• If any two of a, v, s and t are related, then a third variable can be obtained using one of the kinematic equations.
• When performing integration, position and velocity must be known at a given instant (…so the constants or limits can be evaluated)
• Some equations must be used only when ‘a is constant’
Problem solving MUSTS
1. Read the problem carefully (and read it again)
2. Physical situation and theory link
3. Draw diagrams and tabulate problem data
4. Coordinate system!!!
5. Solve equations and be careful with units
6. Be critical. A mass of an aeroplane can not be 50 g
7. Read the problem carefully
A car moves in a straight line such that for a short time
its velocity is defined by
v = (3t2 + 2t) ft/s, where t is in seconds.
Determine its position and acceleration when t = 3 s.
When t = 0, s = o.
Solution:
The car’s velocity is given as a function of time so its
position can be determined from v = ds/dt,
since it relates v, s and t. Note that s=0 when t=0
v= ds = (3t2 + 2t)
dt
Integration over limit t=0, s=0
∫ ds = ∫ (3t2 + 2t) dt
s = t3 + t2
When t= 3s, s= (3)3 + (3)2 = 36 ft
For acceleration
Knowing velocity as function of time, acceleration is
determined from a = dv/dt, since this equation relates
a, v and t
a= dv= d (3t2 + 2t)
dt dt
a= 6t + 2
When t=3s, a=6(3) + 2 = 20 ft/s2
Problem 2
Solution: For velocity using eqn:
a ds= v dv
Integrating over the limit for t=0, s=0 v=0
∫(6+0.02s) ds = ∫v dv
6∫ ds +0.02 ∫ s ds = ∫v dv
6s + 0.02 s2/2 = v2 / 2
v= (12s+0.02s2) 1/2
For s= 2km, s = 2000m,
v = √12(2000)+ 0.02(2000) 2
v = 287.054 m/s
2 2
0 02v v a x x= +
Solution;:
Using the equation of velocity as function of
position
(80)2 = 0 + 2ac ( 500 – 0 )
ac = 6400 / 2 (500)
ac= 6.4 ft/s 2
80
80
0
0 0
2
: 6.4 6.4 [ ]
8012.5 /
6.4
t
Use adt dv dt dv t v
t ft s
= = =
= =
b) For time =?
500 80 2500 80
0 0
0 0
2
: [ ] [ ]2
64006.4 /
2*500
vUse ads v dv ads v dv a S
a ft s
= = =
= =
80
80
0
0 0
2
: 6.4 6.4 [ ]
8012.5 /
6.4
t
Use adt dv dt dv t v
t ft s
= = =
= =
The position of a particle along a straight line is
given by s = ( t3 - 9t2 + 15t) ft, where t is in seconds.
Determine its maximum acceleration and maximum
velocity during the time interval 0< t < 10 sec.
• Solution:
242
18106
1823
151823
135
151018103
152923
15293
15293
max
max
max
max
ft/sa
)(a
t)(a
)t)t(dt
d
dt
dva
ft/sv
)()(v
t)(tv
t)t(tdt
d
dt
dsv
ttts
=
=
=
+==
=
+=
+=
+==
+=
2.5 0.50.333 /
6av
The avargae velocity is found from the beginning and
ending positions and the total travel time.
V m s
The average speed is found from the total distance of
travel devided by the total tr
= =
{(0.5 ( 1.5)) (2.5 ( 1.5))}1 /
6
avel time.
Average speed= m s +
=
AB
C
00.5-1.5 2.5
3 2(6 9*6 15*6) 18At t=6 seconds, S ft
To determine the total distance traveled in 6 s, we
must find when the particle changes its direction. This
occurs when the drivative of the position is zero.
ds
st
= + =
23 18 15 3( 5)( 1) 0
1, 7 5, 25 .
t t t t
at t s ft and t s ft
= + =
= = = =
7 ft
-18 ft
-25 ft
7 ft
-18 ft
-25 ft
Total distance=
7+(7-(-25))-(-25-(-18))=46 ft
Total distance, over time t=6sec
For t=1, s= ( 1-9 +15)=7
t=2, s= 2
t=3, s= -9
t=4, s= -20
t=5, s= -25
t=6, s= -18
Problem 1