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Dynamics MCB 2043Rectilinear Kinematics: Continuous Motion
September 2014 SemesterDereje Engida Woldemichael (PhD, CEng MIMechE)
Lesson Outcomes
At the end of this lecture you should be able to:
Determine the kinematic quantities (position,
displacement, velocity, and acceleration) of a particle
traveling along a straight path.
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Overview
Equation of motion
Position
Velocity
Acceleration
Linear motion with
Constant acceleration
Variable acceleration
Rectilinear Kinematics: Continious Motion
A particle travels along a straight-line
path defined by the coordinate axis 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.
The position of the particle at any
instant, relative to the origin, O, is
defined by the position vectorr, or the
scalar s.
Scalar s can be positive or negative.
Typical units for r and s is meters (m).
The displacement of the particle is
defined as its change in position.
Vector form: r =r - r Scalar form: s = s - s
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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 unit of m/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
1
Acceleration
Acceleration is the rate of change in the velocity of a particle.
It is a vector quantity.
Typical unit is m/s2 .
Note that, the derivative equations for velocity and acceleration can be
manipulated to get a ds = v dv
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).
2
3
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Summary of kinematic relations: Rectilinear motion
Differentiate position to get velocity and acceleration.
v = ds/dt
a = dv/dt
a = v dv/ds
Integrate acceleration for velocity and position.
Note that so and vo represent the initial position and
velocity of the particle at t = 0.
Velocity:
=t
o
v
vo
dtadv =s
s
v
vo
o
dsadvvor =t
o
s
so
dtvds
Position:
1
2
3
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 downward.
These equations are:
tavv co +=yields= t
o
c
v
v
dtadvo
2coo
s
t(1/2) atvss ++=yields=
t
os
dtvdso
)s-(s2a)(vv oc2
o
2+=yields=
s
s
c
v
v oo
dsadvv
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(b) Given a=a(v), develop v-t and s-v relationships
)(vadt
dv====2 ====
t
t
v
vdt
va
dv
00 )(
00 )(
ttva
dvv
v
====
dsvavdv )(====3 ====s
s
v
vdsdv
va
v
00 )(
00 )(
ssdvva
vv
v
====
This gives a relationship between velocity vand time taken t.
This gives the distance travelled sbefore
the velocity vis reached.
(c) Given a=a(s), develop v-s and s-t relationships
dssavdv )(====3 ====s
s
v
v
dssadvv00
)(
====s
sdssavv
0
)(22
0
2
This gives velocity v(s)as a function of distance s.
)(svdt
ds====1 ====
s
s
t
t sv
dsdt
00 )(
====s
s sv
dstt
0 )(0
This gives a relationship between distance sand time taken t.
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Example #1
Plan: Establish the positive coordinate, s, in the direction the
particle is traveling.
Since the velocity is given as a function of time, take a
derivative of it to calculate the acceleration.
Conversely, integrate the velocity function to calculate the
position.
Given: A particle travels along a straight line to the right
with a velocity of v = ( 4 t 3 t2 ) m/s where t is
in seconds. Also, s = 0 when t = 0.
Find: The position and acceleration of the particle
when t = 4 s.
EXAMPLE #1 (continued)
Solution:
1) Take a derivative of the velocity to determine the acceleration.
a = dv / dt = d(4 t 3 t2) / dt =4 6 t
=> a = 20 m/s2 when t = 4 s
2) Calculate the distance traveled in 4s by integrating the
velocity using so = 0:
v = ds / dt => ds = v dt =>
=> s so = 2 t2 t3
=> s 0 = 2(4)2 (4)3 => s = 32 m (in the
direction)
=t
o
s
s
(4 t 3 t2) dtds
o
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EXAMPLE #2
Given:Ball A is released from rest at
a height of 40 m at the same
time that ball B is thrown
upward, 2 m from the ground.
The balls pass one another at a
height of 20 m.
Find:The speed at which ball B was
thrown upward.
Plan: Both balls experience a constant downward acceleration
of 9.81 m/s2 due to gravity. Apply the formulas for
constant acceleration, with ac = -9.81 m/s2.
EXAMPLE #2 (continued)
Solution:
1) First consider ball A. With the origin defined at the ground,
ball A is released from rest ((vA)o = 0) at a height of 40 m
((sA )o = 40 m). Calculate the time required for ball A to drop to
20 m (sA = 20 m) using a position equation.
sA = (sA )o + (vA)o t + (1/2) ac t2
So,
20 m = 40 m + (0)(t) + (1/2)(-9.81)(t2) => t = 2.02 s
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EXAMPLE #2 (continued)
Solution:
2) Now consider ball B. It is throw upward from a height of 2 m
((sB)o = 2 m). It must reach a height of 20 m (sB = 20 m) at the
same time ball A reaches this height (t = 2.02 s). Apply the
position equation again to ball B using t = 2.02s.
sB = (sB)o + (vB)ot + (1/2) ac t2
So,
20 = 2 + (vB)o(2.02) + (1/2)(-9.81)(2.02)2
=> (vB)o = 18.82 m/s
EXAMPLE #3
A car passes you at point 1 travelling at an initial velocity of 6 m/s, and thenaccelerates at a constant rate to reach a velocity of 30 m/s at point 2. This
occurs over an 8 second period.
(a) What is the required constant acceleration during the initial 8 sec period?
(b) Calculate the distance covered by the car in this 8 sec period.
(c) Once the car passes point 2 att= 8 s, the acceleration becomes a function of
time given by . Determine an equation for the velocity of the car
as a function of time v(t) for t>8 s.
1 2
48
1
)( += tta
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(a) Using )( 1212 ttavv ==== 2
12
12 /38
630sm
tt
vva ====
====
====
(b) Using 21212012
)(2
1)( ttattvsss ++++======== ms 14483
2
186
2====++++====
(c) When t>8s from point 2
48
1++++======== t
dt
dva
Ctt
dttadtv ++++++++====++++======== 416)4
8
1(
2
(C is a constant)
To determine C, using the init ial condition @ point 2 (i.e. t=8 s, v=30 m/s)
C++++++++==== 8416
830
2
2====C
2416
1 2 ++++++++==== ttv
Summary Questions
1. A particle moves along a horizontal path with its velocity
varying with time as shown. The average acceleration of the
particle is _________.
A) 0.4 m/s2 B) 0.4 m/s2
C) 1.6 m/s2 D) 1.6 m/s2
2. A particle has an initial velocity of 30 m/s to the left. If it then
passes through the same location 5 seconds later with avelocity of 50 m/s to the right, the average velocity of the
particle during the 5 s time interval is _______.
A) 10 m/s B) 40 m/s
C) 16 m/s D) 0 m/s
t = 2 s t = 7 s
3 m/s 5 m/s
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Summary Questions
4. A particle is moving with an initial velocity of v = 12 m/s
and constant acceleration of 3.78 m/s2 in the same direction
as the velocity. Determine the distance the particle has
traveled when the velocity reaches 30 m/s.
A) 50 m B) 100 m
C) 150 m D) 200 m
3. A particle has an initial velocity of 3 m/s to the left ats0 = 0 m. Determine its position when t = 3 s if the
acceleration is 2 m/s2 to the right.
A) 0.0 m B) 6.0 m
C) 18.0 m D) 9.0 m
References: R.C. Hibbeler, Engineering Mechanics: Dynamics, SI 13th
Edition, Prentice-Hall, 2012.
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