Qassim University
College of Engineering
Mechanical Engineering Dept.
1. 1 Kinematics of Particles
(Rectilinear motion Rectangular coordinates)
Kinematics of Particles
 Kinematics: Branch of dynamics, deals with motion of
bodies without reference to force
 Kinetics: Study relationship between motion of bodies
and Force that cause the motion or forces produced by the
motion
Mechanics
Statics
Dynamics
Kinematics
Kinetics
Kinematics of Particles
 Particle is a body with small physical dimensions
compared to the radius of curvature of the path
 Thus, motion of such particle treated as motion of point
 Only translation is considered, and rotation is negligible
or neglected
 We consider particle motion within a plane for the
following two cases
 Rectilinear motion
 Curvilinear motion
Rectilinear Motion:
Position, Velocity & Acceleration
5
Rectilinear Motion
 Rectilinear motion: Motion along a straight line
 Consider particle P moving along straight line
Note that displacement can be:
• Positive (as in this case)
• Negative
 At time t position P specified at distance S measured from
point O
 At t + ∆t position is P’ specified as measure s + ∆s from O
 Change in position during interval ∆t is called displacement
∆s
Velocity and Acceleration
 Average velocity of particle during interval ∆t defined
as:
s
vav =
t
 Instantaneous velocity defined as:
s ds
v = lim
=
=s
t →0 t
dt
 Similar average and instantaneous acceleration defined
as:
v
aav =
t
v dv
a = lim
=
=v
t →0 t
dt
d 2s
a= 2 =s
dt
Velocity and Acceleration
 Vel. and acc. are vector quantities and for rectilinear
motion direction indicated by + or −
 May establish relations between displacement, velocity
and acceleration as follows:
 Starting with expression for velocity and acceleration:
dv
a=
dt
ds
v=
dt
 We extract the following two expressions:
adt = dv
1
dt = ds
v
Velocity and Acceleration
 Eliminate the dt from the two expressions
adt = dv
1
dt = ds
v
1 
a  ds  = dv
v 
 Then rearrange to have the final expression:
vdv = ads
Velocity and Acceleration
 Thus have three equations for rectilinear motion:
ds
v=
dt
dv
a=
dt
vdv = ads
 Problems in rectilinear motion are solved by integration
of these differential equations
Graphical Representation
 Figure (a) plots disp. s verses time t
and slope of this curve is velocity
ds
v=
dt
 Figure (b) plots vel. verses time t
and slope of this curve is acc.
dv
a=
dt
Graphical Representation
 Area under v-t curve during dt is vdt
which is the displacement ds
ds = vdt
 Thus, net displacement of particle
during interval t1 and t2 is area under
the curve in this duration
s2
t2
s1
t1
 ds =  vdt
s2 − s1 = (area under v - t curve)
Graphical Representation
 Area under a-t curve during dt is adt
which is the velocity dv
dv = adt
 Thus, net change in velocity of
particle during interval t1 and t2 is
area under the curve in this duration
v2
t2
v1
t1
 dv =  adt
v2 − v1 = (area under a - t curve)
Graphical Representation
 Figure (a) plots acceleration a verses position coordinate
s area under curve during displacement ds is ads
vdv = ads
 Thus, net area under curve between positions s1 and s2 is
v2
s2
v1
s1
 vdv =  ads
s2
1 2 2
v2 − v1 ) =  ads
(
s1
2
Special Cases: Constant Acceleration
 Special case of constant acceleration velocity is
determined as:
v
t
v0
0
 dv = a  dt
v = v0 + at
initial conditions at t = 0
s = s0,
v = v0
 Displacement can then be determined as:
s
t
s0
0
 ds =  vdt
t
=  ( v0 + at ) dt
0
1 2
s = s0 + v0t + at
2
Special Cases: Constant Acceleration
 Third equation can be obtained by integrating the
equation vdv=ads
v
s
v0
s0
 vdv = a  ds
v 2 − v02 = 2as
 The three equations are only valid for constant
acceleration
v = v0 + at
1 2
s = s0 + v0t + at
2
v 2 − v02 = 2as
Special Cases: Acc. function of time
 For the case that acc. is a function of time a = f(t)
 We may determine the velocity as:
v
t
v0
0
 dv =  adt
t
=  f ( t ) dt
0
t
v = v0 +  f ( t ) dt
0
 The displacement can then be determined as:
s
t
s0
0
 ds =  vdt
t
s = s0 +  vdt
0
Erratic Motion
Graphing provides a good way
to handle complex motions that
would be difficult to describe
with formulas.
Graphs also provide a visual
description of motion and
reinforce the calculus concepts
of differentiation and
integration as used in
dynamics.
The approach builds on the facts that slope and
differentiation are linked and that integration can be
thought of as finding the area under a curve.