Lec 02 (constant acc 051)

Universiti Malaysia Pahang
Manufacturing Engineering
Dynamics
Lecture # 2

Objective
 To introduce the concepts of position, displacement,
velocity, and acceleration.
 To study particle motion along a straight line.

Rectilinear Kinematics
Section 12.2
 Rectilinear : Straight line motion
 Kinematics : Study the geometry of the motion dealing with
s, v, a.
 Rectilinear Kinematics : To identify at any given instant, the
particle’s position, velocity, and acceleration.
(All objects such as rockets, projectiles, or vehicles will be
considered as particles “has negligible size and shape”
particles : has mass but negligible size and shape

Position
 Position : Location of a
particle at any given instant
with respect to the origin
r : Displacement ( Vector )
s : Distance ( Scalar )

Distance & Displacement
 Displacement : defined as the
change in position.
 r : Displacement ( 3 km )
 s : Distance ( 8 km )
Total length
 For straight-line
Distance = Displacement
s = r
D s = D r
Vector is direction oriented
Dr positive (left )
Dr negative (right)
City
QUT
My Place
X
3km
River
8 km
N

Velocity & Speed
 Velocity : Displacement per unit time
 Average velocity :
 V = Dr / Dt
 Speed : Distance per unit time
 Average speed :
 usp = sT / Dt (Always positive scalar )
 Speed refers to the magnitude of velocity
 Average velocity :
uavg = Ds / Dt

Velocity (con.)
 Instantaneous velocity :
lim
 For straight-line Dr = Ds
dr
dt
r
t
V
t
=
D
D
=
D 
0
ds
dt
v =

Problem
 A particle moves along a straight line such that its
position is defined by s = (t3 – 3 t2 + 2 ) m. Determine the
velocity of the particle when t = 4 s.
ds
= =
dt
v
3t 6t 2 
At t = 4 s, the velocity = 3 (4)2 – 6(4) = 24 m/s

Acceleration
 Acceleration : The rate of change
in velocity {(m/s)/s}
DV =VV
 Average acceleration :
aavg D
 Instantaneous acceleration :
dv
v
D
 If v ‘ > v “ Acceleration “
 If v ‘ < v “ Deceleration”
V
t
D
=
d s
2
2
0
lim
dt
dt
t
a
t
= =
D
=
D 

Problem
 A particle moves along a straight line such that its position is
defined by s = (t3 – 3 t2 + 2 ) m. Determine the acceleration of the
particle when t = 4 s.
ds
v 3 6 2 = = 
dv
 At t = 4
t t
dt
= = 6t  6
dt
a
a(4) = 6(4) - 6 = 18 m/s2

Problem
 A particle moves along a straight line such that its position is defined
by s = (t3 – 12 t2 + 36 t -20 ) cm. Describe the motion of P during the
time interval [0,9]
ds
3 24 36 3( 2)( 6) 2 = = t  t  = t  t 
dt
v
dv
= = 6t  24 = 6(t  4)
dt
a
t 0 2 4 6 9
s -20 12 -4 -20 61
v 36 0 -12 0 63
a -24 -12 0 12 30
Total time = 9 seconds
Total distance = (32+32+81)= 145 meter
Displacement = form -20 to 61 = 81 meter
Average Velocity = 81/9= 9 m/s to the right
Speed = 9 m/s
Average speed = 145/9 = 16.1 m/s
Average acceleration = 27/9= 3 m/s2 to the right

Relation involving s, v, and a
No time t
ds
dv
dt
a 
ds
v
dt =
dt
v 
dv
a
dt =
dv
a
ds
v
=
Position s
Acceleration
ads = vdv
Velocity

Problem 12.18
 A car starts from rest and moves along a straight line with
an acceleration of a = ( 3 s -1/3 ) m/s2. where s is in meters.
Determine the car’s acceleration when t = 4 s.
Rest t = 0 , v = 0
ads = vdv
s v

1
3
s ds vdv
 3
= 
0 0
2
1
3 2
1
3
3
v = 3s
2
(3)
2
s = v
1
3
3s
ds
v = =
dt
1
s 3ds =
3 dt

s 1
t

s 3
ds dt
0 0
 = 
3
2
=
3 3
s 3t
2
3
2
s = (2t)

For constant acceleration
a = ac

Velocity as a Function of Time
dv
dt
ac =
dv a dt c =
t
dv a dt
c
v
 = 
vo
0
v v a t c =  0

Position as a Function of Time
v c = =  0
v a t
ds
dt
t
ds v a t dt
c
s
 =  
so
0
0 ( )
2
1
s s v t a t c =  
0 0 2

velocity as a Function of Position
v dv a ds c =
s
 =
s
c
v
v
v dv a ds
0 0
2 v v a s s c  = 
( )
1
2 v v a s s c =  
2 ( ) 0
2
0
2
1
2
0
2
0

Free Fall
 Ali and Omar are standing at the top of a cliff of height H.
Both throw a ball with initial speed v0, Ali straight down and
Omar straight up. The speed of the balls when they hit the
ground are vA and vO respectively. Which of the following
is true:
(a) vA < vO (b) vA = vO (c) vA > vO
Ali Omar
v0
v0
H
vA vO

Free fall…
 Since the motion up and back down is symmetric, intuition
should tell you that v = v0
We can prove that your intuition is correct:
Omar
v0
2 Equation:  = ( )  =
H
v = v0
v v 2 g H H 0 2
0
This looks just like Omar threw
the ball down with speed v0, so
the speed at the bottom should
be the same as Ali’s ball.
y = 0

Free fall…
 We can also just use the equation directly:
2 Ali :  =  
v v 2( g )0 H 2
0
2 Omar:  = ( ) 
v v 2 g 0 H 2
Ali Omar
v0
v0
y = 0
0
same !!
y = H

Summary
 Time dependent acceleration  Constant acceleration
ds
dt
s(t)
v =
d s
2
2
dt
dv
a = =
dt
ads = vdv
v v a t c =  0
2
1
s s v t a t c =  
0 0 2
2 v v a s s c =  
2 ( ) 0
2
0
This applies to a freely falling object:
2 2 a = g = 9.81m/ s = 32.2 ft / s

Lec 02 (constant acc 051)

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