lec 2 (equation of motion)
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Universiti Tun Universiti Tun Hussein Onn MalaysiaHussein Onn Malaysia
Mechanical EngineeringMechanical EngineeringDynamics DDE2063Dynamics DDE2063
BYBYEn. Khairulnizam Bin OthmanEn. Khairulnizam Bin Othman
ObjectiveObjective
To introduce the concepts of position, displacement, To introduce the concepts of position, displacement, velocity, and acceleration.velocity, and acceleration.
To study particle motion along a straight line.To study particle motion along a straight line. The motion equation application, concept and calculation.The motion equation application, concept and calculation.
Rectilinear KinematicsRectilinear KinematicsSection 12.2Section 12.2
Rectilinear Rectilinear : : Straight line motionStraight line motion Kinematics Kinematics : : Study the geometry of the motion dealing with Study the geometry of the motion dealing with
s, v, a.s, v, a. Rectilinear KinematicsRectilinear Kinematics : To identify at any given instant, the : To identify at any given instant, the
particle’s position, velocity, and acceleration.particle’s position, velocity, and acceleration.
(All objects such as rockets, projectiles, or vehicles will be (All objects such as rockets, projectiles, or vehicles will be considered as particlesconsidered as particles “has negligible size and shape” “has negligible size and shape”
particles : has mass but negligible size and shapeparticles : has mass but negligible size and shape
PositionPosition
PositionPosition : Location of a : Location of a particle at any given instant particle at any given instant with respect to the originwith respect to the origin
rr : Displacement ( Vector ) : Displacement ( Vector )s : Distance ( Scalar )s : Distance ( Scalar )
Distance & DisplacementDistance & Displacement
DisplacementDisplacement : defined : defined as the change in position.as the change in position.
r : Displacement ( 3 r : Displacement ( 3 km )km )
s : Distance ( 8 s : Distance ( 8 km )km ) Total lengthTotal length
For straight-lineFor straight-line Distance = DisplacementDistance = Displacement s = rs = r
s s rr
Vector is direction orientedVector is direction oriented r positive (left )r positive (left )r negative (right)r negative (right)
QUT
City
My PlaceX
3km
River
8 km
N
Velocity & SpeedVelocity & Speed
VelocityVelocity : Displacement per unit : Displacement per unit timetime
Average velocity :Average velocity : V = V = rrtt
SpeedSpeed : : Distance per unit time Distance per unit time Average speed : Average speed : spspssTTt (Always positive t (Always positive
scalar )scalar )
Speed refers to the magnitude of Speed refers to the magnitude of velocityvelocity
Average velocity :Average velocity : avgavg = = r / r / tt
Rectilinear Kinematics: Erratic MotionRectilinear Kinematics: Erratic Motion
dtdsv
Figure show Graph s - t Figure show Graph s - t for a body moving in a straight for a body moving in a straight line. Assuming that the sudden change in velocity at line. Assuming that the sudden change in velocity at time t = 5, 15, 22 s can be carried out. Draw the graph time t = 5, 15, 22 s can be carried out. Draw the graph of v-t ?.of v-t ?.
v =slope of s – t graph v =slope of s – t graph
Graph S-t
05
1015202530354045
0 5 15 22 27
t(s)
s(m
)
Rectilinear Kinematics: Erratic MotionRectilinear Kinematics: Erratic Motion
AnswerAnswer
1
1
1
1
82227
400_2722
6.015223640_2215
6.15152036_155
45
020_50
msvst
msvst
msvst
msvst
dtdsv
Rectilinear Kinematics: Erratic MotionRectilinear Kinematics: Erratic Motion
dtdva
Figure show Graph v - t Figure show Graph v - t for a body moving in a straight for a body moving in a straight line. Assuming that the sudden change in velocity at line. Assuming that the sudden change in velocity at time t = 2, 6, 10 s can be carried out. Draw the graph of time t = 2, 6, 10 s can be carried out. Draw the graph of a - t ?.a - t ?.
a =slope of v – t graph a =slope of v – t graph
Graph v-t
-15
-10
-5
0
5
10
15
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
v(m/s)
t(s)
Rectilinear Kinematics: Erratic MotionRectilinear Kinematics: Erratic Motion
AnswerAnswer
2
2
2
2
7.11016
)10(0_1610
5610
)10()10(_106
0_62
502010_20
msast
msast
msast
msast
dtdva
Rectilinear Kinematics: Erratic MotionRectilinear Kinematics: Erratic Motion
adtv
Change in = Area Under Change in = Area Under Velocity A – t Graph Velocity A – t Graph
adtv
Rectilinear Kinematics: Erratic MotionRectilinear Kinematics: Erratic Motion
vdts
Change in = Area Change in = Area Under Under
Displacement v – t Displacement v – t Graph Graph
vdts
Velocity as a Function of TimeVelocity as a Function of Time
dtdvac
dtadv c
dtadvt
c
v
vo 0
tavv c 0
Position as a Function of TimePosition as a Function of Time
tavdtdsv c 0
dttavdst
c
s
so 0
0 )(
200 2
1 tatvss c
velocity as a Function of Positionvelocity as a Function of Position
s
sc
v
v
dsadvv00
dsadvv c
)(2 0
2
0
2 ssavv c
)(21
21
020
2 ssavv c
Free Fall Free Fall
Ali and Omar are standing at the top of a cliff of Ali and Omar are standing at the top of a cliff of heightheight HH. Both throw a ball with initial speed. Both throw a ball with initial speed vv00, Ali , Ali straightstraight downdown and Omar straightand Omar straight upup. The speed of . The speed of the balls when they hit the ground arethe balls when they hit the ground are vvAA andand vvOO respectively.respectively. Which of the following is true:Which of the following is true:
(a)(a) vvAA < < vvOO (b) (b) vvAA = = vvOO (c) (c) vvAA > > vvOO
vv00
vv00
OmarOmarAliAli
HHvvAA vvOO
Free fall…Free fall…
Since the motion up and back down is symmetric, intuition Since the motion up and back down is symmetric, intuition should tell you that should tell you that v = vv = v00 We can prove that your intuition is correct:We can prove that your intuition is correct:
vv00
OmarOmar
HH
vv = v= v00
0HHg2vv 20
2 )(Equation:Equation:
This looks just like Omar threw This looks just like Omar threw the ball down with speed the ball down with speed vv00, so, sothe speed at the bottom shouldthe speed at the bottom shouldbe the same as Ali’s ball.be the same as Ali’s ball.
y = 0y = 0
Free fall…Free fall…
We can also just use the equation directly:We can also just use the equation directly:
H0)g(2vv 20
2 Ali :Ali :
vv00
vv00
AliAli OmarOmar
y = 0y = 0
H0g2vv 20
2 )(Omar:Omar:same !!same !!
y = Hy = H
SummarySummary
Time dependent Time dependent accelerationacceleration
Constant Constant accelerationacceleration
dtdsv
)(ts
2
2
dtsd
dtdva
dvvdsa
tavv c 0
200 2
1 tatvss c
)(2 020
2 ssavv c
This applies to a freely falling object: 2/81.9ga sm
Question 1Question 1 Initially, the car travels along a straight road
with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 10 m/s in 15 s, determine the constant deceleration of the car.
tavv c 0 atuv
Quiz 1Quiz 1 Initially, the car travels along a
straight road with a speed of 60 m/s. If the brakes are applied and the speed of the car is reduced to 25 m/s in 20 s, determine the constant deceleration of the car.
Question 2Question 2 A truck traveling along a straight road at
speed v1 (20km/h), increases its speed to v2 (120km/h) in 15 s. If its acceleration is constant, determine the distance traveled.
msatuts
smtvva
solution
sthrkmv
hrkmu
67.29121
/852.1
1512020
2
212
Quiz 1Quiz 1 A truck traveling along a straight
road at speed v1 (40km/h), increases its speed to v2 (100km/h) in 3000m. If its acceleration is constant, determine the time traveled.
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