lec 2 (equation of motion)

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

)


AnswerAnswer

1

1

1

1

82227

400_2722

6.015223640_2215

6.15152036_155

45

020_50

msvst

msvst

msvst

msvst

dtdsv


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)


AnswerAnswer

2

2

2

2

7.11016

)10(0_1610

5610

)10()10(_106

0_62

502010_20

msast

msast

msast

msast

dtdva


adtv

Change in = Area Under Change in = Area Under Velocity A – t Graph Velocity A – t Graph

adtv


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.

lec 2 (equation of motion)

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