chapter 1 - kinematicslgb10203

7/29/2019 Chapter 1 - Kinematicslgb10203

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LGB 102 03/ ENGINEERING SCIENCE CHAPTER 1 KINEM ATICS

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CHAPTER 1 KINEM ATICS

A. Introduction Kinematics is the branch of mechanics which studies the motion of objects wi thout consider ing

the fo rces that causes the m otio n.

There are th ree types o f mot ion wh ich are l inear /s t ra igh t mot ion , a rc mot ion andc ircu lar / ro ta t iona l mot ion .

B. Rect i linear M ot ion (Linear M ot ion) Recti l inear mo tion is a mo tion along a stra ight l ine. Involves a concept of d isplacement , veloci ty and accelerat ion. Displaceme nt, s of a bod y is the change in th e posi t ion o f th e body . That is, d isplacement is how

far the ob ject is f rom i ts sta r t ing po in t .- Displacement is the d i f ference bet w een f inal and or ig inal coord inates.- = @ = - Can be n egative or po sit ive value.- Vector quant i ty and the un i t i s m eter (m) .

Distance, x is a tota l length of the pat h w hich always posi t ive value.- Scalar quanti t y and the uni t is met er (m)- = +

Exam ple 1.1:

Im agine a person w alking 70 m t o t he East and t hen tu rning around and w alking back (W est) a d istance

of 30 m . Determine t he d isp lacement and d istance the person w a lk.

Solutions:

D isplacem ent , s = 70 m 30 m = 40 m

Distance, x = 70 m + 30 m = 100 m


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Velocity, v is the speed of an o bject and also the direct ion t he object is travel l ing.- Vector quant i ty and the un i t i s m eter per second (ms-1 )- Average Ve locity, vav e [vector quant i ty and t he un i t i s m s-1 ]

=

=

Can be n egative or po sit ive value.

- Average Spee d, [sca lar quant i ty and the u n i t i s m s-1 ] =

=

Can be n egative or po sit ive value.

- Instan tane ous velocity , i ts veloci ty at a part icular instant.v = rate of change of d isplacem ent =

ve loci ty is constant , both m agni tude and d i rect ion of ve loc ity do not change

- Instan tane ous spee d = rate of d istance travel led. Accelerat ion , a is the change in veloci ty (any change in speed @ d irect ion).

- Vector q uant i ty and t he un i t i s m eter per second square (ms-2 )- Can be n egative or po sit ive value.

- Average Accelerat ion , a av e =

=

- Instan tane ous Accelerat ion , i ts accelerat ion at a part icular instant. a = rate of change of veloci ty =

C. M ot ion Graph Constant accelerat ion m otio n can be character ized by mo t ion equa t i ons and by m ot ion graphs. M otio n graphs can te l l us how far a body has travel led, how fast i t is m oving and al l the speed

changes ther e have been.

The graphs of d istance, veloci ty and accelerat ion as funct ions of t im e below wer e calculated forone-dim ensional mot ion using the m otio n equations in a spreadsheet.

A considerab le amount o f in fo rmat ion about the mot ion can be ob ta ined by examin ing theslope of the var iou s graphs. The slope of the graph o f posi t ion as a funct io n of t im e is equal tothe ve loc ity a t t ha t t im e, and t he slope o f th e graph o f ve loci ty as a funct ion o f t ime is equa l to

the accelerat ion .

M ot ion o f a bo dy can be i l lust ra ted bya) Displacement t im e graphb ) Veloci ty t im e graphc) Accelerat ion t im e graph

Slow ing dow n

(decelerat ion, a)

Speeding up

(accelerat ion, a)

Accelerat ionVelocit


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Distan ce - t im e graph (s-t graph)

Distance o f a body f rom star t po in t is m easured. Here are 4 exam ples o f t he m ot ion o f a car represented

by 4 s- t graph

Case 1: A car is tr avell ing at constant speed Case 2: A car is tr avell ing w ith increasing velocity

Case 3: A car is tr avell ing w ith decreasing velocit y Case 4: A car is at rest (stat ionar y)


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Velocity - t im e graph (v-t graph)

V-t graph gives the veloci ty of a m oving object at d i f fer ent t im e. Here are 4 v- t graphs representing the

m ot ion o f 4 cars:

Accelerat ion - t im e graph (a-t graph)

The a-t graph gives the accelerat ion of a moving object at d i f ferent t imes. Here are 3 examples of a- t

g raph represent ing the m ot ion o f 3 d i f fe ren t cars.

Case 1: From th is graph, we know th at t he speed is increasing and

the s- t graph should a lso be increasing with a concave downward

shape.

Case 2: From th is graph, we know that the object should e i ther

travel in constant speed or at r est.


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Case 3: From this graph, the v-t graph and s-t graph are all

increasing with concave dow nw ard shape.

Example 1 .2 :

The accelerat ion t im e graph of a car wh ich star t s fro m rest is as shown .

Determine:

a) The veloci ty o f t he car after i ) 10 s and i i ) 30 sb) Sketch th e veloci ty t im e graph and m ark values of t he veloci ty at t im e 10 s and 30 s.c) From the veloci ty- t im e graph, calculate t he distance travel led by the car in 30 s.

Solutions:

a ) Velocity = area under graph(a- t ) t=10 s , 2 (20) = 40 m s-1

t = 30 s , (0 ) (0 ) = 0 m s-1

b )

Time, s

Ve lo ci t , m s-1

10 20 30

40

20

-2 0

Time, s

Accelerat ion , m s-

10 20 30

4

2

-2


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c) Distan ce = area u nde r graph (v-t )Distance =

( ) ( ) +

( ) ( ) =

D . M ot ion w ith Constant Accelerat ion W hen an objects m oves w ith constant accelerat ion, th e instant accelerat ion at any point in a

t im e interval is equal to t he value of average accelerat ion over th e enti re t im e interval .

Suppose u = in i t ia l veloci ty, v = f inal veloci tyUn i fo rm acce lera t ion , =

=

v = u + at

Displacemen t, s = average velocity x t im e

=1

2( + )

=1

2( + ( + ) )

= + 12

Velocity, v = u + 2as

Basic equation fo r uni form ly accelerated m ot ion; = +

= ( + )

v = u + 2as

v = u + at


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Example 1 .3 :

A car star t s from rest and accelerates uni form ly. After 10.0 s, its d isplacem ent is 25.0 m . Calculate

a) The veloci ty of the car after 10.0 s.

b) The accelerat ion of t he car after 10.0 s.

c) The displacement in t he next 10.0 s i f the car cont inues i ts m otio n w ith t he sam e accelerat ion.

Solutions:

Given in the quest ion, u = 0 m / s, t 1 0 = 25 .0 m ;

a ) Using = ( + ) b) Using v = u + a t

. =

( + ) 5 m / s = 0 m / s + a (10s)

V = 5 m / s a = 0 .5 m / s 2

c) S = S20 S10Using = +

= ( ( ) +

( . ) ( )

=

S = 100 m 25 m = 75 m

E. Freely Falling Bod ies A body is said to be in f ree fa l l i f i ts fa l ls under t he act ion of gravi ty w i tho ut a ir resistance. For poin ts close to t he surface of the Earth, t he accelerat ion of free fa l l g (9.81 m s-2 ) is constan t

and assum ed vert ical ly dow nw ards.

Since the accelerat ion of free fa l l is constant , we can use the equat ions of m ot ion und erconstant accelerat ion .

g = 9.81 m s-2

=

= ( + )

v = u - 2gs v = u - gt


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Example; bal l thr ow n vert ical ly upw ards ( ) , and bal l tow ards the groun d ( )

Example 1 .4 :

A person throw s a ball upw ard into t he air wi th an in i t ia l veloci ty of 15.0 m / s. Calculate

a) How h igh i t goesb) How long the bal l is in the air befor e i t comes back to h is hand.

Solutions

a ) v = u - 2gs(0) = (15) - 2(9.81 ) s

19 .62 = 225S = 11.4 7 m

b) =

( + ) 22 . 94 =

( )

t = 3 .06 s

a -g -a g

(9 .81 m s-2 ) ( -9.81 m s-2 ) ( -9.81 m s-2 ) (9.81 m s-2 )

chapter 1 - kinematicslgb10203

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