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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 )
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