chapter-2 - kinematics of linear motion

5/26/2018 Chapter-2 - Kinematics of Linear Motion

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Chapter 2 : Kinematics of Linear Motion[ 5 Hours ]

2.1 Linear Motion

2.2 Uniformly acceleratedmotion

2.3 Free Falling Body

Matriculation Physics SF016 1

.

2.0 Introduction

Kinematics

Description of the motion of objectswithout consideration of whatcauses the motion.

2

Learning Outcomes

At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:

Define and distinguish betweenDefine and distinguish between

ii) distance and displacement,) distance and displacement,ii) speed and velocity,ii) speed and velocity,

iii) instantaneous velocity, average velocity andiii) instantaneous velocity, average velocity and

2.1 Linear Motion

uniform velocityuniform velocity

iv) instantaneous acceleration, averageiv) instantaneous acceleration, average

acceleration and uniform accelerationacceleration and uniform acceleration

Sketch graphs of displacementSketch graphs of displacement--time, velocitytime, velocity--time andtime andaccelerationacceleration--time.time.

Determine the distance travelled, displacement, velocityDetermine the distance travelled, displacement, velocityand acceleration from appropriate graphs.and acceleration from appropriate graphs.

3

Linear motion motion of an object along astraight line path.

Distance, d-- the total length of travel in moving from one

location to another.-- scalar quantity.

4

-- always positive.

Displacement, s-- straight line distance from the initial

position to the final position of an object.

-- Vector quantity

-- can be positive, negative or zero.


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5

InitialPosition

finalPosition

6

Distance travelled = 200mDisplacement = 120 m, in the direction of Northeast

Example

An air plane flies 600 km north and then 400 kmto the east.

N E

Final

400 km

7

Total distance traveled d = 600 + 400 = 1000 km

Displacement, s = = 721.11 km(magnitude)

22 400600 +

Begin

N

600 km

Test your understanding

1. You walk from your house to friends house thento the grocery shop. Calculate:

(i) the distance traveled.(ii) the displacement.

8

2. An athlete runs four laps of a 400 m track.What is the athletes total displacement?


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Speed, v-- Rate of change in distance

-- S.I. unit : m s1 ; scalar quantity.

t

dv ==

distancethattraveltotime

traveleddistance

9

Average Speed, ( )

-- total distance traveled divided by the total

time elapsed in traveling that distance.

t

dvspeedAverage

=,

v

Velocity, v-- tells us how fast object is moving & in which

direction it is moving.

10

-- is the rate of change in displacement.

-- vector quantity ;

tsvvelocity ==

timetravelntdisplaceme,

SI unit : m s1

Average velocity, vav

changefor thetakentime

ntdisplacemeinchange=avv

12

12

tt

ss

t

s

vav

=

=

11

ns an aneous ve oc y, v-- velocity at a specified position or instant of

time along the path of motion.

-- commonly referred as `velocity at point A `or ` velocity at time t `.

dt

ds

t

sv

t

=

=

lim0

-- equal to the gradient at any point on thecurve of a displacement time (s t) graph.

Slope BE = average velocity

Slope at C = Instantaneous velocity

12

s

t


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Example 1

An insect crawls along the edge of a rectangularswimming pool of length 27m & width 21m. If itcrawls from corner A to corner B in 30 min.

(a) What is its average speed ?(b) What is the magnitude of its average velocity ?

13

Solution

Given : Length, L = 27m; Width, W = 21m

Total distance traveled, d = 27 + 21 = 48 m

Displacement, s = m21.342127 22 =+

14

(a)t

dvspeedAverage

=,

)60(3048=

10267.0 = sm

(b)

t

svvelocityAverage av

=

,

)60(30

21.34=

1019.0 = sm

15

Acceleration, a

-- time rate of change of velocity.

takentimeyin velocitchangeonaccelerati =

-- An acceleration may due to:1) change in speed (magnitude),

2) change in direction or3) change in both speed and direction.

-- Velocity is vector quantity, a change invelocity may thus involve either or bothmagnitude & direction.

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Quick Test

A car is traveling at 30 km h1 to the north.Then it turns to the west without changing itsspeed. Is the car accelerating?

Answer : YES !Reason : there is a change in direction


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-- Deceleration : object is slowing down (directionof acceleration is opposite to the direction ofthe motion or velocity).

17 18

Car in figure (a) & (d) accelerating

Car in figure (b) & (c) decelerating

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Average acceleration

-- change in velocity divided by the time taken

to make the change.

t

v

tt

vva

=

==

12

12

changethemaketotime

yin velocitchange

-- vector quantity.

20

Instantaneous acceleration

-- acceleration at a particular instant of time.

2

2

0lim dt

sddtdv

tva

t

===

-- un : m s


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For linear motion, + and signs is used to

indicate direction of motion, velocity &

acceleration.

With horizontal direction we may take :

to the right as +

to the left as

With vertical direction we ma take :

21

upward as + ; downward as

The sign convention you choose isentirely up to you. It doesnt matter aslong as you keep the same signconvention for the entire calculation.

Example 2

A car travels in a straight line along a road. Itsdistance, s is given as a function of time, t by theequation :

32 12.04.2)( ttts =

(a) Calculate the average velocity of the car for

22

the time interval, t = 0 s and t = 10 s.

(b) Calculate the instantaneous velocity of the carat t = 5 s.

(c) Calculate the instantaneous acceleration ofthe car at t = 5 s.

Solution

Given : 32 12.04.2)( ttts =

(a) At t1= 0 s, s1 = 0 m

At t2=10 s,

s2 = 2.4(10)2 0.12 (10)3 = 120 m

Average velocity,12 ss

vav

= 0120

=

23

12 tt 010

112 = smvav

(b) Instantaneous velocity,

dt

dsv =

)12.04.2( 32 ttdt

d=

236.08.4 ttv =

At t = 5s,2)5(36.0)5(8.4 =v

115 = smv

(c) Instantaneous acceleration,

24

dta = )36.08.4( tt

dt=

ta 72.08.4 =

At t = 5s, )5(72.08.4 =a

22.1 = sma


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Graphical Methods

Displacement time (s - t) graphs

25

Instantaneous velocity = gradient of (s-t) graphdt

dsv =

velocity time (v - t) graphs

26

acceleration, a = = gradient of (v - t) graph

Displacement of the object = shaded area

under the (v - t) graph

dt

dv

Example 3 ( PSPM Session 2008/09 )

27

Figure 5 shows a graph of displacement x against

time t of an object moving along x-axis. Calculate

(a) Average velocity for the time interval, 1 s to 4 s.

(b) Average speed for the time interval, 1 s to 4 s.

(c) Instantaneous velocity at t = 2.5 s.

(d) Instantaneous acceleration at t = 5.5 s.

28

(a)

t

svaverage

=

14

2)3(

=

1sm67.1 =

(b)

t

xvaverage

=

14

342

++=

1sm3 =


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(c)s2.5at tgradient|

5.2===

=

dt

dxv

st

234)3(

=

29

sm7 =

(d)

dtdva = 2sm0 =

Uniform Linear Motion

-- motion with constant velocity

-- acceleration, a = 0 m s2

The displacement increases by equal amounts in

30

Test your concept

1. Can you accelerate a body without speeding upor slowing down? Is it possible?

2. A car is traveling at 30 km h1 to the north.Then it turns to the west without changing itsspeed. Is the car accelerating?

31

3. How would you draw a displacement time graphfor a stationary object?

4. What would the gradient of a distance timegraph represent?

5. What does the area a speed - time graphrepresent?

6. The distance time graph in figure belowrepresents the motion of an ant in 7 seconds.Describe its motion.

32


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7. A body moves along the x-axis. Assume that apositive sign represents a direction to the right.The velocity, vof the body is related to time, t

through the equation

232 tv =-1

33

w ere v an are measure n an srespectively. t= 0 when x= 0.Determine(a) the displacement

(b) the accelerationat the instant of time t= 1 s

Learning Outcomes

At the end of this chapter, students should beAt the end of this chapter, students should be

able to:able to: Derive and apply equations of motion withDerive and apply equations of motion with

uniform acceleration:uniform acceleration:

2.2 Linear Motion2.2 Uniformly accelerated motion

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atuv +=

2

2

1atuts +=

asuv 222 +=

atuv +=

Uniform acceleration means the accelerationdoes not depend on time or always constant.

velocity changes at a uniform rate.

constant==dt

dva

35

A v - t plot is a straight line whose gradient isequal to acceleration.

= =

36

Acceleration, a = constant

value


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Kinematics Equation for uniform acceleration

Assume a car has uniform acceleration & considerthe motion between X and Y :

37

u= initial velocity ( velocity on passing X )v= final velocity ( velocity on passing Y )a= accelerations= displacement ( in moving from X to Y )t = time taken ( to move from X to Y )

velocity time graph for the car

38

Acceleration, a = gradient of graph v t

t

uva

= uvat =

rearranged to give:

(1)tauv +=

Distance traveled, s = area under the graph

= area of trapezium

(2)tvus )(1

+=

39

Substitute (1) into (2) :

ttauus )(2

1++=

2

2

1tatus += (3)

From (1) : v = u + at , get an expression for t :

a

uvt

=

Substitute into (2) :

))((1 uvuvs +=

40

a

uvuv

2

))(( +=

222 uvas =

asuv 222 += (4)


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Example 4

The driver of a pickup truck going 100 km h1

applies the brakes, giving the truck a uniform

deceleration of 6.50 m s2 while it travels 20.0m.

(a) What is the speed of the truck in kilometersper hour at the end of this distance ?

(b) How much time has elapsed ?

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Solution

Given :a = 6.50 m s2 ( deceleration )

s = 20.0 mu = 100 km hour 1

1sm78.27

s)60(60

m)1000(100 ==

(a) Final velocity , v = ?

asuv 222 +=

42

(b) Assume : time elapsed, t

atuv +=

43

Example 5

A park ranger driving on a back country road

suddenly sees a deer frozen in his headlights.

The ranger, who is driving at 11.4 m s1

immediately applies the brakes and slows withan acceleration of 3.8 m s2.

44

a) If the deer is 20.0 m from the rangers

vehicle when the brakes are applied, how

close does the ranger come to hitting the

deer?

b) How much time is needed for the rangersvehicle to stop?


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Solution

45

s

Given: u = 11.4 m s1 ; a = 3.80 m s2

1st find the distance traveled before stopped

asuv 222 +=From:

The distance between the stopped vehicle & deer:

46

(b) Time needed to stop = ?

atuv +=From:

Example 6

A toy car moves with an acceleration of 2 m s2

from rest for 2.0 s. It then moves with constantvelocity for another 3.0 s. It finally comes to restafter another 1.0 s.

(a) Sketch a velocity-time graph to shown themotion of the toy car.

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(b) What is the velocity of the toy car after first2 seconds ?

(c) Calculate the deceleration of the car.

(d) What is the total displacement of the car forthe whole journey.

(e) Sketch the acceleration-time graph for themotion of toy car.

Solution

(a) v(m s1)

4

DecelerationDecelerationDecelerationDecelerationV decreasesV decreasesV decreasesV decreasesfromfromfromfrom 4444 m sm sm sm s1111totototo

0000 m sm sm sm s1111 inininin 1111 ssss

48

(b) Using : atuv +=

t (s)

2 5 60


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(c) Using : atuv +=

49

(d) Displacement, s = area under the graph

(e) Acceleration time graph

a(m s2)

50

t (s)

2 5 60

4

Follow up exercise

The speed limit in a school zone is 40 km h1. A

driver traveling at this speed sees a child run

onto the road 13 m ahead of his car. He applies

the brakes and the car decelerates at a uniformrate of 8.0 m s2. If the drivers reaction time is

1

51

. s, w t e car stop e ore tt ng t e c

(a) Is it possible for an object moving at non zero

velocity has a zero acceleration? Explain.

(b) A car is capable of accelerating at 0.60 m s2.

Calculate the time needed for this car to go froma speed of 5.5 m s1 to a speed of 8.0 m s1.

2

Figure 2 shows a displacement time graph of a car movingalong a straight road.Copy and complete Table 1 by stating any change ( increase/ decrease / constant / zero / no change ) in the distance,speed and acceleration of the car for each zone.

3

52

Zone Distance Speed Acceleration

A

B

C


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Show that4 atuv +=

5 Show that2

2

1atuts +=

6Show that asuv 2

22+=

7 An object moves along the x-axis. When it is at the centre ofcoordinate, its velocity is and its acceleration is .-1sm6

-2

53

Determine(a) its position at t = 2.0 s(b) Its velocity at t= 3.0 s

sm.

8 An object moves along a straight line with constantacceleration. Its initial velocity is After 5.0 s, thevelocity becomes Determine the distance travelledduring the third second.

.sm02 -1

.sm04 -1

Learning Outcomes

At the end of this chapter, students should beAt the end of this chapter, students should beable to :able to :

2.3 Free Falling Body

Describe free falling body.Describe free falling body.

Solve problems on free falling body.Solve problems on free falling body.

54

Free fall motion is linear vertical motion underthe sole influence of gravity.

The only force acting on the

object is the pull of gravity.

Assumption : free falling

55

objects do not encounter airresistance

All free falling objects ( onEarth ) always acceleratedownwards with an

acceleration a = g

Value of g = 9.81 m s2.

g (vector quantity) is given a minus () signindicating that it is always directed downward.

Replace a with g into kinematics equation :

Free Fall motion Equationv = u g t

v2

= u2

2 g ss = u t g t2

s = u + v t

56

where

s : vertical displacementu : initial velocityv : final velocityt : time interval

g = 9.81 m s2

Value of s, u, v may be (+) or () depending onthe direction of motion.


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Upwards Journey:displacement : +velocity : +acceleration : g

Downwards Journey:Above release point:displacement : +velocity :

Below release point:Displacement : velocity : Acceleration : g

(Referencelevel/origin)

Graphs of free fall objects position, velocity &acceleration as functions of time.

Motion graphs for an object thrownvertically upwards and then falling

back to the ground.

58

Example 7

A student drops a ball from the top of a tall

building, it takes 2.8 s for the ball to reach the

ground.

(a) What was the balls speed just before hitting

the ground ?

b What is the hei ht of the buildin ?

59

Solution

Given :u = 0 m s1 (dropped) ;t = 2.8 s ;

free fall motion, g = 9.81 m s2

gtuv =

)8.2)(81.9(0=v

147.27 = smv

* ( Minus sign indicates that v is downward )

(a)

60

2

2gtuts =

2)8.2)(81.9(

2

10=s

downwardisntdisplaceme:ve-

46.38 ms =

(b)


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Example 8

A boy throws a stone straight upward with an

initial speed of 15 m s1. What maximum height

will the stone reach before falling back down ?

SolutionAt maximum height, objects velocity is zero foran instant (v = 0 m s1)

61

gsuv 222 =

s)81.9(2)15()0( 22 =

62.19

225=s m47.11=

Example 9

A stone is thrown vertically downward at aninitial speed of 14 m s1 from a height of 65 mabove the ground.(a) How far does the stone travel in 2 s ?

(b) What is its velocity just before it hits theground ?

62

Given : u = 14 m s1 ; g = 9.81 m s2 ; t = 2 s

(a) Using free fall equation :

2

2

1

gtuts =

(b) Assume the velocity just before hitting theground = v

gsuv 222 =

63

Example 10

A small pebble is thrown upward from a cliffwith an initial velocity 20 m s-1. Calculate(a) Maximum height reached.(b) Time taken to reach a point 25 m below the

initial point.

Solution

64


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(a) At max height, v = 0 m s1

From: gsuv 222 =

65

(b) Given : s = 25 m ( below initial point )

2

2

1gtuts =

From:

02520905.42

= tt

From :

a

acbb

t 2

42

=

)25)(905.4(4)20()20(2

66

)905.4(2=

81.9

84.2920=

only)valueve(1.5 += st

Follow up exercise

1 An object is thrown vertically upwards from a point

on the ground with speed u. Neglect air resistance.

Determine

(a) the maximum height reached by the object

(b) the time taken to return to the starting point in

67

2 A ball is thrown vertically upwards from the top of a

building at a speed of 15 m s1 . If the height of the

building is 200 m, determine

(a) the time taken by the ball to reach the ground

(b) the velocity when the ball reaches the ground

Learning Outcomes

At the end of this chapter, students should beAt the end of this chapter, students should be

able to:able to:

Describe projectile motion.Describe projectile motion.

2.4 Projectile Motion

Solve problems on projectile motion.Solve problems on projectile motion.

68


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Projectile motion refers to the motion of anobject projected into the air at an angle.

69

A motion where object travels at uniform velocityin horizontal direction; at the same timeundergoing acceleration in downward directionunder the influence of gravity.

-- 2 dimensional motion.

-- consists of horizontal

and vertical motion.These 2 components ofmotion MUST be

70

-- Assumptions of projectilemotion:

1) free fall acceleration, g isconstant and is alwaysdirected downward.

2) Neglect air resistance.

.

(2) Horizontal Projection(1) Projection at an angle

71

Consider an object thrown with a velocity u at anangle relative to the horizontal.

sx( Horizontal displacement )

72

-- the object move upward or downward it alsomoving horizontally.

ay

= g

ax= 0 m s 2

sy ( vertical displacement )


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-- The initial velocity, u is resolved into horizontaland vertical components :

Horizontal & vertical motions are

independent and discussedseparately in calculation.

-- the path of motion : parabolic arc

73

sin

cos

uu

uu

y

x

=

=

-- In vertical ( y axis ), gravity acts downwards(ay = g) object does free fall motion.

-- y-component of the velocity changes with time.

tguv yy =

The vertical displacement, sy is given by:

2

21 tgtus yy =

and yyy sguv 222=

ssssyyyy also knownalso knownalso knownalso knownas heightas heightas heightas height

74

Gravitational force does not act horizontally -- noacceleration in horizontal direction ( ax = 0 ), objecttravels in horizontal direction with uniform velocity.

Horizontal component of the velocity, vx is constant.

xx uv =

tauv xxx +=From:0

With ax = 0, the horizontal displacement, sxisgiven by:

tus xx =

The magnitude of the instantaneous velocity v

2

2

1tatus xxx +=

0

75

at any point is given by :

22yx vvv +=

x

y

v

v=tan linehorizontalfromis

Direction of the velocity :

vxvy = 0

For projectile returns to the same verticallevel at which it was launched

76

Maximum Height, H

Range, R


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=

Using : vy2 = uy

2 2g sy

0 = ( u sin )2 2gH

guH

2sin

22

= [ Maximum Height ]

At the maximum height (H),vy = 0.

77

H

maximum height, H

Using :

vy = uyg t

0 = ( u sin ) g tH

g

utH

sin= [ time to reach maximum height ]

Let T = time of flight

When the body lands on the ground, sy = 0

Using :

2

2

1gttus yy =

21)sin(0 gTTu =

78

g

uT

sin2=

)sin

(2 g

uT

= HtT 2=

* As the path is symmetrical, time in going up is

equal to the time in coming down.

Range, R is the maximum horizontal displacement

traveled.

Horizontal displacement, sx = R when t = T

Using :

sx = uxt

R = u cos (T)

79

)(cosg

uR =

g

u )cossin2(2 =

g

uR

2sin2=

Maximum range, Rmax at a

particular speed is obtained

when :

12sin =

= 45

= 902

80


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Object projected horizontally

81

Projectile travels to

the right as it fallsdownward.Initially, we ONLY

have ux , uy= 0

Quick Test

Can you differentiate between free fall and projectile

motion ?

82

u

Horizontal displacement sx

83

Verticaldisplacement,sy

Equations in Projectile Motion

Horizontal (x motion) Vertical (y motion)

cosuux =

0=xa

sinuuy =

gay =

84

tus xx =

xx uv = tguv yy =

2

2

1

tgtus yy =

yyy sguv 222=


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Procedure for Solving ProjectileProcedure for Solving ProjectileProcedure for Solving ProjectileProcedure for Solving Projectile

Motion ProblemsMotion ProblemsMotion ProblemsMotion Problems

1.1.1.1.Separate the motion into theSeparate the motion into theSeparate the motion into theSeparate the motion into the xxxx

(horizontal) part and(horizontal) part and(horizontal) part and(horizontal) part and yyyy(vertical)(vertical)(vertical)(vertical)

85

part.part.part.part.

2. Consider each part separately using2. Consider each part separately using2. Consider each part separately using2. Consider each part separately using

the appropriate equations for x andthe appropriate equations for x andthe appropriate equations for x andthe appropriate equations for x and

y motion.y motion.y motion.y motion.

Example 11

A cannonball is fired with an initial velocity of 30.0 ms1 at an angle of 35 to the horizontal.(a) What is the maximum height reached by the ball ?(b) What is its range ?

Solution

Given : u = 30.0 m s1 ; = 35 ; ay = g = 9.81 m s2

86

u= 30 m s1

=35

R

H

(a) Maximum height ( comp y ) ;At maximum height, vy = 0

From u = 30 m s1 & = 35 , resolved u into x & y comp.

ux = u cos 35 = 30 cos 35= 24.6 m s1

uy = u sin 35 = 30 sin 35 = 17.2 m s1

Using :

2 = 2

87

y y y

(0)2 = ( 17.2 )2 2 ( 9.81 )(sy)

62.19

84.295=ys m1.15=

Or use equation

g

uH

2

sin22 =

)81.9(2

)35(sin)30( 22= m1.15=

(b) Range = ? Max. horizontal displacement (comp x)

R = sx = ux t ;to find R, must know value of time of flight, T .

Time of flight, T = 2 tup

At max. height , vy = 0

From : vy = uy gt

=

88

. . up

stup 753.181.9

2.17==

Total flight time, T = 2 tup = 2(1.753) = 3.506 s

Range, R = Sx(max) = ux ( T )= 24.6 (3.506)

= 86.25 m

C id d t l


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Example 12

A ball is projected from a height of 25.0 m above the

ground. It is thrown with an initial horizontal velocity of

8.25 m s1.

(a) How long is the ball in flight before striking the

ground ?(b) How far from the building does the ball strike the

ground ?

c What is thevelocit of the ball ust before it strikes

89

the ground ?

Solution

sx = ?

t = ?

v = ?

(a) From : Sy = uyt g t2

( 25 ) = (0)t (9.81)(t2)

x-motion y-motion

sx= ? sy= 25.0 m

ux= 8.25 m/s uy = 0

ax = 0 ay = 9.81 m/s2

Consider x and y separately :

90

st 26.281.9

)25(2==

(b) )(tus xx =

m6.18

)26.2)(25.8(

=

=

(c) In horizontal, -- constant velocity motion.

x- component of the velocity is unchanged

vx= ux = 8.25 m s1

In vertical, g acts on the object, so velocity

changed with time.

From :

vy = uy g t

= (0) 9.81( 2.26 )

xv

91

= 22.17 m s1

Velocity, v = 22 yx vv +22 )17.22()25.8( +=

166.23 = ms

x

y

v

v=tan

25.8

17.22= = 59.69Direction of v :

v is 23.66 m s1 at an angle 69.59 below +x axis.

yv

Example 13

A hockey player hits a slap shot in practice ( with

no goalie present ) when he is 15.0 m directly in

front of the net. The net is 1.20 m high and the puck

is initially hit at an angle of 5 above the ice with aspeed of 35.0 m s1.

92

(a) Make a sketch of the situation using x - y

coordinates, assuming that the puck is at the

origin at the time it is hit. Be sure to locate the net

in the sketch and show its height.

(b) Determine if the puck makes it into the net.

R l d i &


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Solution

(a) Sketching

sy

93

Logical Thinking :

For the puck goes into the net,

sy 1.20 m

initially : = 5 ; u = 35.0 m s1 ; sx = 15 m

(b)

sx= 15 m

From : sx = ux t

sx 0.15

Resolved u into x & y component:

ux = u cos 5 = 34.9 m s1

uy = u sin 5 = 3.05 m s1

In order to find sy, we must calculate tusing sx ;

94

ux.

9.34

From : sy = uy (t) g t2

= (3.05)(0.43) (9.81)(0.43) 2

= 1.31 0.906

= 0.40 m

sy 1.2 m puck goes into the net.

Exercise (Past Year Question Session 2004/05 Q1)

VB = 1.96 m/s

95

gure s ows a s a onary o ec on a smoo a e aheight h above the floor. The object moves horizontallya distance of 1.6 m from A to B with uniformacceleration 1.2 m s2. Then the object is projectedfrom B and fall onto the floor in 0.5 s. Calculate(a) The velocity of the object at B

(b) The value of h [ 4 marks]

Ans : (a) 1.96 ; (b) 1.23

Solution

Motion can be divided into 2 :1.Linear motion on horizontal table A B2.Projectile motion from B & fall on ground

Given : sAB = 1.6 m, a = 1.2 ms-2

time for projectile motion, t = 0.5 s

(a) Using Linear Kinematics Equation

96

asuv 222 +=

)6.1)(2.1(2)0(22+=v

84.32=v

196.1 = msv

R1

GJU1


25/27

(b) h, relate with comp y. At point B, uy = 0

Using Projectile motion equation:

* Remember : often motion of an object is divided intosegments, each with a different acceleration. When solvingproblems, it is important to realize that the final velocityfor one segment is the initial velocity for the nextsegment.

97

2

2

1tgtus yy =

2

)5.0()81.9(2

1)5.0()0( = h

mh 23.1=

R1

GJU1

Mid Term Examination Session 2005/06

A stone is thrown upward from the roof of abuilding with velocity 15 m s1 at an angle of 30

to the horizontal. The height of the building is40.0 m. Calculate

a The maximum hei ht of the stone

98

from the ground.

(b) The magnitude of the velocity ofthe stone just before it strikes theground.

Answer(a) 42.87 m(b) 31.78 m s1

Test your understanding

(1) (a) State ONE similarity between free fall andprojectile motion.

(b) An airplane moving horizontally with aconstant velocity of 115 m s1 at an altitude of

1050 m. The plane released on aid parcel thatfalls to the ground.

99

(i) What are the horizontal and verticalcomponents of the parcels initialvelocity?

(ii) How long does the parcel take to hit theground?

(iii) Calculate the velocity of the parcel justbefore it hits the ground.

100

() ()


26/27

()

101

()

102

A

C

B

D

A transport plane travelling horizontally at aconstant velocity of 50 m s1 at an altitude of 300 mreleases a parcel when directly above a point X onlevel ground. Calculate

a. the flight time of the parcel,

b. the velocity of impact of the parcel,

c. the distance from X to the point of impact.

-2

()

103

.

300 m

d

1sm50 =u

X

A basketball player who is 2.00 m tall is standingon the floor 10.0 m from the basket, as in Figure2.13. If he shoots the ball at a 40.0 angle abovethe horizontal, at what initial speed must he throwso that it goes through the hoop without strikingthe backboard? The basket height is 3.05 m.

()

(Giveng = 9.81 m s-2)

104Figure 2.13Figure 2.13


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Matriculation Physics SF016 105

End of Chapter 2

chapter-2 - kinematics of linear motion

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