kinetic and kinematic.docx

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Kinetics and Kinematics

Motion

For an object moving with constant accelerationand initial velocity, the following equations connect the displacementand the velocity after a time.

atuv +=

2

2

1 atutS +=

tvuS )(2

1+=

2

2

1atvtS =

aSuv 222 +=

Exercise:

1 particle passes through a point moving at ! ms"1and continues with a constant velocity for # s until it reaches a

point $. t $, it accelerates at a constant rate for % s until it passes through a point & moving with a velocity of #ms "

1. Find the distance covered and the acceleration from $ to &.

!

stone slides across a sheet of ice and passes through a point moving with velocity of 1! ms "1and a point $ ! s

later. ssuming the acceleration is constant and that $ is !' m.

Find

(a) the deceleration,

(b) the velocity at $,

(c) *he distance travelled by the stone from until it comes to rest.

1 | P a g e

+here,

=unitial velocity (m-s)

=aacceleration (m-s!)

=vFinal velocity (m-s)

=ttime (s)


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/

%

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#

Motion under gravity (vertical motion)

ll objects when dropped fall towards the earth in a vertical line with the same constant acceleration, provided thatthere is no air resistance.

gtuv =

2

2

1gtutS =

tvuS )(2

1+=

2

2

1gtvtS +=

gSuv 222 =

Exercise:1

3 | P a g e

+here,

=unitial velocity (m-s)

=vFinal velocity (m-s)

=t*ime (s)

=Sisplacement (m)


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!

/

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%

#

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2

Motion Graphs

Displacement time graph ( x-t)

" *he gradient of the x"t graph represent the velocity of theobject.

Velocity- time graph (v-t)

" *he gradient of the v"t graph represent the acceleration ofthe object." *he area under the graph represent the displacement of

the graph.

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0raph analysis:

3bject static

4ero velocity

0raph analysis:

&onstant velocity

0raph analysis:

&onstant velocity

0raph analysis:

&onstant acceleration

0raph analysis:

ncreasing velocity0raph analysis:

ncreasing acceleration

Exercise:

1

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!

/

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%

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#

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General motion in a straight lineFor an object moving in a straight line, if x denotes the displacement from a fixed point 3 of the line at time t, v denotesthe velocity and a denotes the acceleration, then

dt

dxv=

( )dtav =

2

2

dt

xd

dt

dva ==

( )= dtvx

Exercise:

1

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!

/

%

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#

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2

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5

6

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1'

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11

1!

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kinetic and kinematic.docx

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