KINEMATICS USING VECTOR ANALYSIS Standard Competency Analyzes the nature phenomenon and its regularity within the scope of particle’s Mechanics Base Competency Analyzes linier, circular and parabolic motions using vector analysis Learning Objectives After completing this chapter, all students should be able to: 1 Analyzes the quantity of displacement, velocity and

acceleration on linier motion using vector analysis 2 Applies the vector analysis of position, displacement,

velocity and acceleration vectors on linier motion equations 3 Calculates the velocity from position’s function 4 Calculates the acceleration from velocity’s function 5 Determines the position from the function of velocity and

acceleration References [1] John D Cutnell dan Kenneth W. Johnson (2002). Physics 5th Ed with

Compliments. John Wiley and Sons, Inc. [2] Sunardi dan Etsa Indra Irawan (2007). Fisika Bilingual SMA/MA untuk

SMA/MA Kelas XI. CV Yrama Widya

x

y

rr

i

j

rx

ry

All object’s motions are described in terms of - position (x) and displacement (Δr), - velocity (v) and - acceleration (a) Where all the descriptions are considered occur in Cartesian or xy-coordinate. The discussion is categorize as follows - 1 Dimensional motion or LINIER MOTION:

object moves on either x-axis or y-axis

- 2 Dimensional motion or PLANE MOTION: object moves on xy-axis

- 3 Dimensional motion or SPACE MOTION: object moves on xyz-axis

In describing motion, Physics is using Vector Analysis and some basic Calculus (differential and integral’s concept) Motion Description: Position and Displacement A particle position within Cartesian coordinate describes as

or r = xi + yj

In general term it writing as

ji yxr +=r

( )j i yxr ±±=r

vector position: vector that describes a position of a particle in Cartesian coord r

r

i , j unit vector: vector that describes unit scale of an axis

rx , ry vector component: projection of a vector position on x and y-axis

ji yxr +=r

description a vector position on a plane (xy-plane)

Vector position is a vector that describe the position of a particle. It has vector components in x and y-direction and written as r = rx + ry The value of vector position is written as

22 yxr += This value is a magnitude of vector position r based on vector components When a symbol is typed in italic format, such r, it indicates that the symbol is a scalar quantity. Otherwise, when it typed in bold format, such r, it indicates that the symbol is a vector quantity. Vector Displacement

Consider a particle that moves randomly in arbitrary path on xy-plane. The term of vector displacement is the difference of vector positions where in unit vector is

Δr vector displacement is vector difference of r2 and r1

x

y

1rr

i

j 2rr

12 rrrrrr

−=Δ

arbitrary

ji

j)(i)(

)ji()ji(

1212

1122

12

yx

yyxx

yxyx

rrr

Δ+Δ=

−+−=

+−+=

−=Δrrr

Please note, the typing of vector position in bold, r is similar to the symbol of r in vector notation, r

r.

vy

vx

v

Re-writing the equation Δr = Δxi + Δyj where Δr = rfinal − rinitial or Δr = r2 − r1 Δx = xfinal − xinitial or Δx = x2 − x1 Δy = yfinal − yinitial or Δr = y2 − y1 The value of vector displacement is similar to the value of vector position, i.e.,

22 yxr Δ+Δ=Δ

Motion Description: Velocity Velocity is another motion description which indicate how fast or how slow is the object moves. It is very common in our daily life term, where usually describes how fast an object moves.

If an object moves in an arbitrary plane (xy-coordinate) then the direction is given as

x

y

v

v=θtan

Velocity Vector

The position vector r is a vector that goes from the origin of the coordinate system to a given point in the system. The change in position Δr (delta-r) is the difference between the start point (r1) to end point (r2). We define the average velocity (vav) as: vav = (r2 - r1) / (t2 - t1) = Δr / Δt or

tr

vΔΔ

=rr

Taking the limit as Δt approaches 0, we achieve the instanta-neous velocity v. In calculus terms, this is the derivative of r with respect to t, or dr/dt which is written as

dtrd

vr

r=

If r = atn then 1−= ntnadtrdr

As the difference in time reduces, the start and end points move closer together. Since the direction of r is the same direction as v, it becomes clear that the instantaneous velocity vector at every point along the path is tangent to the path.

Description of particle’s velocity onn Cartesian coordinate is described using calculus

Velocity Components

The useful trait of vector quantities is that they can be broken up into their component vectors. The derivative of a vector is the sum of its component derivatives, therefore:

dtdx

vx =

dtdy

vy =

The magnitude of the velocity vector is given by the Pythagorean Theorem in the form:

22yx vvv +== v

12

12

ttrr

tr

v−−

=ΔΔ

=rrr

r

ji

ji

)ji(

yx vvdtdy

dtdx

dtyxd

dtrd

v

+=

+=

+==

rr

The direction of v is oriented theta degrees counter-clockwise from the x-component, and can be calculated from the following equation:

x

y

v

v=θtan

Motion Description: Acceleration Various changes in a particle’s motion may produce an acceleration. When an acceleration is build, it brings consequences, i.e., * The magnitude of the velocity vector may change * The direction of the velocity vector may change (even if the

magnitude remains constant) * Both may change simultaneously

v = 0 for an i t t

Acceleration Vector

Acceleration is the change of velocity over a given period of time. Similar to the analysis above, we find that it's Δv/Δt. The limit of this as Δt approaches 0 yields the derivative of v with respect to t.

In terms of components, the acceleration vector can be written as:

dtdv

a xx = or 2

2

dtxd

ax =

dt

dva y

y = 2

2

dtyd

ay =

The magnitude and angle of the net acceleration vector are calculated with components in a fashion similar to those for velocity. Description of particle’s acceleration on Cartesian coordinate is described using calculus Due to Then

12

12

ttvv

tv

a−−

=ΔΔ

=rrr

r

ji

ji

)ji(

yx

yx

yx

aadt

dv

dtdv

dt

vvd

dtvd

a

+=

+=

+==

rr

2

2

dtxd

dtdx

dtd

ax =⎟⎠⎞

⎜⎝⎛=

2

2

dtyd

dtdy

dtd

ay =⎟⎠⎞

⎜⎝⎛=

ji 2

2

2

2

dtyd

dtxd

a +=r

If x and y component of a is perpendicular, then

If their components made an angle of θ, then

Example [1] Consider a particle in a Cartesian coordinate (xy-

coordinate) where is initially positioned on P1 (4, −1) was then moved to P2 (8, 2) within 1 second. Finds (a) initial and final of vector position, (b) its vector displacement and its value (c) its average velocity and its value

Known variables: initial position: P1 (4, −1) final position: P2 (8, 2) time elapsed: Δt = 1 s Asked: (a) r1 and r2

(b) Δr and Δr (c) vv and

r

Answer

(a) r1 = 4i − j (b) Δr = 4i + 3j

r2 = 8i + 2j 22 34 +=Δr = 5 units

(c) tr

vΔΔ

=rr

= 4i + 3j v = 5 units

22yx aaaa +==

r

x

y

a

a=θtan

Example [2] A particle moves on a circular track r = 2t + t3 with r in

meter and t in second. Calculate the velocity of particle when (a) t = 0, (b) t = 2s

Known variables: r = 2t + t3 Δt = 0 s Δt = 2 s Asked: (a) vo

(b) v2 Answer

2

3

32

2

tdt

ttdtrd

v

+=

+==

r

(a) vo = 2 + 3t2 = 2 m/s

(b) v2 = 2 + 3t2 = 14 m/s Example [3] Given velocity components at time t, i.e., vx = 2t and

vy = (t2 + 4) where t is in second and v is in m/s. Determine its average acceleration between t = 1 s and t = 2 s along with its direction.

Answer

)4(

22 +=

=

tv

tv

y

x

4)1(

)1(22 +=

=

y

x

v

v

4)2(

)2(22 +=

=

y

x

v

v

2

12

1

2

12

1

m/s31

58

m/s21

24

2

2

=−

=−

−=

ΔΔ

=

=−

=−

−=

ΔΔ

=

tt

vv

t

va

tt

vv

tv

a

yyyy

xxxx

Exercises [1] A particle moves from point A (1,0) to point B (5,4) on xy-

plane. Write down the displacement vector from A to B and determine its value

[2] A tennis ball moves on xy-plane. The coordinat position of

point X and Y of the ball is describe by an equation such that x = 18t and y = 4t − 5t2 and a relevant constant. Write down an equation for vector position r with respect to time t using unit vector i and j.

[3] Position of a particle due to time change on xy-plane was

described by vector position r(t) = (at2 + bt)i + (ct + d)j with a, b, c, and d are constants of similar units. Determine its displacement vector between t = 1 second and t = 2 seconds, and define the value of its displacement.

[4] Position of a particle describes by an equation such as r =

2t − 5t2 with r in meter and t in second. Determine (a) the initial velocity of particle (b) the velocity of particle at t = 2 seconds (c) its maximum distance that can be reached by the

particle in positive direction [5] An object moves with velocity of 20 m/s by the direction of

210o counter clockwise related to x-axis. Determine the-x and y components of such velocity.

j)m/s3(i)m/s2(ji

:onaccelerati vector average22 +=+= xx aaa

r

22222 m/s6,332

:magnitude

=+=→+= aaaa yx

o3,56 23

tan :direction =→== θθx

y

a

a

Motion Description Using Differential Concept Consider a motion of two men (blue and red clothes as shown on the picture). They both on xy-plane, where the blue clothe man is walking and the red clothe man is running. The dotted line showed their real motion and short bold line showed instant and short range motion. The relevance of motion on xy-plane or Cartesian coordinate with linier line as follows - in textbook format, distance denote as d or s - in Cartesian coordinate, distance denote as x if the motion

lies on horizontal line or x-axis and denotes as y if it lies on vertical line or y-axis

In general, any linier motion is described by equation

distance = velocity x time

)1(.or. tvxtvs ==

→=dtdxv

For motion with small path or short distance and refer as instant distance, the equation modify as instant velocity means that it has constant velocity Consider if the distance of motion become shorter and shorter and even in very tiny distance where it can be realized. This condition refers as infinity small path. Term refer as the rate of change of position (read as) velocity is the differential of position over

the time (mean) rate of change of position is equal to

velocity The differential concept indeed is very accurate way to describe a very tiny or even infinity path. Similar to equation (1c), we will define differential term of acceleration with respect to velocity. Recall that v = vo + at. If vo = 0, then Form is similar to term velocity in differential way. It reads as acceleration is the differential of velocity over the time and means that rate of change of velocity is equal to acceleration.

instant (distance) = instant (velocity x time)

)1().( atvx Δ=Δ

)1(. btvx Δ=Δ

)1(or. cdtdx

vdtvdx ==

→dtdx

dtadvtavtav =→Δ=Δ→= ..

→=dtdv

a

In term of double rate of change, the differential of velocity over the time can be re-write as Acceleration is double differential of position. In other way around, it could be conclude that Most Important Differential Rules Example [1] A vector position of particle describes as

where t is in second and r in meter. Determine (a) its instantaneous velocity at t = 2 s, and (b) the magnitude and direction of (a)

Answer

2

2

dtxd

dtdtdx

da →

⎟⎠⎞

⎜⎝⎛

=

Differentiating position could yielding velocity

Differentiating velocity could yielding acceleration

1. −=

=

n

n

xndxdy

xy

1..

constant :;.

−=

=

n

n

tnadtdx

atax

j)42(i)42( 32 tttr +++=r

tty

tx

52

423

2

+=

+=

j)42(i)42( 32 tttr +++=r

ji yxr +=r

(a) (b) Integral: Opposed to Differential Concept When both sides of differential form of position being integralled, it can be found that This is the integral form of position. Related to the concept of differential of motion, position of an object could be traced back by perform integral operation on velocity.

As for reminder, integral is the way to turned back a differential equation into “its original” equation

Simillary, when integralling on both sides gives

dtvdx =

∫∫ ∫ =→= dtvxdtvdx

dtadv =

∫∫ ∫ =→= dtavdtadv

m/s)56()52(

m/s4)42(

3

2

+=+==

=+==

tttdtd

dtdy

v

ttdtd

dtdx

v

y

x

j29i8 s 2 if

j)56(i)4(ji 2

+=→=

++=+=

vt

ttvvv yxr

r

m/s08,30298

:magnitude2222 =+=→+= vvvv yx

In a short diagram,

Most Important Integral Rules Eq. Function: Solution:

“original” equation

tvx .=differentialled

dtvdx .=

integralled

dtvx .∫=Ctvx += .yielded

similar to its original equation

Integralling velocity yields position

Integralling acceleration yields velocity

“original” equation

tav .=differentialled

dtadv .=

integralled

dtav .∫=Ctav += .yielded

similar to its original equation

2

1

1

2

1

11

x

x

n

x

x

n

xn

y

dxxy

−

+=

= ∫

Cxn

ay

xay

n

n

++

=

=

+

∫1

1

Example [1] A particle moves on xy-plane at initial position (2,4) m.

Their velocity components are Determine (a) the equation of its position and (b) its position at t = 3 s

Answer (a) Initial position (2,4) means xo = 2m and yo = 4 m

Vector position of the particle is

(b) Particle’s position at t = 3 s Example [2] An object moves from rest with acceleration of

Determine object’s velocity at t = 4 s

Answer

)34( and5 2y tvtvx +==

( )( )m5,22

52

52

2

22

1

0

0

t

t

dtt

dtvxx

t

t

xo

+=

+=

+=

+=

∫

∫

( )( )

( )m44

344

344

3

33

1

0

2

0

tt

tt

dtt

dtvyy

t

t

yo

++=

++=

++=

+=

∫

∫

ji yxr +=r

( ) ( ) j44i5,22 32 tttr ++++=r

( ) ( )( ) ( ) j43i5,24

j44i5,22 32

+=++++= tttr

r

j6i)46( 2 +−= tar

[ ]

j24i32

j4.6i)4.44(

s4j6i)4(

j6i)46(0

0

22

6

22

6

+=+−=

=→+−=

+−+=

=→+=

∫∫

v

tttt

dtt

vdtavv oo

r

rrrr

m/s402432

: vel) s(object' is velocity its of value2222 =+=→+= vvvv yx

Integral as an area beneath the curve

The curve path which is known as displacement is defined as

From the figure, displacement sample Δt is not appropriate for the curve path, i.e, it is not depicted the real path of the curve therefore we could pick an infinity small displacement. Mathematics provide the infinity situation by adding limit.

In limit form, Δv tends to be constant where the value is v. Form Σ lim is simplify by the symbol of integral (means: summation of infinity small parts). The equation is then end up as

t

v (t) x(t)

to tΔt

Δv

t

v (t) x(t)

to t Δt

Δv

∑ ΔΔ= tvtx )(

∑ ΔΔ=→Δ

tvtxt 0lim)(

∫=t

to

dtvtx )(

It can then be concluded that the integration of area under the curve (path) which is likely the integration of velocity will provide the whole object’s distance which represented by the function of x(t). Exercises [1] An object is throwing out to xy-plane such as its vector

velocity was described as v = 50 m/s i + (100 m/s − 10 m/s2 t)j, The positive direction of y-axis to be vertical direction. At t = 0, the object is on its origin (0, 0). (a) determine the vector position of object as a function of

time (b) its position when t = 2 s (c) the maximum height that can be reach by the object

[2] Determine the particle’s position as a function of time if the

particle velocity is (a) v = 2t + 6t2

tv

ttvb

y

x

5sin

53)( 23

21

=

+=

(c) v = 4ti + 3j The particle is initially at its origin (0, 0)

[3] An object is moving on xy-plane with vector velocity of

v = { (3 − 3t2)i + 2tj } m/s Determine the value of its diplacement of the object between t = 1 s and t = 2 s.