1. 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

2. 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
r
r = xi + y j or r = xi + yj
r
In general term it writing as r = (± x i ± y j)
y r vector position: vector that describes a
r position of a particle in Cartesian coord
ry i,j unit vector: vector that describes
r unit scale of an axis
r
rx , ry vector component: projection of a
j vector position on x and y-axis
r
x r = x i + y j description a vector position
i rx on a plane (xy-plane)

3. 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 = r x + ry
The value of vector position is written as
r = x2 + y2
This value is a magnitude of vector position r based on vector
components
Please note, the typing of vector position
in bold, r is similar to the symbol of r
r
in vector notation, r .
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
arbitrary randomly in arbitrary path on xy-
y plane. The term of vector
r r r
Δr = r2 − r1 displacement is the difference of
r vector positions where in unit
r1 vector is
r
r2 Δr vector displacement is vector
j difference of r2 and r1
r r r
x Δr = r2 − r1
i = ( x2 i + y2 j) − ( x1 i + y1 j)
= ( x2 − x1 ) i + (y2 − y1 ) j
= Δx i + Δy j

4. 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.,
Δr = Δx 2 + Δy 2
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.
v
If an object moves in an arbitrary plane (xy-
coordinate) then the direction is given as
vy
vy
tan θ =
vx vx
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
r r
Δr
v =
Δt

5. 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
r
r dr
v =
dt
r
dr
If r = at n
then = a n t n −1
dt
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
r r r r
r Δr r2 − r1 r dr d( x i + y j)
v = = v = =
Δt t2 − t1 dt dt
dx dy
= i+ j
dt dt
= v x i + vy j
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:
dx
vx =
dt
dy
vy =
dt
The magnitude of the velocity vector is given by the
Pythagorean Theorem in the form:
2 2
v = v = vx + vy

6. The direction of v is oriented theta degrees counter-clockwise
from the x-component, and can be calculated from the
following equation:
vy
tan θ =
vx
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

7. 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:
dv x d2x
ax = or ax =
dt dt 2
dv y d 2y
ay = ay =
dt dt 2
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
r r r r
r Δv v2 − v1 r dv d(v x i + v y j)
a= = a= =
Δt t2 − t1 dt dt
dv x dv y
= i+ j
dt dt
= ax i + ay j
Due to
d ⎛ dx ⎞ d 2 x d ⎛ dy ⎞ d 2y
ax = ⎜ ⎟= ay = ⎜ ⎟=
dt ⎝ dt ⎠ dt 2 dt ⎝ dt ⎠ dt
2
Then
r d2x d 2y
a= i+ j
dt 2 dt 2

8. If x and y component of a is perpendicular, then
r 2 2
a= a = ax + ay
If their components made an angle of θ, then
ay
tan θ =
ax
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
r
(c) v and v
Answer
(a) r1 = 4i − j (b) Δr = 4i + 3j
r2 = 8i + 2j Δr = 42 + 32
= 5 units
r r
Δr
(c) v =
Δt
= 4i + 3j
v = 5 units

9. 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
r
dr 2t + t 3
v = =
dt dt (a) vo = 2 + 3t2 = 2 m/s
= 2 + 3t 2
(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
v x = 2t v x = 2(1) v x = 2(2)
v y = (t 2 + 4) v y = (1)2 + 4 v y = (2)2 + 4
Δv x v x 2 − v x 1 4 − 2
ax = = = = 2 m/s2
Δt t2 − t1 1
Δv y v y2 − v y 1 8−5
ay = = = = 3 m/s2
Δt t2 − t1 1

10. average vector acceleration :
r
a = ax i + ax j = (2 m/s2 ) i + (3 m/s2 ) j
magnitude :
2 2
a= ax + ay → a = 22 + 32 = 3,6 m/s2
a 3
direction : tan θ = y = → θ = 56,3o
ax 2
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.

11. 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
s = v.t or x = v. t (1)

12. For motion with small path or short distance and refer as
instant distance, the equation modify as
instant (distance) = instant (velocity x time)
Δx = Δ(v. t ) (1a)
instant velocity means that it has constant velocity
Δx = v. Δt (1b)
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.
dx
dx = v. dt or v = (1c)
dt
Term dx → refer as the rate of change of position
dt
dx
v = → (read as) velocity is the differential of position over
dt 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
v = a.t → Δv = a.Δt → dv = a dt
dv
Form a = → is similar to term velocity in differential way.
dt
It reads as acceleration is the differential of velocity over the
time and means that rate of change of velocity is equal to
acceleration.

13. In term of double rate of change, the differential of velocity
over the time can be re-write as
⎛ dx ⎞
d⎜ ⎟
⎝ dt ⎠ d2x
a= →
dt dt 2
Acceleration is double differential of position.
In other way around, it could be conclude that
Differentiating position Differentiating velocity
could yielding velocity could yielding acceleration
Most Important Differential Rules
y = xn x = a.t n ; a : constant
dy dx
= n.x n −1 = a.n.t n −1
dx dt
Example
[1] A vector position of particle describes as
r
r = (2t 2 + 4) i + (2t 3 + 4t ) j
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 r
r = x i+y j
x = 2t 2 + 4
r y = 2t 3 + 5t
r = (2t 2 + 4) i + (2t 3 + 4t ) j

14. dx d
(a) v x = = (2t 2 + 4) = 4t m/s
dt dt
dy d
vy = = (2t 3 + 5t ) = (6t + 5) m/s
dt dt
r
v = v x i + v y j = (4t ) i + (6t 2 + 5) j
r
if t = 2 s → v = 8 i + 29 j
(b) magnitude :
2 2
v = v x + vy → v = 82 + 292 = 30,08 m/s
Integral: Opposed to Differential Concept
When both sides of differential form of position dx = v dt being
integralled, it can be found that
∫ dx = ∫ v dt → x = ∫ v dt
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, dv = a dt when integralling on both sides gives
∫ dv = ∫ a dt → v = ∫ a dt

15. In a short diagram,
Integralling velocity yields position
“original”
equation differentialled
x = v.t dx = v.dt
similar to
its original integralled
equation
x = v.t + C x = ∫ v.dt
yielded
Integralling acceleration yields velocity
“original”
equation differentialled
v = a. t dv = a. dt
similar to
integralled
its original
equation
v = a. t + C v = ∫ a. dt
yielded
Most Important Integral Rules
x2
∫x
n
y = dx
∫ax
n
Eq. Function: y =
x1
a x2
Solution: y = x n +1 + C 1
n +1 y = x n −1
n +1 x1

16. Example
[1] A particle moves on xy-plane at initial position (2,4) m.
Their velocity components are v x = 5t and v y = (4 + 3t 2 )
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
t t
x = xo + ∫ v x dt y = y o + ∫ v y dt
0 0
t t
= 2 + ∫ 5t dt ∫ (4 + 3t ) dt
2
=4+
0 0
= 2 + 5 12 t ( 2
) (
= 4 + 4t + 3 13 t 3 )
(
= 2 + 2,5 t 2
)m (
= 4 + 4t + t 3 m )
r
Vector position of the particle is r = x i + y j
r = (2 + 2,5t 2 ) i + (4 + 4t + t 3 ) j
r
(b) Particle’s position at t = 3 s
r
( ) (
r = 2 + 2,5t 2 i + 4 + 4t + t 3 j )
= (24,5) i + (43) j
Example
[2] An object moves from rest with acceleration of
r
a = (6t 2 − 4) i + 6 j
Determine object’s velocity at t = 4 s
Answer r r r r
v = v o + ∫ a dt → vo = 0
=0+ ∫ [(6t − 4) i + 6 j] dt
= (6 2 t 2 − 4t ) i + 6t j →t =4s
r
v = (6 2 42 − 4.4) i + 6.4 j
= 32 i + 24 j
value of its velocity is (object' s vel) :
2 2
v = v x + vy → v = 322 + 242 = 40 m/s

17. Integral as an area beneath the curve
v (t) x(t)
Δv
t
to Δt t
The curve path which is known as displacement is defined as
x(t ) = ∑ Δv Δt
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.
x(t ) = ∑ lim Δv Δt
Δt → 0
v (t) x(t)
Δv
t
to Δt t
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
x(t ) = ∫ v dt
to

18. 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
1 3
(b) v x = 3t 2
+ 5t 2
v y = sin 5t
(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.