Text Book: An Introduction to Mechanics by Kleppner and Kolenkow Chapter 1: Vectors and Kinematics -Explain the concept of vectors. -Explain the concepts of position, velocity and acceleration for different kinds of motion. References: Halliday, Resnick and Walker Berkley Physics Volume-1

1. Chapter – 1
Vectors and Kinematics
Text Book: AN INTRODUCTION TO MECHANICS
by Kleppner and Kolenkow
Dr. Virendra Kumar Verma
Madanapalle Institute of Technology and Science
(MITS)

2. Scalars and Vectors
A scalar quantity is a quantity that has only magnitude.
A vector quantity is a quantity that has both a magnitude
and a direction.
Scalar quantities
Length, Area, Volume,
Speed,
Mass, Density
Temperature, Pressure
Energy, Entropy
Work, Power
Vector quantities
Displacement, Direction,
Velocity, Acceleration,
Momentum, Force,
Electric field, Magnetic field
Volume
Velocity

3. Vector notation
Vector notation was invented by a physicist, Willard Gibbs
of Yale University.
By using vector notation, physical laws can often be
written in compact and simple form.
For example, Newton’s second law
In old notation,
F 
ma
x x
F 
ma
y y
F ma
z z
In vector notation,

 

F ma

4. Equal vectors
If two vectors have the same length and the same
direction they are equal.
B C
 The vectors B and C are equal.
B = C
 The length of a vector is called its magnitude.
e.g. Magnitude of vector B = |B|

5. Unit vectors
 A unit vector is a vector that has a magnitude of exactly 1.
Ex: The unit vectors point along axes in a right-handed
coordinate system.
 If the length of a vector is one unit.
e.g. The vector of unit length parallel to A is Â.
A = |A|Â.

6. Algebra of vectors
Multiplication of a Vector by a Scalar:
If we multiply a vector a by a scalar s, we get a new
vector. Its magnitude is the product of the magnitude of a
and the absolute value of s.

7. Algebra of vectors
Addition of two Vectors:

8. Algebra of vectors
Subtraction of two Vectors:

9. Scalar Product (“Dot” Product)

The scalar product of the vectors and is defined as a
 
a b  abcos

b
The above equation can be re written as
 
a b  (a cos )(b)  (a)(bcos )
   
 
a b (b)(Projection of a on b)
OR
 
a b  (a)(bcos )
   
 
a b (a)(Projection of b on a)

10. Scalar Product (“Dot” Product)
If a b  0, then

 The commutative law applies to a scalar product, so we can write
   
  
a b b a
When two vectors are in unit-vector notation, we write their dot
product as
  ( ˆ  ˆ  ˆ)  ( ˆ  ˆ  ˆ)
a b a i a j a k b i b j b k
x y z x y z
a b a b a b
  
x x y y z z
 
 
a = 0
or b = 0

or cos θ = 0 ( a
is perpendicular to ) b

Note:
 
2 a a  a

11. Example 1.1 Law of Cosines
  
 
C A B
     
    
C C (A B) (A B)
    
2 2 2
C A B A B
2 cos
  
This result is generally expressed in terms of the angle φ
2 cos 2 2 2 C  A  B  AB
[cos  cos(  )  cos ]

12. Example 1.2 Work and the Dot product
The work W done by a force F on an object is the displacement d of
the object times the component of F along the direction of d. If the
force is applied at an angle θ to the displacement,
W F d
or
 ( cos )
 
W  F 
d

13. Vector Product 

(“Cross”  
Product)
The vector 
product of A
and B
, written A

B , produces a third
vector whose magnitude is
C
    
  
C B A
C A B A B sin
  
    
 
A  B   (B 
A)

14. Vector Product (“Cross” Product)
When two vectors are in unit-vector notation, we write their cross
product as
( ˆ ˆ ˆ) ( ˆ ˆ ˆ)
a  b  a i  a j  a k  b i  b j 
b k
x y z x y z
( a b b a )i ˆ ( a b b a ) ˆ j ( a b b a ) k
ˆ
     
y z y z z x z x x y y x
ˆ ˆ ˆ
i j k
a a a
x y z
b b b
x y z

 


 
Note: If a and b
are parallel or antiparallel, .
a b  0

15. Example 1.4 Area as a Vector

Consider the area of a quadrilateral formed by two vectors, and
The area of the parallelogram A is given by
A = base ×height
= C D sinθ
 
If we think of A as a vector, we have
C

D
C D
 
  
 
A C D
Magnitude of A is the area of the parallelogram, and the vector
product defines the convention for assigning a direction to the area.

16. Example 1.5 Vector Algebra
Let
A = (3,5,-7)
B = (2,7,1)
Find A + B, A – B, |A|, |B|, A · B, and the cosine of the
angle between A and B.
Example 1.6 Construction of a perpendicular Vector
Find a unit vector in the xy plane which is perpendicular
to A = (3,5,1).

17. Base Vectors
Base vectors are a set of orthogonal (perpendicular) unit
vectors, one for each dimension.

18. Displacement and the Position Vector
( )ˆ ( ) ˆ ( ) ˆ 2 1 2 1 2 1
  
S x x i y y j z z k
     

ˆ ˆ ˆ
S S i S j S k
x y z

The values of the
coordinates of the initial
and final points depend
on the coordinates
system, S does not.

19. Displacement and the Position Vector

r  xiˆ  yˆj  zkˆ
  
' 
r r R
 
 
2 1 S r r
 
r R r R
 
   
( ' ) ( ' )
2 1
r r
 
' '
2 1
A true vector, such as a displacement S, is independent of coordinate system.

20. Velocity and Acceleration
Motion in One Dimension
The average velocity ‘v’ of the point between two times, t1
and t2, is defined by
x t x t
( ) 
( )
2 1t t
2 1
v


( ) 1x t ( ) 2 x t
The instantaneous velocity ‘v’ is the limit of the average
velocity as the time interval approaches zero.
x t   t 
x t
t
v
t 

 
( ) ( )
lim
dx
v 
0 dt

21. Velocity and Acceleration
In a similar fashion, the instantaneous acceleration is
v t   t 
v t
t
a
t 

 
( ) ( )
lim
0
dv
dt
a 

22. Motion in Several Dimensions
The position of the particle
At time t1, At time t2,

( , ) 1 1 1 r  x y

( , ) 2 2 2 r  x y
The displacement of the particle between times t1 and t2 is
 
( , ) 2 1 2 1 2 1 r  r  x  x y  y

23. Motion in Several Dimensions
We can generalize our example by considering the position at
some time t, and some later time t+Δt. The displacement of the
particle between these times is
  
    
r r (t t) r (t)
This above vector equation is equivalent to the two scalar
equations
x  x(t  t)  x(t)
y  y(t  t)  y(t)

24. Motion in Several Dimensions
The velocity v of the particle as it moves along the path is
defined to be
lim
dr
dt
r
t
v
t







 0
dx
dy
dt
x
y
t
v
dt
t
v
lim
 
lim
t
y
t
x








0
 
0
In 3D, The third component of velocity
dz
dt
z

z 
t
v
lim
t


 0

25. Motion in Several Dimensions

v  ˆ  ˆ  k
ˆ
dz
dt
j
dy
dt
i
dx
dt
dr
dt


 ˆ  ˆ  ˆ
v v i v j v k x y z
The magnitude of ‘v’ is
1
2
2 2 2 ( ) x y z v v  v  v
Similarly, the acceleration a is defined by
dv
a   x  y 
z
d r
dt
k
dv
dt
j
dv
dt
i
dv
dt
dt



2
ˆ ˆ ˆ


26. Motion in Several Dimensions
Let the particle undergo a displacement Δr in time Δt. In the time Δt→0,
Δr becomes tangent to the trajectory.

dr
  

v t
t
dt
r
 

Δt→0, v is parallel to Δr
The instantaneous velocity ‘v’ of a particle is everywhere tangent
to the trajectory.

27. Motion in Several Dimensions
Similarly, the acceleration a is defined by
dv
a   x  y 
z
d r
dt
k
dv
dt
j
dv
dt
i
dv
dt
dt



2
ˆ ˆ ˆ


28. Example 1.7 Finding v from r
The position of a particle is given by

r A(e iˆ e ˆj) t t  
Where α is a constant. Find the velocity, and sketch the trajectory.

dr
dt
A( e iˆ e ˆj)
v

t t     

==> t
v  
A  e   y x v A e   t
Solution:

29. The magnitude of ‘v’ is
1
2
1
2
2 2
v  v 
v
( )
x y
t t
 A  e 2  
e
 2
 ( )
To sketch the trajectory, apply the limiting cases.
At t = 0, we get At t = ∞, we get

r (0)  A (i ˆ 
ˆj
)
v 
A i j
(0)  (ˆ 
ˆ)

t t
 
  
e and e
In this limit
0
   
ˆ ˆ
t t
r  Ae i and v 
Ae i
r and v x
vectors pointed along the axis.
 

30. Example 1.7 Finding v from r
Note:
1. Check
 
r(0) 
v(0)
2. Find the acceleration.
3. Check
4. Check
 
a(0) 
v(0)
 
a(t) r (t) 


v
r
a

5. Check the direction of , and at t = 0 and t  ∞.

31. Example 1.8 Uniform Circular Motion
Consider a particle is moving in the xy
plane according to

r r(cost iˆ  sint ˆj)
[ (cos sin )] 2 2 2 1/ 2
 
r r t i t
 
r
constant

The trajectory is a circle.

32. Example 1.8 Uniform Circular Motion

r r(cost iˆ  sint ˆj)
r d
dt
v




v r(sint iˆ  cost ˆj)
v r  r   t t  t t
0
( sin cos cos sin ) 2

 


v is perpendicular to .

r
v  r  constant.

33. Example 1.8 Uniform Circular Motion
dv
dt
a



2 ( cos ˆ sin ˆ)

a  r    t i 

t j

r

2
 
The acceleration is directed radially inward, and is
known as the centripetal acceleration.
Note:
1. Check
2. Check
 
a 
v
r a
 


34. Kinematical equations

( ) 1v t

( ) 0 v t
Suppose - velocity at time tand - velocity at time t1 0.
Dividing the time interval (t-t) in n parts,
 t  (t 
t ) / n 101 0     
v t v t v t t v t t v t
( )  ( )   (   )   (  2  ) 
( )
1 0 0 0 1     
v ( t )  v ( t )  a ( t   t )  t  a ( t  2  t )  t  a ( t )

t
1 0 0 0 1
For n∞ (Δt0), and the sum becomes an integral:
t
1
v t v t a t dt
( ) ( ) ( ) 1 0
  
0
t
 

35. Kinematical equations
t
1
 
v t v t a t dt
( ) ( ) ( ) 1 0
  
0
t
The above result is the same as the formal integration of
 
dv ( t ) 
a ( t )
dt
 
dv ( t ) 
a ( t )
dt
  
v ( t )  v ( t ) 
a ( t )
dt
1 0
t
1
( ) ( ) , [ ( ) intial velocity]
1 0 0 0
0
1
0
1
0
1
0
   


 
v t v a t dt v t v
t
t
t
t
t
t
t
    

36. Kinematical equations
If acceleration a is constant and t0 = 0, we get
  
v t v at
  0 ( )
   
r ( t ) r ( v at )
dt
2
   
0 0
0
or
r t r v t at
0 0
1
2
( )
t
   
  

37. More about the derivative of a vector
Consider some vector A(t) which is function of time. The change in
A during the interval from t to t + Δt is
  
    
A A(t t) A(t)
We define the time derivative of A by
  
A t   t 
A t
t
dA
dt
t 

 
( ) ( )
lim
0

Ad
 is a new vector.
dt
Depending on the behaviour of A:

dA
 The magnitude of can be large or small.
dt

dA
 The direction of can be point in any direction.
dt

38. More about the derivative of a vector
Case 1:
 
A ||
A
direction unaltered but
 
A A
change the magnitude to  
.


Case 2:
 
A  A
change in direction but leave
the magnitude practically unaltered
Case 3: In general

direction. and magnitude both in change will A
Case 3

39. More about the derivative of a vector
 
ΔA A
is always perpendicular to ,
A
must rotate, since its magnitude cannot change.


 
ΔA A
is always parallel to ,

A
direction of same and its magnitude change.


40. More about the derivative of a vector
2

A A
2 sin
For    1, sin   / 2  

/ 2.weget

2

A 2
A

A A
  
and

A




Taking the limit 0,
d
A
dt
dA
d dt A
/ is called the angular velocity of .
dt
t
t
A
t






 




 

 

41. More about the derivative of a vector
.
  
A A A
    
For sufficiently small,


A A
  


||

dividing by and taking the limit,
d
A
dA
dt

dA

dA

 
dA dt A d dt
/ is zero if does not rotate ( / 0),
and / is zero if is constanti n magnitude.
||
||
||
dA dt A
dt
dt
dt
t
A A
 




  






42. More about the derivative of a vector
 
Let be the rotating vector . Then and

d
r
  

 

 
dr
v r
t r
d
r
dt
dt
dt
A r t


or
( )

43. Some formal identities
 
cA
( )
 
 
A B
( )
    
 
A B
dc
( )

dB
dt

dB
dt
 
A B

dA
dt
A c
 
B A
 
B A
dt

dA
dt

dA
    
dt
d
dt
d
dt
d
dt
In the second relation, let . Then
2

A A
( ) 2

 
dA/dt A

dA
dt
d
dt
 
we can see again if is perpendicular to ,

A
the magnitude of is constant.

44. Motion in Plane Polar Coordinates
2 2
r x y
y
x
 
arctan



45. Motion in Plane Polar Coordinates
 The lines of constant x and of constant y are straight and perpendicular to
each other.
 The lines of constant θ and constant r are perpendicular.

46. Motion in Plane Polar Coordinates
The direction of ˆ and ˆ
vary with position,
whereas i ˆ and ˆ j
have
fixed directions.
r 
ˆ ˆcos ˆ sin
r  i 
j
  
 
ˆ i ˆsin ˆ j
cos
  

47. Velocity in Polar Coordinates
d
dr
dt

dr
 

rr r
rr
dt
dt
v
ˆ
ˆ
( ˆ)
 

Recall that in cartesian coordinates -
v ( xi ˆ yj ˆ) xi ˆ yj
ˆ
d
dt
dr
dt
 


    
We know that
ˆ ˆ cos ˆ sin
r  i 
j
 
d
j
d
i
ˆ (cos ) ˆ (sin )
 
 
dt
dt
 
ˆ sin ˆ cos
i j
  
   
( i ˆ sin ˆ j
cos )
  
  
ˆ [ ˆ sin ˆ cos ˆ]
ˆ
i j
   
    
dr
dt
 


v  rr  r
ˆ ˆ

48. Velocity in Polar Coordinates
ˆ ˆ

v  rr  r
The first term on the right is the component of the velocity

directed radially outward.
The second term is in the tangential (i.e., ˆ) direction.
 
 
1 .  constant ,  0 and v 
rr
ˆ
velocity is radial.
i e
. ., the motion in a fixed radial direction.

2. constant, 0 and ˆ
r  r  v 
r
velocity is tangential.
i e
. ., the motion lies on the arc of the circle.


 
 

49. Motion in Plane Polar Coordinates
We know that
ˆ ˆ sin ˆ cos
d
j
    

d
i
ˆ (sin  ) ˆ (cos 
)
dt
  
dt
 
ˆ cos ˆ sin
i j
  
   
(i ˆ cos ˆ j
sin )
  
  
ˆ [ ˆ cos ˆ sin ˆ]
ˆ
r i j r
d
dt
i j
     


 

r
ˆ ˆ
dr
dt
d
dt
ˆ
ˆ


 



 

50. Acceleration in Polar Coordinates
 
( ˆ 
ˆ)
d
d
  
d

dv
ˆ ˆ ˆ ˆ ˆ
   
dt
r r r r
dt
rr r
rr r
dt
dt
a
     

    
Substitute the value of drˆ / dt and dˆ / dt, we get

    
a rrˆ r ˆ r ˆ r ˆ r rˆ 2     

a  r  r r  r  r
(  ) ˆ (  2  )ˆ  2   

51. Velocity and Acceleration in Polar Coordinates
 
ˆ ˆ
v  rr 
r

ˆ ˆ
 v r v
 

r
 
Radial | Tangential term

    
( ) ˆ ( 2 ) ˆ 2

a  r  r r  r 
r
   
ˆ ˆ
 a r a
 

r
 
Radial | Tangential term

52. References
1. Fundamentals of Physics by Halliday, Resnick and Walker
2. Berkeley Physics Course Volume-1
please contact me via email for any further suggestions/comments.
Email: virendrave@gmail.com

53. Thank you