Lecture 3.1
Scalars and Vectors,
Kinematics in Two and Three Dimensions
1. Scalars and Vectors
Physics is a quantitative science, where everything can be described in mathematical terms. As
soon as the system of units has been established, physical quantities can be presented as numbers,
reflecting the magnitude of the quantity. However, many physical quantities have not just
magnitudes but directions as well.
Think about examples of such physical quantities.
Other physical quantities may only have the magnitude but not a direction.
Think about examples of those quantities.
For instance, if we are talking about motion of a car, it is not just moving, but it is moving in a
certain direction, such as to the North or to the West or in somewhere else. So, in order to describe
its motion by means of displacement or velocity, we need not only the magnitude of these quantities,
but also we have to specify their direction. Such quantities are known as vectors. On the other hand
this car also has mass which only has the magnitude but has no direction. This type of quantity is
called scalar. Even though a scalar has no direction but it still has a sign. This may not be true for
mass which, as far as we know, is always positive. But there are a lot of other scalars, such as charge
or temperature that can be negative as well as positive.

We shall represent a vector quantity as a letter with an arrow, for instance A , and we will use

the same letter but without the arrow for the magnitude of this vector A  A . To represent vectors
in the picture, we will also use arrows with lengths proportional to the magnitudes of these vectors.


There are several vectors shown in the picture below. Vector A has magnitude larger than vector B
and it is longer in the picture.


Let vectors A and B show how car moves from one place to another place. If this car first moves


along vector A and then along vector B , the final position of the car is the same as if it just goes




along vector C . This is why we will call this vector C to be the sum of vectors A and B
  
C  A B
As you can see, vector summation is not the same as algebraic summation. The magnitude of vector



C does not equal to the algebraic sum of the magnitudes of vectors A and B . This is because we
are adding not just absolute values but also directions. The picture suggests a practical way of how
one can add two vectors geometrically. You have to draw both vectors in the same scale with tail of
the second vector at the head of the first vector. Then the vector-sum of those two vectors is a vector
which extends from the tail of the first vector to the head of the last vector.
Since vectors can be shifted without changing their absolute value and direction, we can add
them in any order. This gives us the same commutative and associative laws as in the case of the
algebraic addition
   
A  B  B  A,
 
 
 
(3.1.1)
A B  D  A B D .

One can also introduce the  A vector, which has the same magnitude A but the opposite direction



compared to vector A . This means that A   A  0 , and now not only we can add vectors but we
d
i
d
i
d i
can also subtract them. For instance in the picture
  
B  C  A.
2. Vectors and their components.
The geometrical method of adding vectors, described above, helps us to visualize the picture,

but it is not too much of the use, if we want to find the exact magnitude of vector C and specify its
direction.
First of all we need to clarify, what we mean by direction. In a same way as we can obtain
vector's magnitude by comparing its length to a standard unit, we can define its direction by finding
the angle which it makes with some standard direction. The choice of the standard direction, as well
as the choice of units, is somewhat arbitrary and depends on each particular problem. Sometimes to
choose the right standard directions means to solve the problem. For instance, if all the vectors in the
problem are oriented along the same direction, this direction will be the perfect choice for the
standard direction. In general, when we have two vectors, as in the picture, they form a plane
surface. So, this problem becomes two-dimensional problem. To solve it, we have to choose two
standard directions on this surface. Usually they are chosen to be perpendicular to each other and

form the x-y coordinate system. In this coordinate system we choose i and j (other notation x and

y is also in use) to be basis (unit) vectors in the direction of axes x and y. Vector i is a
dimensionless vector of the unit length pointing in the positive x-direction. Vector j is a
dimensionless vector of the unit length pointing in the positive y-direction. Then we can resolve any
vector into components in such a way that
A  Ax i  Ay j .
(3.1.2)

If  is the angle between axis x and direction of vector A then
Ax  A cos ,
(3.1.3)
Ay  A sin  .

Since x and y are perpendicular, the magnitude of vector A can be determined based on
Pythagorean theorem, and its direction using trigonometry as
A
Ax2  Ay2 ,
tan  
Ay
Ax
.
Since we live in 3-dimensional space, if we add more than 2 vectors, in general, they are not
necessarily in the same x-y plane. This is why the third axis z must be introduced perpendicular to x-

y plane. The unit vector in the positive direction of axis z is usually called k ( z notation is also in
use). It is customary to use right-handed coordinate system, such as if you place your right hand
around z-axes in a way that your fingers would sweep vector i into vector j through the smaller
angle between them, your outstretched thumb points in the direction of k . Any vector can then be
resolved in 3 components.
3. Kinematics in two and three dimensions
During our previous meetings we have been talking about objects moving along the straight
line. In reality, however, it rarely happens when something moves along the straight path. For
instance, if we are talking about traffic, only few highways are really straight. If we consider a
motion of the ball during a football game, we can almost never treat it as an object moving along the
straight line. This is why we have to study more complicated cases: two and three dimensional
motion, like a car moving on a curved highway or the ball’s motion during any sport game.
4. Position and Displacement
Think what we need to change in our approach in order to be able to describe position of the
object in two or three dimensional case.
To describe position of the object in three-dimensional space, we have to introduce a
coordinate system, which now has not only one axis x, but three axes x, y and z. It will be the righthanded coordinate system, as it was described before in my lecture about vectors. I have already
mentioned that the choice of the system very much depends on the problem. You have to pick the
origin and directions of the axes in order to obtain the simplest possible description of the problem.

Position of the object can now be defined by means of the position vector (radius vector) r which
connects the origin of the coordinate system with the point where the object is located. As any other
vector the position vector can be resolved into components
r  xiˆ  yjˆ  zkˆ
(3.1.4)
The coefficients x, y and z are also known as the object's rectangular coordinates. For some
problems with specific symmetry it is more convenient to use not rectangular but spherical,
cylindrical or some other coordinates.


As the object moves, its position is changing from the initial position r1 to the final position r2 .

We shall define displacement vector, r , as change the of position, in a same way as we did in one
dimension, so
  
r  r2  r1 ,

(3.1.5)
 

r  x2iˆ  y2 ˆj  z2 kˆ  x1iˆ  y1 ˆj  z1kˆ ,
r   x2  x1  iˆ   y2  y1  ˆj   z2  z1  kˆ,
(3.1.6)
r  xiˆ  yjˆ  zkˆ.
5. Average Velocity and Instantaneous Velocity
Our second step in description of one-dimensional motion, was the answer to the question:
How fast the object is moving? In a same way, now we have to discuss this question for the motion
in several dimensions. We still may use the same definition of the average velocity

displacement r

.
vavg 

time interval t
(3.1.7)
Let us notice that displacement and velocity are vectors. Displacement only depends on the original
and final positions of the object but not on the actual path of the object from the original position to
its final position. Average velocity also depends on the choice of the time interval for which it is
taken. Following equation 3.1.6 we can rewrite equation for the average velocity as
vavg 
xiˆ  yjˆ  zkˆ x ˆ y ˆ z ˆ

i
j
k.
t
t
t
t
(3.1.8)
Think about graphical interpretation of average velocity in three-dimensional case.
The instantaneous velocity at some moment in time is defined as
v  lim
t 0
r dr
.

t dt
(3.1.9)
This means that instantaneous velocity has tangential direction to the object's path at every point of
its trajectory. So we have
v


d ˆ ˆ
dx
dy ˆ dz ˆ
xi  yj  zkˆ  iˆ 
j  k  vxiˆ  v y ˆj  vz kˆ ,
dt
dt
dt
dt
(3.1.10)
x dx
y dy
z dz
, v y  lim
and vz  lim


 .
t 0 t
t 0 t
t 0 t
dt
dt
dt
where velocity vector components are vx  lim
This velocity vector shows us the instantaneous direction of travel.
6. Average Acceleration and Instantaneous Acceleration
However, most probably the object under consideration is not moving with constant velocity.
So, we also have to take into account how fast its velocity is changing. This means that we have to
introduce acceleration as we did in the one-dimensional case. If the velocity is changing as the
object travels from point 1 to point 2 during the time interval t , one can define the average
acceleration as
 

change in velocity v2  v1 v

.
aavg 


time interval
t
t
(3.1.11)
Acceleration is also a vector and average acceleration depends on the time interval for which it is
taken. If this time interval shrinks to zero, we have instantaneous acceleration




v dv d dr
d 2r

a  lim


 2 .
t 0 t
dt dt dt
dt
FG IJ
H K
(3.1.12)
Think what is the main difference between acceleration in 3-dimensional case compared to the
1-dimensional case.
Since velocity and acceleration are vectors, the object has acceleration not only if it changes
speed (magnitude of velocity), but also if it only changes the direction of motion. One can also
rewrite equation 3.1.12 as
dv y
dv
 d
j  dvz k  a xi  a y j  az k ,
a
v xi  v y j  vz k  x i 
dt
dt
dt
dt
e
j
where components of acceleration are a x 
(3.1.13)
dv y d 2 y
dv x d 2 x
dvz d 2 z
,
and
 2
az 
 2 .
 2 ay 
dt
dt
dt
dt
dt
dt
Example 3.1.1. (Sample problem from the book) A rabbit runs across a parking lot, on which
set of coordinate axes, strangely enough been drawn. The rabbit's position are given by
x  0.31t 2  7.2t  28,
y  0.22t 2  9.1t  30,
y(m)
x(m)
where coordinates are given in meters and time in seconds.
The picture shows trajectory of this rabbit.
(a) At time t=15s, what is rabbit's position vector?
We can represent this vector in the unit vector notation, which is

r  xi  yj,
where
b g b gb g b gb g
y  yb15g  b0.22gb15g  b9.1gb15g  30  57m,
2
x  x 15  0.31 15  7.2 15  28  66m,
2
The
magnitude
of
this
vector
will
be
b g b g
F yI
F 57 IJ  41
 y  b66mg  b57mg  87m and it makes angle   tan G J  tan G
H xK
H 66 K

r  66m i  57m j.
so
r  x2
2
2
1
2
1
o
with respect to axis x. "Minus" sign means clockwise angle.
(b) What is rabbit's velocity at the same time t=15s?
Let us use definition of velocity to find the components of velocity vector at this time.
d
d
i
dx d

0.31t 2  7.2t  28  0.62t  7.2,
dt dt
dy d
vy 

0.22t 2  9.1t  30  0.44t  9.1.
dt dt
vx 
i
So for t=15s, we have
b g b gb g
 v b15g  b0.44gb15g  9.1  2.5 m s .
v x  v x 15  0.62 15  7.2  2.1m s ,
vy
y
In unit vector notation the velocity will be

v  2.1m s i  2.5 m s j ,
b
g b
g
which has the magnitude of v  vx2  v 2y 
  tan 1
FG v IJ  tan FG 2.5IJ  130
H 2.1K
Hv K
y
1
o
b2.1m sg  b2.5m sg
2
2
 3.3 m s and it makes angle
with axis x. This angle is negative, because it is in the
x
clockwise direction with respect to axis x. It shows the direction of motion at time t=15s, which is
tangential to the trajectory at that moment.
Exercise What is the average velocity of the rabbit for the first 15s of his trip?
(c) What is rabbit's acceleration at time t=15s?
Let us use our definition of acceleration to find the components of the acceleration vector at
this time. From the velocity equation, which we have derived last time, one has
b
b
g
dv x d

0.62t  7.2  0.62 m s2 ,
dt
dt
dv y d
ay 

0.44t  9.1  0.44 m s2 .
dt
dt
ax 
g
So acceleration stays constant all the time during the rabbit's trip and for t=15s, we have that in unit
vector notation acceleration will be

a  0.62 m s2 i  0.44 m s2 j ,
d
i d
i
it has the magnitude a  a x2  a 2y 
  tan 1
FG a IJ  tan FG 0.44 IJ  145
H 0.62 K
Ha K
y
1
o
e0.62 m s j  e0.44 m s j
2 2
2 2
 0.76 m s2 and it makes angle
with positive x-direction.
x
7. Two-dimensional motion with constant acceleration
We have already studied one-dimensional motion with constant acceleration. All of the
equations which we have obtained in that case are still valid but have to be extended into 2
dimensions, so

a  const ,
a x  const ,
(3.1.14)
a y  const .
  
v  v0  at ,
v x  v0 x  a x t ,
(3.1.15)
v y  v0 y  a y t .
1
  
r  r0  v0t  at 2 ,
2
1
x  x0  v 0 x t  a x t 2 ,
2
1
y  y0  v 0 y t  a y t 2 .
2
(3.1.16)