4. Distance and displacement
(displacement as an example of a vector)
Example 1: The distance between points A and B
is equal to the distance between A and C.
B
A
In contrast, the displacement from point A to point B
is not equal to the displacement from A to C.
C
d AB  dCA


d AB  d AC
Example 2: For the motion around a closed loop (from A to A)
the displacement is zero, but the distance is not equal to zero.
A
Distance - fundamental physical quantity measured in units of length.
Displacement - physical quantity that should be described by both its
magnitude (measured in units of length) and direction.
Distance is an example of a scalar quantity.
Displacement is an example of a vector quantity.
Scalars have numerical value only (one number).
Vectors have magnitude and direction (at least two numbers).
1
5. Vectors
•A vector has magnitude as well as direction
•Some vector quantities: displacement, velocity, force, momentum
•A scalar has only magnitude and sign
•Some scalar quantities: mass, time, temperature

a
Geometric presentation:

a-
a – bold font

Magnitude (length of the vector):
a a
Notations:
letter with arrow;
Some properties:
 

A  B  C

A

B

C
2
5a. Vector addition (geometric)
Two vectors:

b

c
  
a b  c

a

d
Several vectors
   
a b c  d
Subtraction
  
a b  c

b

a

b

a

c

a

c

b

b

b

c

b

a

c
3
Question 1: Which of the following arrangements will produce the largest
resultant when the two vectors of the same magnitude are added?
A
B
C
Question 2: A person walks 3.0 mi north and then 4.0 mi west.
The length and direction of the net displacement of the person are:
1) 25 mi and 45˚ north of east
2) 5 mi and 37˚ north of west
3) 5 mi and 37˚ west of north
4) 7 mi and 77˚ south of west
β = 37˚<45˚
ϴ= 53˚> 45˚
β
Question 3: Consider the following three vectors:
What is the correct relationship between the three vectors?
  
1. C  A  B
  
2. C  A  B



3. C  

4. C  
 
A B
 
A B


 
A B
4
5b. Vectors and system of coordinates
2D:
  
r  rx  ry  rx , ry   x, y 
y

r

ry
y  ry

rx
3D:

r
x
x  rx
   
r  rx  ry  rz  rx , ry , rz   x, y, z 
y

r
x
z
5
6. Average speed and velocity
a) Average speed
Definition:
d d final  dinitial
v

t
t final  tinitial
(total distance
over total time)
b) Average velocity
Definition:
x-component
of velocity:

 
 r rfinal  rinitial
v

t t final  tinitial
(total displacement
over total time)
x x final  xinitial
vx 

t t final  tinitial
6
7. Instantaneous speed and velocity
(Speed and velocity at a given point)
a) Instantaneous speed
Definition:
d
v  lim
t 0 t
b) Instantaneous velocity
Definition:

r

v  lim
t 0 t
The magnitude of instantaneous velocity
is equal to the instantaneous speed
x
v x  lim
t  0  t

v v
In contrast, the magnitude of average velocity
is not necessarily equal to the average speed
7
6. Geometric interpretation
a) One dimensional uniform motion (v = const)
x  x0  vt
x
v
t
x
 tan 
t
x

x2
t 2

x1
t1
t1
t2
t3
t4
t
Velocity is equal to the slope of the graph (rise over run): distance over time.
Question: The graph of position versus time for a car is given above.
The velocity of the car is positive or negative?
8
b) Motion with changing velocity
C
x
B
x
v x  lim
t  0  t
x
A
t
t
Instantaneous velocity is equal to the slope of the line tangent to the graph.
(When Δt becomes smaller and smaller, point B becomes closer and closer
to the point A, and, eventually, line AB coincides with tangent line AC.)
Question: The graph of position versus time for a car is given above. The
velocity of the car is positive or negative? Is it increasing or decreasing?
9
8. Acceleration
•Acceleration shows how fast velocity changes
•Acceleration is the rate at which velocity is changing - “velocity of velocity”

 r
v
t

r

v  lim
t 0 t

v v

 v
a
t

v

a  lim
t 0 t

a a
10
Example: The speed of a bicycle increases from 5 mi/h to 10 mi/h.
In the same time the speed of a car increases from 50 mi/h to 55 mi/h.
Compare their accelerations.
Solution:
We denote the time interval as Δt. Then the acceleration of the bicycle is:
10 mi / h  5mi / h 5mi / h
a

t
t
and the acceleration of the car is:
55mi / h  50 mi / h 5mi / h
a

t
t
Hence, the acceleration of the bicycle is equal to the acceleration of the car.
11
9. Motion with constant acceleration
2ax  x0   v 2  v02
at 2
x  x0  v0 t 
2
v  v0  at
Example 1:
x 0  2m
v 0  2m / s
a  3m / s 2
t  2s
x? v?
x v0  v
v

t
2

3m / s 2s 
x  2m  2m / s 2s  
2
2


v  2m / s  3m / s 2 2s  
2
 x  12m
v  8m / s
Example 2:
x  10m
v 0  2m / s
a  3m / s 2
v?
2ax  x0   v 2  v02  2ax  v02  v 2  v  2ax  v02



v  2 3m / s 10m  2m / s
2

2 2
 v  8m / s
12
Question 1: If the velocity of a car is non-zero, can the acceleration of the
car be zero?
A) Yes
B) No
C) It depends
Question 2: If the velocity of a car is zero, can the acceleration of the car be
non-zero?
A) Yes
B) No
C) It depends
Question 3: The graph of position versus time for a car is given below.
What can you say about the velocity of the car over time?
x
A)
B)
C)
D)
t
It speeds up all the time
It slows down all the time
It moves at constant velocity
Sometimes it speeds up and
sometimes it slows down
E) Not really sure
13