Transcript lecture02

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.

In contrast, the displacement from point A to point B is not equal to the displacement from A to C.

B A

d d

AB AB

 

d d

CA AC

C 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.

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 A

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

Geometric presentation :

a

a

- letter with arrow;

a

– bold font

Magnitude (length of the vector):

a

a

Some properties:

A

 

B

  

C

A

B

C

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5a. Vector addition (geometric) Two vectors:

a

b

  

c c

 

a b

Several vectors

a

b

  

c

d

d

 

a

Subtraction

a

b

  

c b

 

a

c

b

b

c

b

c

 

a b

 

b

 

a

c

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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 

A

 

B

Question 3: Consider the following three vectors: What is the correct relationship between the three vectors?

1 .

2 .

C

C

  

A

A

  

B

B

3 .

4 .

C

C

      

A

A

  

B

B

  4

5b. Vectors and system of coordinates 2D:

r

 

r

x

 

r y

r x r y

r

y r

y

r y y r

x

3D:

y r

  

r x

 

r y

 

r z

 

r x

,

r y

,

r z

  

x

,

y

,

z

r

x z r

x

r x x

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6. Average speed and velocity a) Average speed Definition:

v

 

d

t

d final t final

d initial

t initial

b) Average velocity Definition:

v

  

r

 

t

t r

final final

 

r initial

t initial

x-component of velocity:

v x

 

x

t

x final t final

x initial

t initial

(total distance over total time) (total displacement over total time)

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7. Instantaneous speed and velocity

(Speed and velocity at a given point)

a) Instantaneous speed Definition:

v

 lim 

t

 0 

d

t

b) Instantaneous velocity Definition:

v

  lim

t

 0 

r

 

t v x

 lim 

t

 0 

x

t

The magnitude of instantaneous velocity is equal to the instantaneous speed

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

x

x

0 

vt x

= const)

v

 

x

t

 

x

1 

x

t

 tan 

t

1 

t

1

t

2

t

3 

t

2

t

4 

x

2 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?

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b) Motion with changing velocity

x

C A B 

x v x

 lim 

t

 0 

x

t

t 

t

Instantaneous velocity is equal to the slope of the line tangent to the graph.

(When Δ

t

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?

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8. Acceleration

•Acceleration shows how fast velocity changes •Acceleration is the rate at which velocity is changing - “velocity of velocity”

v

  

r

 

t v

v

  lim 

t

 0

v

 

r

 

t

a

a a

   

v

 

t

lim 

t

a

 0 

v

 

t

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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 Δ

t

. Then the acceleration of the bicycle is :

a

 10

mi

/

h

 5

mi

/ 

t h

 5

mi

/

h

t

and the acceleration of the car is:

a

 55

mi

/

h

 

t

50

mi

/

h

 5

mi

/ 

t h

Hence, the acceleration of the bicycle is equal to the acceleration of the car.

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9. Motion with constant acceleration

x v

 

v x

0 0 

v

0

t

at

2 2 

at v

2

a

x

  

x

t x

0   

v

0 2

v

2 

v

v

0 2

Example 1:

x

0  2

m v

0  2

m

/

s t a

  3

m

/

s

2 2

s x

 ?

v

 ?

Example 2:

x v

0  10

m

 2

m

/

s a

 3

m

/

s

2

v

 ?

v x

  2

m

  2

m

/

s

2

m

/

s

  3

m

/

s

2      3

m

/

s

2    2 

x

2 

v

  12

m

8

m

/

s

2

a

x

x

0  

v

2 

v

0 2

v

  2

a

x

v

0 2 

v

2 

v

 2

a

x

v

0 2 2  3

m

/

s

2   10

m

   2

m

/

s

2  2 

v

 8

m

/

s

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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 t

A) It speeds up all the time B) It slows down all the time C) It moves at constant velocity D) Sometimes it speeds up and sometimes it slows down E) Not really sure 13