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(1)

INTRO TO PHYSICS

(2)

PHYSICS

PHYSICS

►PHYSICS IS THE STUDY OF THE PHYSICS IS THE STUDY OF THE

FUNDAMENTAL LAWS OF NATURE. ALL

FUNDAMENTAL LAWS OF NATURE. ALL

PHYSICAL PHENOMENA IN THE UNIVERSE.

PHYSICAL PHENOMENA IN THE UNIVERSE.

►IT IS A COMBINATION OF MATHEMATICAL IT IS A COMBINATION OF MATHEMATICAL

EQUATIONS AND EXPERIMENTAL

EQUATIONS AND EXPERIMENTAL

OBSERVATION.

(3)

VECTORS

VECTORS

(4)

VECTORS

VECTORS

►A NUMBER WITH ITS UNITS IS REFERRED A NUMBER WITH ITS UNITS IS REFERRED

TO AS A SCALAR. IT CAN BE POSITIVE,

TO AS A SCALAR. IT CAN BE POSITIVE,

NEGATIVE, OR ZERO. (.5 mi IS SCALAR).

NEGATIVE, OR ZERO. (.5 mi IS SCALAR).

►SOMETIMES, SCALAR ISN’T ENOUGH TO SOMETIMES, SCALAR ISN’T ENOUGH TO

ADEQUATELY DESCRIBE A PHYSICAL

ADEQUATELY DESCRIBE A PHYSICAL

QUANTITY- IN MANY CASES, A DIRECTION

QUANTITY- IN MANY CASES, A DIRECTION

IS NEEDED AS WELL. (.5 mi NORTHWEST).

(5)

VECTORS

VECTORS

► A VECTOR IS A MATHEMATICAL QUANTITY WITH A VECTOR IS A MATHEMATICAL QUANTITY WITH BOTH DIRECTION AND MAGNITUDE. MAGNITUDE

BOTH DIRECTION AND MAGNITUDE. MAGNITUDE

IS THE DISTANCE.

IS THE DISTANCE.

► TO FIND THE COMPONENTS OF A VECTOR WE TO FIND THE COMPONENTS OF A VECTOR WE NEED TO SET UP A COORDINATE SYSTEM. IN

NEED TO SET UP A COORDINATE SYSTEM. IN

TWO DIMENSIONS WE CHOOSE AN ORIGIN “O”

TWO DIMENSIONS WE CHOOSE AN ORIGIN “O”

AND A POSITIVE DIRECTION FOR BOTH THE X

(6)

VECTORS

VECTORS

►LOOK AT THE BOARD FOR A DESCRIPTION LOOK AT THE BOARD FOR A DESCRIPTION

OF A VECTOR AND ITS COMPONENTS AND

OF A VECTOR AND ITS COMPONENTS AND

GIVEN THE COMPONENTS, FINDING ITS

GIVEN THE COMPONENTS, FINDING ITS

MAGNITUDE AND DIRECTION…….

(7)

VECTORS

VECTORS

► An ant leaves its nest at the origin and, after An ant leaves its nest at the origin and, after

foraging for some time, is at a location given by

foraging for some time, is at a location given by

the vector r. this vector has a magnitude r = 1.50

the vector r. this vector has a magnitude r = 1.50

m and points in a direction theta = 25

m and points in a direction theta = 25 oo above the above the

x axis. r can be defined by saying that it extends a

x axis. r can be defined by saying that it extends a

distance r

distance rxx in the x direction and a distance r in the x direction and a distance ryy in in the y direction.

(8)

VECTORS

VECTORS

►LOOK AT THE BOARD FOR A VISUAL LOOK AT THE BOARD FOR A VISUAL

DESCRIPTION OF THE ANT PROBLEM…

(9)

VECTORS

VECTORS

►WE CAN FIND rx AND ry USING TRIG WE CAN FIND rx AND ry USING TRIG

RELATIONS:

RELATIONS:

►rrxx = r cos 25 = r cos 25oo

(10)

VECTORS

VECTORS

►WE CAN SAY THAT THE ANTS FINAL WE CAN SAY THAT THE ANTS FINAL

DISPLACEMENT IS EQUIVALENT TO WHAT

DISPLACEMENT IS EQUIVALENT TO WHAT

IT WOULD BE IF THE ANT HAD SIMPLY

IT WOULD BE IF THE ANT HAD SIMPLY

WALKED 1.36 m IN THE X DIRECTION AND

WALKED 1.36 m IN THE X DIRECTION AND

THEN .634 m IN THE Y DIRECTION.

(11)

VECTORS

VECTORS

►WITH THE COMPONENTS OF r, AS WITH THE COMPONENTS OF r, AS

DETERMINED PREVIOUSLY, USE THEM TO

DETERMINED PREVIOUSLY, USE THEM TO

CALCULATE THE MAGNITUDE r AND THE

CALCULATE THE MAGNITUDE r AND THE

ANGLE THETA. LOOK AT THE NEXT

ANGLE THETA. LOOK AT THE NEXT

DRAWING ON THE BOARD…

(12)

VECTORS

VECTORS

►THE DRAWING SHOWS THAT rx, ry, AND r THE DRAWING SHOWS THAT rx, ry, AND r

FORM A RIGHT TRIANGLE WITH r AS THE

FORM A RIGHT TRIANGLE WITH r AS THE

HYPOTENUSE. WE USE THE PYTHAGOREAN

HYPOTENUSE. WE USE THE PYTHAGOREAN

THEOREM…

THEOREM…

(13)

VECTORS

VECTORS

►SQUARE ROOT OF ((1.36 m)SQUARE ROOT OF ((1.36 m)22 + (.634 m) + (.634 m)22))

►r = 1.50 mr = 1.50 m

►Second, use any two sides of the triangle to Second, use any two sides of the triangle to

obtain angle theta….

(14)

VECTORS

VECTORS

►Theta = sinTheta = sin-1-1 (.634 m / 1.5 m) (.634 m / 1.5 m)

►.634/1.5 = .423 m.634/1.5 = .423 m

►Hit the 2Hit the 2ndnd button then sin on the calculator button then sin on the calculator

to get sin

to get sin-1-1

(15)

VECTORS

VECTORS

►Theta = cosTheta = cos-1-1 (1.36 m / 1.5 m) (1.36 m / 1.5 m)

►coscos-1-1(.907) = 25 degrees(.907) = 25 degrees

►Theta = tanTheta = tan-1-1 (.634 m / 1.36 m) (.634 m / 1.36 m)

(16)

VECTORS

VECTORS

PROBLEMPROBLEM: a man wants to find the height of a : a man wants to find the height of a cliff. He stands with his back to the base of the

cliff. He stands with his back to the base of the

cliff and walks away from it for 5.00 x 10

cliff and walks away from it for 5.00 x 1022 ft. At ft. At

this point he lies on the ground and measures the

this point he lies on the ground and measures the

angle from the horizontal to the top of the cliff. If

angle from the horizontal to the top of the cliff. If

the angle is 34

the angle is 3400, a) how high is the cliff b) what is , a) how high is the cliff b) what is

the straight line distance from the man to the top

the straight line distance from the man to the top

of the cliff? (TRY TO ILLUSTRATE THIS)

of the cliff? (TRY TO ILLUSTRATE THIS)

(17)

VECTORS

VECTORS

►Based on the drawing, the opposite side of Based on the drawing, the opposite side of

the triangle is the height of the cliff (h), the

the triangle is the height of the cliff (h), the

adjacent side is the distance from base of

adjacent side is the distance from base of

the cliff to the man (o) origin, b = 5.00 x

the cliff to the man (o) origin, b = 5.00 x

10

1022 ft, and finally, the hypotenuse is the ft, and finally, the hypotenuse is the

distance (d) from the man to the top of the

distance (d) from the man to the top of the

cliff.

(18)

VECTORS

VECTORS

►PART (A) SOLUTION:PART (A) SOLUTION:

►USE tan theta = h/b to solve for the height USE tan theta = h/b to solve for the height

of the cliff, h.

of the cliff, h.

►h = b tan thetah = b tan theta

►h = (5.00 x 10h = (5.00 x 1022 ft) tan 34 ft) tan 3400

(19)

VECTORS

VECTORS

►TWO WAYS TO SOLVE FOR (B).TWO WAYS TO SOLVE FOR (B).

►d = square root of hd = square root of h22 + b + b22

►d = square root of ((337)d = square root of ((337)22 + (5 x 10 + (5 x 1022))22))

(20)

VECTORS

VECTORS

►Use cos theta = b/dUse cos theta = b/d

►d = b / cos thetad = b / cos theta

►d = 5.00 x 10d = 5.00 x 1022 ft / cos 34 ft / cos 3400

(21)

VECTORS

VECTORS

►HOW TO DETERMINE THE CORRECT SIGNS HOW TO DETERMINE THE CORRECT SIGNS

FOR THE X AND Y COMPONENTS OF A

FOR THE X AND Y COMPONENTS OF A

VECTOR:

VECTOR:

►LOOK AT THE DRAWING ON THE BOARD LOOK AT THE DRAWING ON THE BOARD

AND CONSIDER THE RIGHT TRIANGLE

AND CONSIDER THE RIGHT TRIANGLE

FORMED BY A

(22)

VECTORS

VECTORS

►To determine the sign of ATo determine the sign of Axx, start at the tail , start at the tail

of the vector and move along the x axis

of the vector and move along the x axis

toward the right angle. If you are moving

toward the right angle. If you are moving

toward the positive x direction, then A

toward the positive x direction, then Axx is is

positive (A

positive (Axx is > 0). If you are moving in the is > 0). If you are moving in the negative x direction, then Ax is negative

negative x direction, then Ax is negative

(A

(23)

VECTORS

VECTORS

►For the y component, start at the right For the y component, start at the right

angle and move toward the tip of the arrow,

angle and move toward the tip of the arrow,

Ay is positive or negative depending on

Ay is positive or negative depending on

whether you are moving in the positive or

whether you are moving in the positive or

negative y direction.

negative y direction.

►Look at the following drawings on the board Look at the following drawings on the board

to assign the correct signs….

(24)

VECTORS

VECTORS

►Be careful when using your calculator to Be careful when using your calculator to

determine the direction angle, theta,

determine the direction angle, theta,

because you may need to add 180

because you may need to add 18000 to get to get

the correct angle, as measured

the correct angle, as measured

counterclockwise from the positive x axis.

counterclockwise from the positive x axis.

For example, if A

For example, if Axx = -.50 m and A = -.50 m and Ayy = 1 m, = 1 m, your calculator will give you the following

your calculator will give you the following

result:

(25)

VECTORS

VECTORS

►Theta = tanTheta = tan-1-1 (1 m / -.50 m) (1 m / -.50 m)

►= tan= tan-1-1 (-2) (-2)

(26)

VECTORS

VECTORS

►Does this angle correspond to the specified Does this angle correspond to the specified

vector? The way to check is to sketch vector

vector? The way to check is to sketch vector

“A” The drawing is similar to what I’m about

“A” The drawing is similar to what I’m about

to put on the board….

to put on the board….

►The direction angle of vector “A” should be The direction angle of vector “A” should be

between 90

between 9000 and 180 and 18000. to obtain the correct . to obtain the correct

angle, add 180

angle, add 18000 to the calculators result. to the calculators result.

(shown on next slide)

(27)

VECTORS

VECTORS

►Theta = -63Theta = -6300 + 180 + 18000 = 117 = 11700

(28)

VECTORS

VECTORS

►PROBLEMPROBLEM: The vector “B” has components : The vector “B” has components

B

Bxx = -2.10 m and B = -2.10 m and Byy = -1.70 m Find the = -1.70 m Find the direction angle, theta, for this vector.

direction angle, theta, for this vector.

►tantan-1-1 ((-1.70 m) / (-2.10 m)) = 39 ((-1.70 m) / (-2.10 m)) = 3900

(29)

VECTORS

VECTORS

►In many situations the direction of a vector In many situations the direction of a vector

“A” is given by the angle theta, measured

“A” is given by the angle theta, measured

relative to the x axis. Look at the drawing

relative to the x axis. Look at the drawing

on the board. In these cases know that:

on the board. In these cases know that:

(30)

VECTORS

VECTORS

►On the other hand, sometimes we are given On the other hand, sometimes we are given

the angle between the vector and the y

the angle between the vector and the y

axis. Look at the drawing on the board. If

axis. Look at the drawing on the board. If

we call this angle theta’ (look on board),

we call this angle theta’ (look on board),

then it follows that:

then it follows that:

(31)

VECTORS

VECTORS

►ProblemProblem: If a vectors direction angle relative : If a vectors direction angle relative

to the x axis is 35

to the x axis is 3500, then its direction angle , then its direction angle

relative to the y axis is 55

relative to the y axis is 5500 (35 + 55 =90). (35 + 55 =90).

Find the components of a vector “A” of

Find the components of a vector “A” of

magnitude 5.2 m in terms of:

magnitude 5.2 m in terms of:

(32)

VECTORS

VECTORS

►A) A)

►AAxx = (5.2 m) cos 35 = (5.2 m) cos 3500 = 4.3 m = 4.3 m ►AAyy = (5.2 m) sin 35 = (5.2 m) sin 3500 = 3.0 m = 3.0 m

►B)B)

(33)

VECTORS

VECTORS

►ADDING VECTORS:ADDING VECTORS:

►You find a treasure map. To find the You find a treasure map. To find the

treasure, the map says to go to the tree in

treasure, the map says to go to the tree in

the backyard, march 5 paces north, then 3

the backyard, march 5 paces north, then 3

paces east. If these two displacements are

paces east. If these two displacements are

represented by vectors “A” and “B” (look at

represented by vectors “A” and “B” (look at

drawing on board), the total displacement

(34)

VECTORS

VECTORS

►Vector “C” is the vector sum of “A” and “B” Vector “C” is the vector sum of “A” and “B”

that is, vector “C” = vector “A” + vector “B”

that is, vector “C” = vector “A” + vector “B”

►To add vectors “A” and “B”, place the tail of To add vectors “A” and “B”, place the tail of

vector “B” at the head of vector “A”

(35)

VECTORS

VECTORS

►What if the instructions to find the map What if the instructions to find the map

were more complicated? 5 paces N, 3 paces

were more complicated? 5 paces N, 3 paces

E, then 4 paces SE. (look at drawing on

E, then 4 paces SE. (look at drawing on

board)

board)

(36)

VECTORS

VECTORS

►PROBLEMPROBLEM: Vector “A” has a magnitude of : Vector “A” has a magnitude of

5 m and a direction angle of 60

5 m and a direction angle of 6000 above the x above the x

axis. Vector “B” has a magnitude of 4 m and

axis. Vector “B” has a magnitude of 4 m and

a direction angle of 20

a direction angle of 2000 above the x axis. above the x axis.

These two vectors and their sum, vector “C”

These two vectors and their sum, vector “C”

are shown in my drawing (look on board)

are shown in my drawing (look on board)

►What is the length and direction angle of What is the length and direction angle of

vector “C”?

(37)

VECTORS

VECTORS

►EXACT RESULTS CAN BE OBTAINED BY EXACT RESULTS CAN BE OBTAINED BY

ADDING VECTOR “A” AND VECTOR “B” IN

ADDING VECTOR “A” AND VECTOR “B” IN

TERMS OF THEIR COMPONENTS. LOOK AT

TERMS OF THEIR COMPONENTS. LOOK AT

MY NEXT TWO DRAWINGS SHOWING THE

MY NEXT TWO DRAWINGS SHOWING THE

COMPONENTS OF “A” AND “B” AND THE

COMPONENTS OF “A” AND “B” AND THE

COMPONENTS OF “C”

(38)

VECTORS

VECTORS

►First, to add vectors, you add the First, to add vectors, you add the

components:

components:

►AAxx = (5 m) cos 60 = (5 m) cos 6000 = 2.50 m = 2.50 m ►AAyy = (5 m) sin 60 = (5 m) sin 6000 = 4.33 m = 4.33 m

►BBxx = (4 m) cos 20 = (4 m) cos 2000 = 3.76 m = 3.76 m

(39)

VECTORS

VECTORS

►NextNext, adding component by component , adding component by component

yields the components vector “C” = vector

yields the components vector “C” = vector

“A” + vector “B”.

“A” + vector “B”.

►CCxx = A = Axx + B + Bxx = 2.5 m + 3.76 m = 6.26 m = 2.5 m + 3.76 m = 6.26 m

(40)

VECTORS

VECTORS

►FinallyFinally, now we can find precise values for , now we can find precise values for

“C”, the magnitude of vector “C”, and its

“C”, the magnitude of vector “C”, and its

direction angle theta.

direction angle theta.

►C = the square root of CC = the square root of Cxx22 + C + Cyy22

(41)

VECTORS

VECTORS

►Direction angle theta:Direction angle theta: ►theta = tantheta = tan-1-1 (Cy / Cx) (Cy / Cx)

►TanTan-1-1 ((5.70 m / 6.26 m)) ((5.70 m / 6.26 m))

(42)

VECTORS

VECTORS

►Subtracting Vectors: We want to determine Subtracting Vectors: We want to determine

the vector “D” where vector D equals

the vector “D” where vector D equals

vector “A” – vector “B”. (look on board)

vector “A” – vector “B”. (look on board)

►To find vector “D”, start by rewriting it as To find vector “D”, start by rewriting it as

follows:

follows:

(43)

VECTORS

VECTORS

►vector “D” is the sum of vector “A” and vector “D” is the sum of vector “A” and

vector “- B”. Now the negative of a vector

vector “- B”. Now the negative of a vector

has a simple graphical interpretation. –”B”

has a simple graphical interpretation. –”B”

reverses its direction.. (Look on the board)

reverses its direction.. (Look on the board)

►To subtract vector “B” from vector “A”, To subtract vector “B” from vector “A”,

reverse the direction of “B” and add it to

reverse the direction of “B” and add it to

“A”. (look on board)

(44)

VECTORS

VECTORS

►vector “D” = vector “A” – vector “B”vector “D” = vector “A” – vector “B”

►DDxx = A = Axx – B – Bxx

►DDyy = A = Ayy – B – Byy

►Once the components of vector “D” are Once the components of vector “D” are

found, its magnitude and direction angle can

(45)

VECTORS

VECTORS

►Ax = 5m cos 60 = 2.50 mAx = 5m cos 60 = 2.50 m ►Bx = 4m cos 20 = 3.76 mBx = 4m cos 20 = 3.76 m

(46)

VECTORS

VECTORS

►Ay = 5m sin 60 = 4.33 mAy = 5m sin 60 = 4.33 m ►By = 4m sin 20 = 1.37 mBy = 4m sin 20 = 1.37 m

(47)

VECTORS

VECTORS

►TO FIND D (MAGNITUDE)TO FIND D (MAGNITUDE)

►The square root of ((-1.26)The square root of ((-1.26)22 + (2.96) + (2.96)22))

(48)

VECTORS

VECTORS

►TO FIND THETA:TO FIND THETA:

►TanTan-1-1(D(Dyy / D / Dxx))

►TanTan-1-1(2.96 / -1.26)(2.96 / -1.26)

►= -66.9 degrees + 180 degrees= -66.9 degrees + 180 degrees

= 113 degrees

(49)

VECTORS

VECTORS

►Unit vectors (look on board to see how they Unit vectors (look on board to see how they

are written) are defined to be dimensionless

are written) are defined to be dimensionless

vectors of unit magnitude pointing in the

vectors of unit magnitude pointing in the

positive x and y directions.

positive x and y directions.

►The x unit vector ^x is a dimensionless The x unit vector ^x is a dimensionless

vector of unit length pointing in the positive

vector of unit length pointing in the positive

x direction.

(50)

VECTORS

VECTORS

►The y unit vector ^y is a dimensionless The y unit vector ^y is a dimensionless

vector of unit length pointing in the positive

vector of unit length pointing in the positive

y direction.

y direction.

►Look at the diagram on the board. It shows Look at the diagram on the board. It shows

^x and ^y on a two dimensional coordinate

^x and ^y on a two dimensional coordinate

system.

system.

►Since unit vectors have no physical Since unit vectors have no physical

dimensions like mass, length, or time – they

(51)

VECTORS

VECTORS

►MULTIPLYING UNIT VECTORS BY A MULTIPLYING UNIT VECTORS BY A

SCALAR: (REMEMBER)

SCALAR: (REMEMBER)

►SCALAR: A QUANTITY SPECIFIED BY SCALAR: A QUANTITY SPECIFIED BY

NUMERICAL VALUE ONLY…

NUMERICAL VALUE ONLY…

►VECTOR: QUANTITIES THAT REQUIRE VECTOR: QUANTITIES THAT REQUIRE

BOTH A NUMERICAL VALUE AND A

(52)

VECTORS

VECTORS

►Multiplying a vector by 3 increases its Multiplying a vector by 3 increases its

magnitude by a factor of 3, but does not

magnitude by a factor of 3, but does not

change its direction. Multiplying by -3

change its direction. Multiplying by -3

increases the magnitude by a factor of 3

increases the magnitude by a factor of 3

and reverses the direction of the vector.

and reverses the direction of the vector.

Look at the diagram on the board….next

Look at the diagram on the board….next

slide for explanation…..

(53)

VECTORS

VECTORS

►Multiplying a vector by a positive scalar Multiplying a vector by a positive scalar

different from 1 will change the length of

different from 1 will change the length of

the vector but leave its direction the same.

the vector but leave its direction the same.

If the vector is multiplied by a negative

If the vector is multiplied by a negative

scalar its direction is reversed…

(54)

VECTORS

VECTORS

►In the case of unit vectors (^x and ^y) – In the case of unit vectors (^x and ^y) –

which have a magnitude of 1 and are

which have a magnitude of 1 and are

dimensionless - multiplication by a scalar

dimensionless - multiplication by a scalar

results in a vector with the same magnitude

results in a vector with the same magnitude

and dimensions as the scalar…example on

and dimensions as the scalar…example on

next slide…..

(55)

VECTORS

VECTORS

►If a vector “A” has the scalar components If a vector “A” has the scalar components

Ax = 5 m and Ay = 3 m, it can be written as

Ax = 5 m and Ay = 3 m, it can be written as

vector A = (5m)^x + (3m)^y

vector A = (5m)^x + (3m)^y

(56)

VECTORS

VECTORS

►We refer to the quantities (5m)^x and We refer to the quantities (5m)^x and

(3m)^y as the x and y

(3m)^y as the x and y vector vector components

(57)

VECTORS

VECTORS

►In general, an arbitrary two-dimensional In general, an arbitrary two-dimensional

vector “A” can always be written as the sum

vector “A” can always be written as the sum

of a vector component in the x direction and

of a vector component in the x direction and

a vector component in the y direction.

a vector component in the y direction.

(58)

VECTORS

VECTORS

►This represents an equivalent way of This represents an equivalent way of

representing the vector components of a

representing the vector components of a

vector (look on board)….

vector (look on board)….

►In this case we see that the vector In this case we see that the vector

components are the projection of a vector

components are the projection of a vector

onto the x and y axes. The sign of the

onto the x and y axes. The sign of the

vector components is positive if they point

vector components is positive if they point

in the positive x or y direction, and negative

in the positive x or y direction, and negative

if they point in the negative x or y direction.

(59)

VECTORS

VECTORS

►Note that vector addition and subtraction is Note that vector addition and subtraction is

straightforward with unit vector notation:

straightforward with unit vector notation:

►See the board….See the board….

(60)

VECTORS

VECTORS

► In intro chapter (one-dimensional kinematics), we In intro chapter (one-dimensional kinematics), we talked about four different one-dimensional

talked about four different one-dimensional

vectors: position, displacement, velocity, and

vectors: position, displacement, velocity, and

acceleration. Each of these quantities had a

acceleration. Each of these quantities had a

direction associated with it, indicated by its sign;

direction associated with it, indicated by its sign;

positive meant in the positive direction, negative

positive meant in the positive direction, negative

meant in the negative direction. Now we consider

meant in the negative direction. Now we consider

these vectors again, this time in two dimensions,

these vectors again, this time in two dimensions,

where the possibilities for direction are not so

(61)

VECTORS

VECTORS

►POSITION VECTORS:POSITION VECTORS:

►Look on the board: a two dimensional Look on the board: a two dimensional

coordinate system…

coordinate system…

►Position is indicated by a vector from the Position is indicated by a vector from the

origin to the location in question. We refer

origin to the location in question. We refer

to the position vector as “r” as indicated in

(62)

VECTORS

VECTORS

►The position vector “r” points from the The position vector “r” points from the

origin to the current location of an object.

origin to the current location of an object.

The x and y vector components of vector “r”

The x and y vector components of vector “r”

are x^x and y^y respectively.

are x^x and y^y respectively.

►Position vector “r”Position vector “r” ►SI unit meter, mSI unit meter, m

(63)

VECTORS

VECTORS

►DISPLACEMENT VECTOR:DISPLACEMENT VECTOR: ►Look on board:Look on board:

►Initially you are at a location indicated by Initially you are at a location indicated by

the position vector r

the position vector rii, later at the final , later at the final

position vector r

position vector rff. Your displacement vector . Your displacement vector ^r, is the change in position.

(64)

VECTORS

VECTORS

►Displacement vector ^rDisplacement vector ^r

►Look on board….Look on board…. ►^r = r^r = rff – r – rii

►SI unit meterSI unit meter

(65)

VECTORS

VECTORS

►LOOK AT THE DIAGRAM ON BOARD..LOOK AT THE DIAGRAM ON BOARD.. ►The final position is equal to the initial The final position is equal to the initial

position plus the change in position. ^r

position plus the change in position. ^r

extends from the head of r

extends from the head of rii to the head of to the head of

r

(66)

VECTORS

VECTORS

►Average velocity vector Average velocity vector is defined as the is defined as the

displacement vector ^r divided by the

displacement vector ^r divided by the

elapsed time ^t. (look on board….)

elapsed time ^t. (look on board….)

(67)

VECTORS

VECTORS

►PROBLEM: A dragonfly is observed initially PROBLEM: A dragonfly is observed initially

at the position r

at the position rii = (2 m)^x + (3.5 m)^y. = (2 m)^x + (3.5 m)^y. Three seconds later it is at the position

Three seconds later it is at the position

r

rff = (-3 m)^x + (5.5 m)^y. what was the = (-3 m)^x + (5.5 m)^y. what was the

dragonfly’s average velocity during this

dragonfly’s average velocity during this

time?

(68)

VECTORS

VECTORS

►To help visualize average velocity vector vTo help visualize average velocity vector vavav, ,

look at the drawing on the board. The next

look at the drawing on the board. The next

slide will give an explanation….

(69)

VECTORS

VECTORS

►A particle moving in two dimensions along A particle moving in two dimensions along

the path shown. If the particle is at point P1

the path shown. If the particle is at point P1

at time t1, and at P2 at time t2, its

at time t1, and at P2 at time t2, its

displacement is indicated by the vector ^r.

displacement is indicated by the vector ^r.

The average velocity is parallel to ^r. On

The average velocity is parallel to ^r. On

average, you have moved in the direction of

average, you have moved in the direction of

^r during the time from t1 to t2.

(70)

VECTORS

VECTORS

►By considering smaller and smaller time By considering smaller and smaller time

intervals, it is possible to calculate the

intervals, it is possible to calculate the

instantaneous velocity vector…

instantaneous velocity vector…

►Look on the board for its definition and Look on the board for its definition and

diagram…the diagram is explained on the

diagram…the diagram is explained on the

following slides…

(71)

VECTORS

VECTORS

►The diagram shows that the instantaneous The diagram shows that the instantaneous

velocity at a given time is tangential to the

velocity at a given time is tangential to the

path of the particle at that time. In addition,

path of the particle at that time. In addition,

the magnitude of the velocity vector is the

the magnitude of the velocity vector is the

speed of the particle. Thus, the

speed of the particle. Thus, the

instantaneous velocity vector tells you both

instantaneous velocity vector tells you both

how fast a particle is moving and in what

(72)

VECTORS

VECTORS

►The instantaneous velocity vector “v” is The instantaneous velocity vector “v” is

obtained by calculating the average velocity

obtained by calculating the average velocity

vector over smaller and smaller time

vector over smaller and smaller time

intervals. In the limit of vanishingly small

intervals. In the limit of vanishingly small

time intervals, the average velocities

time intervals, the average velocities

approach the instantaneous velocity. Notice

approach the instantaneous velocity. Notice

that the instantaneous velocity vector points

that the instantaneous velocity vector points

in the direction of motion at any given time.

(73)

VECTORS

VECTORS

►PROBLEM: Find the speed and direction of PROBLEM: Find the speed and direction of

motion for a rainbow trout whose velocity is

motion for a rainbow trout whose velocity is

“v” = (3.7 m/s)^x + (-1.3 m/s)^y.

“v” = (3.7 m/s)^x + (-1.3 m/s)^y.

►How have we found magnitude and How have we found magnitude and

direction??? Look on the board…..

(74)

VECTORS

VECTORS

►ACCELERATION VECTORS:ACCELERATION VECTORS:

►The average acceleration vector is defined The average acceleration vector is defined

as the change in the velocity vector divided

as the change in the velocity vector divided

by the scalar change in time. Look on the

by the scalar change in time. Look on the

board for the formula….

(75)

VECTORS

VECTORS

►An example is on the board…where we An example is on the board…where we

show the initial and final velocity vectors

show the initial and final velocity vectors

corresponding to two different times.

corresponding to two different times.

►Also see on the board, how the change in Also see on the board, how the change in

velocity is defined…..

(76)

VECTORS

VECTORS

►The next couple of slides will explain the The next couple of slides will explain the

following drawing on the board..

following drawing on the board..

►The average acceleration vector aThe average acceleration vector aavav = ^v/^t = ^v/^t

points in the direction of the change in

points in the direction of the change in

velocity vector ^v. we obtain ^v by moving

velocity vector ^v. we obtain ^v by moving

v

vff so that its tail coincides with the tail of v so that its tail coincides with the tail of vii, ,

and then drawing the arrow that connects

(77)

VECTORS

VECTORS

►You can see in the diagram that aYou can see in the diagram that aavav is the is the

direction of ^v. a

direction of ^v. aavav need not point in the need not point in the direction of motion, and in general it

direction of motion, and in general it

doesn’t.

doesn’t.

(78)

VECTORS

VECTORS

►ProblemProblem: The initial velocity of a car is vi = : The initial velocity of a car is vi =

(12 m/s)^x and that 10 seconds later its

(12 m/s)^x and that 10 seconds later its

final velocity is vf = (-12 m/s)^y. Note that

final velocity is vf = (-12 m/s)^y. Note that

the speed is 12 m/s in each case, but the

the speed is 12 m/s in each case, but the

velocity is different because the direction

velocity is different because the direction

has changed.

has changed.

(79)

VECTORS

VECTORS

►Just as we did with velocity, we go small Just as we did with velocity, we go small

time intervals, ^t --->0, we can define

time intervals, ^t --->0, we can define

instantaneous acceleration.

instantaneous acceleration.

References

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