INTRO TO PHYSICS
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.
VECTORS
VECTORS
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).
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
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…….
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.
VECTORS
VECTORS
►LOOK AT THE BOARD FOR A VISUAL LOOK AT THE BOARD FOR A VISUAL
DESCRIPTION OF THE ANT PROBLEM…
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
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.
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…
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…
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….
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
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)
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)
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.
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
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))
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
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
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
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….
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:
VECTORS
VECTORS
►Theta = tanTheta = tan-1-1 (1 m / -.50 m) (1 m / -.50 m)
►= tan= tan-1-1 (-2) (-2)
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)
VECTORS
VECTORS
►Theta = -63Theta = -6300 + 180 + 18000 = 117 = 11700
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
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:
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:
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:
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)
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
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”
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)
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”?
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”
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
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
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
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))
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:
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)
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
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
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
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))
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
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.
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
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
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…..
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…
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…..
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
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
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.
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.
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….
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
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
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
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.
VECTORS
VECTORS
►Displacement vector ^rDisplacement vector ^r
►Look on board….Look on board…. ►^r = r^r = rff – r – rii
►SI unit meterSI unit meter
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
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….)
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?
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….
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.
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…
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
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.
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…..
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….
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…..
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
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.
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.
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.