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ball1b2 VECTORS1.VECTORS2<VECTORS3JVECTORS4XVECTORS5dACCEL1pACCEL2|ACCEL3HARMONY1HARMONY2HARMONY3CRAZYPCRAZYVCRAZYABIGBALLBIGBALL1BIGBALL2BIGBALL3BIGBALL4L\RESOLV1\
\L\RESOLV3\
\L\RESOLVED\
\L\INCLTEST\
\L\FINCLIN1\
\L\V5OLD\
35
\L\BIGBALL4\
\L\BIGBALL5\
\L\ADDV1\
\L\ADDV2\
\L\ADDV3\
\L\ADDV4\
\L\ADDV5\
\L\ADDV6\
\L\ADDV7\
\L\ADDV7B\
\L\ADDV8\
\L\ADDV9\
\L\ADDV10\
\L\ADDV10B\
\L\ADDV11\
\L\ADDV12\
\L\ADDV12B\
\L\ADDV12C\
\L\ADDV12D\
\L\FRICTN1\
\L\FRICTN2\
\\L\ball1\
\L\b2\
\L\VECTORS1\
\L\VECTORS2\
\L\VECTORS3\
\L\VECTORS4\
\L\VECTORS5\
\L\ACCEL1\
\L\ACCEL2\
\L\ACCEL3\
\L\HARMONY1\
\L\HARMONY2\
\L\HARMONY3\
\L\CRAZYP\
\L\CRAZYV\
\L\CRAZYA\
\L\BIGBALL\
\L\BIGBALL1\
\L\BIGBALL2\
\L\BIGBALL3\ts learn about the laws of physics.\
\DCW\
\SWT\\\6\\
\DTNNW\ When an object moves, it moves from
one place to another, over some period
of time.\
\SA\
\DFA\BALL1\8\8\
\SAS\50\1\1\-1\
\SAL\26\10\0\0\0\
\DA\
\DT\\
\DTN\ When an object moEW\
\BL\FORCE\
\L\VELOCITY LENGTH\
\SCH\2\2\
\DTCNW\Length and Time\
\SCH\1\1\
\DTN\ The central theme of all physics is
the movement of matter.
Without movement, there would be no
study of physics.\
\DTN\ By studying motion, physicis Time\
\SMO\Speed\
\SMO\Velocity and Vectors\
\SMO\Vocabulary Review\
\SMO\Review Questions\
\SMO\Return to Main Menu\
\DM\
\BMC\1\VELOCITY LENGTH\
\BMC\2\VELOCITY SPEED\
\BMC\3\VELOCITY VELOCITY\
\BMC\4\VELOCITY VOCAB\
\BMC\5\VELOCITY REVIMore on This Topic\
\SMO\Leave Force & Motion\
\DM\
\BMC\1\VELOCITY\
\BMC\2\ACCELERATION\
\BMC\3\FRICTION\
\BMC\4\EXPERIMENT\
\BMC\5\VOCAB\
\BMC\6\REVIEW\
\BMC\7\MORE\
\BS\NETWORK\
\L\VELOCITY\
\SM\
\SMT\Velocity & Vectors\
\SMO\Length andINCLTESTBFINCLIN1MV5OLDScientists have arbitrarily defined
the meter as the unit of length and
distance.
If you stretch your arm out to the
side and face forward, the distance
from the tip of your nose to the tip of your fingers is probably about one
meter.\
\DTN\3@``0UUU000```@
@```000TPP@@6CJ
*
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*
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*|`@@xCuU?~p3MMl`x`|azj* x@>p?}uU|`@@``pp88?~z*}UU)~,??*%qS`}u@~><`x@9yY~FF~p|`
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//{k+}uU
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\DTN\
Distance and time are basic physical quantities.
They are basic characteristics of
the universe, not the creations of
people.
The measurement of these quantities, however, requires some arbitrary units
of measurement.\
\DTN\ ves, you can ask
how far did it move?\
\DTN\ In physics, you measure this
distance in meters.\
\DCW\
\DTN\ You can also ask how long did it
take for the object to move?\
\DTN\ In physics, you measure this time in seconds.\
\DCWucations, Inc.\
\XDEL\90\
\L\FORCE\
\SM\
\SMT\1: FORCE & MOTION\
\SMO\Velocity and Vectors\
\SMO\Acceleration and Force\
\SMO\Friction and Vectors\
\SMO\Experiment with Falling Bodies\
\SMO\Vocabulary Review\
\SMO\Review Questions\
\SMO\Mon Feb 4 13:03:34 1991
\DCW\
\SCH\2\\
\SCT\0\4\
\DTCNW\FORCE\
\DTCNW\ & MOTION \
\SCH\1\\
\SCT\6\11\
\DTCNW\Robert Stickgold, Ph.D.\
\SCT\8\16\
\DTCNW\Copyright (c) 1990\
\SCT\8\17\
\DTCNW\Queue, Inc.\
\SCT\4\18\
\DTCNW\Intentional Ed Similarly, they have defined the
second as the standard unit of time.
If you listen to your heartbeat, the time interval between beats is about
one second.\
\DTN\ The unit meters is usually
abbreviated m.
The unit seconds is usually
abbreviated s.\
\DCW\
\DTNNW\ On your monitor, this ball is moving about two-tenths of a meter, or 0.2 m.\
\SFF\0\\
\SA\
\DFA\BALL1\268\8\
\SAS\50\1\1\-1\
\SAL\26\-10\0\0\0\
\DA\
\DT\\
\DTN\ The dot on the letter i is about
t\0\0\
\DA\
\GCR\
\XERASE\248\5\267\20\
\DCW\
\DTNNW\ Now the ball is moving twice as fast as it was before.\
\XDEL\3\
\SA\
\DFA\BALL1\8\8\
\SAS\50\1\1\-1\
\SAL\13\20\0\0\0\
\DA\
\GCR\
\SFF\0\\
\SA\
\DFA\BALL1\272\8\
\SAS\50\1\1\-1\
\SALs\
\DTNW\
\
\DT\ 2 s\
\SCH\1\1\
\DCW\
\DTNNW\ It takes the ball the same amount of time to move in each direction.
This means that the ball's speed is
the same in each case.\
\SFF\0\\
\SA\
\DFA\BALL1\8\8\
\SAS\50\1\1\-1\
\SAL\26\10\0\DA\
\DTNNW\ To get the ball's speed, we divided
the distance it moved by the time it
took:\
\SCH\2\3\
\DTNNW\ 0.2 m\
\DTNW\
------ = 0.1 m/s\
\DT\
2 s\
\SCH\1\1\
\DCW\
\DTNNW\Or:\
\SCH\2\3\
\DTNNW\ 200 mm
\
\DTNW\------- = 100 mm/how fast the ball is moving.
In this case, the speed is measured
in meters per second.\
\DCW\
\DTNNW\ To determine a speed, you must
divide a distance by a time.\
\SFF\0\\
\SA\
\DFA\BALL1\268\8\
\SAS\50\1\1\-1\
\SAL\26\-10\0\0\0\
fast does an object move?\
\DCW\
\SWT\\\6\\
\DTNNW\ Assume that this ball moves about
two-tenths of a meter in about 2
seconds.\
\SA\
\DFA\BALL1\8\8\
\SAS\50\1\1\-1\
\SAL\26\10\0\0\0\
\DA\
\DTN\ The speed of the ball is a measure
of and "kilo" means one
thousand.\
\SQANS\1000\
\SQLAST\No, there 1000 meters in a kilometer.\
\SQA\T\=\1000\Right, "kilometer" means one thousand meters.\
\DQ\
\SQSEND\
\BESC\
\L\VELOCITY SPEED\
\SCH\2\2\
\DTCNW\Speed\
\SCH\1\1\
\DTN\ How ond.\
\SQANS\1000\
\SQLAST\No, there are 1000 ms in a second.\
\SQA\T\=\1000\Right. "milli" means one-thousandth.\
\DQ\
\BQESC\
\DCW\
\SQ\1\Y\
\SQM\2\
\SQF\S\4\
\SQT\1 km is equal to ____ m?\
\SQH\Remember that "milli" means
one-thousandtheter.\
\SQANS\1000\
\SQLAST\No, there are 1000 mm in a meter.\
\SQA\T\=\1000\Right. "milli" means one-thousandth.\
\DQ\
\BQESC\
\DCW\
\SQ\1\Y\
\SQM\2\
\SQF\S\4\
\SQT\1 s is equal to ____ ms?\
\SQH\Remember, there are a thousand ms in
a sec\SQANS\200\
\SQLAST\No, there are 200 ms in 0.2 s.
(0.2 x 1000 = 200 ms).\
\SQA\T\=\200\Right. (0.2 x 1000 = 200 ms).\
\DQ\
\BQESC\
\DCW\
\SQ\1\Y\
\SQM\2\
\SQF\S\4\
\SQT\1 m is equal to ____ mm?\
\SQH\Remember, there are a thousand mm in
a m\
\SQA\T\=\200\Right. (0.2 x 1000 = 200 mm).\
\DQ\
\BQESC\
\DCW\
\SQ\1\Y\
\SQF\S\4\
\SQM\3\
\SQT\0.2 s is equal to ____ ms.\
\SQH\Remember, there are 1000 ms in a
second.\
\SQH\If there are 1000 ms in a second,
there are 100 ms in 0.1 s.\
Q\1\Y\
\SQF\S\4\
\SQM\3\
\SQT\0.2 m is equal to ____ mm.\
\SQH\Remember, there are 1000 mm in a
meter.\
\SQH\If there are 1000 mm in a meter,
there are 100 mm in 0.1 m.\
\SQANS\200\
\SQLAST\No, there are 200 mm in 0.2 m.
(0.2 x 1000 = 200 mm). Assume that this ball moves about
two-tenths of a meter in about 2
seconds.\
\SFF\0\\
\SA\
\DFA\BALL1\8\8\
\SAS\50\1\1\-1\
\SAL\26\10\0\0\0\
\DA\
\SFF\0\\
\SA\
\DFA\BALL1\268\8\
\SAS\50\1\1\-1\
\SAL\26\-10\0\0\0\
\DA\
\DCW\
\SQS\1\Y\
\Ss = 0.001 s\
\DTN\ Similarly, when working with large
quantities of meters or seconds,
scientists add the prefix kilo- to the
unit to indicate thousands of units.
1 kilometer = 1 km = 1,000 m
1 kilosecond = 1 ks = 1,000 s\
\DCW\
\DTNNW\wo-thousandths of a meter, or 0.002 m.\
\DTN\ When working with small fractions of a meter or a second, scientists add the prefix milli- to the unit to indicate
one-thousandths of the unit.
1 millimeter = 1 mm = 0.001 m
1 millisecond = 1 m\13\-20\0\0\0\
\DA\
\SQS\1\Y\
\DCW\
\SQ\1\Y\
\SQM\3\
\SQF\S\1\
\SQT\If the ball moved 200 mm in 2 s and
is now going twice as fast, how long
does it take the ball to move the 200 mm?
a. 0.5 s c. 2 s e. 20 s
b. 1.0 s d. 4 s f. 200 s\
\SQH\The ball was initially going 100
mm/s. Think about how fast it is
going now.\
\SQH\If it's moving twice as fast, it
should take only half as long.\
\SQANS\b\
\SQLAST\No, since it's moving twice as fast,
it will take only half aselocity mean the same thing.
But physicists use these terms in
different ways.\
\DCW\
\SWT\\\9\\
\DTNNW\ A ball moves across the screen in
each direction at the same speed of 200 mm/s.\
\SA\
\DFA\BALL1\8\5\
\SAS\50\1\1\-1\
\SAL\13\20\0\0\0 taken.\
\XPLOT\43\91\
\XDRAW\200\-82\
\DTN\ This block moved a total of 8 m in 4 s, for an average speed of 2 m/s.\
\BESC\
\L\VELOCITY VELOCITY\
\SCH\2\2\
\DTCNW\Velocity & Vectors\
\SCH\1\1\
\DTN\ In common usage, the words speed and v a
speed of 1 m/s.\
\SCH\\3\
\XPLOT\113\84\
\XDRAW\81\-33\
\SCH\\1\
\DTN\ At 2 sec, the block is moving at a
speed of 2 m/s.\
\DCW\
\DTN\ The average speed of the block is
obtained by dividing the total distance traveled by the total timeck at any given moment is the slope of the curve at that point.\
\XPLOT\56\94\
\XDRAW\83\-17\
\DTN\ You can measure this slope by
drawing a line tangent to the curve at
that point and measuring its slope.\
\DTN\ At 1 s, the block is moving attance traveled by
a block of wood sliding down an
inclined plane.\
\DTN\ As it slides down, its speed
increases.
This is seen in the increasing slope of the curve as it moves to the right.\
\DCW\
\DTN\ The instantaneous speed of the bloVT\meters\
\SGYMAX\8\
\SGXMAX\4\
\SGVL\2\0\2\
\SLGHL\0\1\
\SLGLX\0\.25\.50\.75\1\1.25\1.50\1.75\2\2.25\2.50\2.75\3\3.25\3.50\3.75\4\
\SLGLY\0\0\.1\.3\.5\.8\1.1\1.5\2\2.5\3.1\3.8\4.5\5.3\6.1\7.0\8.0\
\DLG\
\SWT\M\M\15\M\
\DTN\ This shows the disow far the ball
moved at each moment during its travel
at the faster speed.
Its speed was twice as great and the slope of the line is therefore twice as great.\
\SWS\1\
\DCW\
\SWT\2\35\M\13\
\SLG\1\
\SLGL\1\2\\0\
\SGHT\seconds\
\SGDF\2\1\
\SG\M\M\M\10\
\SLG\2\
\SLGL\1\2\\0\
\SLGL\2\1\\0\
\SGHT\sec\
\SGDF\1\1\
\SGVT\mm\
\SGYMAX\200\
\SGXMAX\3\
\SGVL\3\0\100\
\SLGHL\0\1\
\SLGLX\0\2\
\SLGLY\0\200\
\SLGLX\0\1\
\SLGLY\0\200\
\DLG\
\SWT\M\M\12\M\
\DCW\
\DTN\ The new line shows h200\
\DLG\
\SWT\\\12\M\
\DTN\ This graph shows how far the ball
has moved at each moment in its travel
at the slower speed.
The ball moved 200 mm in 2 s.\
\DTN\ The slope of the line gives the
speed of the ball.\
\SWS\1\
\DCW\
\SWT00 mm)/(200 mm/s) = 5 s.\
\SQA\T\=\c\Right! (1000 mm)/(200 mm) = 5\
\DQ\
\SQSEND\
\SWS\1\
\DCW\
\SWT\M\M\M\10\
\SLG\1\
\SLGL\1\2\\0\
\SGHT\sec\
\SGDF\1\1\
\SGVT\mm\
\SGYMAX\200\
\SGXMAX\3\
\SGVL\3\0\100\
\SLGHL\0\1\
\SLGLX\0\2\
\SLGLY\0\s c. 5 s e. 20 s
b. 2 s d. 10 s f. 50 s\
\SQH\Remember that 1 meter = 1000 mm.\
\SQH\Think - if it goes 200 mm/s, then to
determine how many seconds it needs
to go 1000 mm, you must divide
1000/200.\
\SQANS\c\
\SQLAST\No, (10m/s, twice
the original speed of 100 mm/s.\
\SQA\T\=\d\Right! 2 x 100 mm/s = 200 mm/s\
\SQA\F\=\b\No, it's going faster now.\
\DQ\
\BQESC\
\DCW\
\SQ\1\Y\
\SQF\S\1\
\SQT\At 200 mm/s, how long would it take
the ball to move 1 meter?
a. 1 and
now is moving twice as fast, what is
the new speed of the ball?
a. 10 mm/s c. 100 mm/s e. 400 mm/s
b. 50 mm/s d. 200 mm/s f. 800 mm/s\
\SQH\The old speed was 200 mm per 2
seconds, or 100 mm/s.\
\SQANS\d\
\SQLAST\No, the new speed is 200 m long, or 1
s.\
\SQA\T\=\b\Right, if it's moving twice as fast,
it will take only half as long.\
\SQA\F\=\d\No, if it's going faster, it will
take LESS time.\
\DQ\
\BQESC\
\DCW\
\SQ\1\Y\
\SQF\S\1\
\SQM\3\
\SQT\If the ball was moving 100 mm/s\
\SAL\13\-20\0\0\0\
\DA\
\DTN\ However, the velocities of the ball
on the two legs of the trip are not the same.\
\DTN\ The ball first moves with a velocity of 200 mm/s to the right and then with
a velocity of 200 mm/s to the left.\
\DTN\ A velocity is a speed and a
direction.\
\SWS\1\
\DCW\
\SWT\\\16\\
\DTNW\ A ball travels a circular path at a
constant speed.\
\SA\
\DFA\BALL1\130\0\
\SAS\30\\1\3\
\SAL\10\11\-1\1\1\
\SAL\10\-2\-1\10\-1\
\SAL\10\-11\1\-1\-1\
\SAL\1\
\SWT\M\M\17\M\
\DFB\VECTORS4\56\0\
\XPLOT\97\38\
\XDRAW\8\-8\0\3\0\-3\-3\0\
\XPLOT\182\30\
\XDRAW\8\8\0\-3\0\3\-3\0\
\XPLOT\194\90\
\XDRAW\-8\8\0\-3\0\3\3\0\
\XPLOT\104\98\
\XDRAW\-8\-8\3\0\-3\0\0\3\
\SWT\\\17\\
\DTN\ A ball moves around a direction in
which the object is moving.\
\SQANS\down\
\SQLAST\No, the arrow points down, so the
object must be moving down.\
\SQA\T\CONTAINS\down\Right. The arrow points down, so the object is moving down.\
\DQ\
\BQESC\
\SQSEND\
\SWS\1\
\DCW3\
\DF\VECTORS3\1\1\1\0\
\SWT\\\R2\\1\
\DCW\
\SQ\1\Y\
\SQF\S\6\
\SQT\This vector represents the velocity
of an object moving in what
direction?\
\SQH\Enter UP, DOWN, LEFT, or RIGHT.\
\SQH\Remember that the direction of the
arrow indicates therrow indicates the direction in
which the object is moving.\
\SQANS\up\
\SQLAST\No, the arrow points up, so the
object must be moving up.\
\SQA\T\CONTAINS\up\Right. The arrow points up, so the
object is moving up.\
\DQ\
\BQESC\
\SWS\1\
\SFF\0\SWS\1\
\SQS\1\Y\
\SFF\0\\
\DF\VECTORS2\1\1\1\0\
\SWT\\\R2\\1\
\SQ\1\Y\
\SQF\S\5\
\SQT\This vector represents the velocity
of an object moving in what
direction?\
\SQH\Enter UP, DOWN, LEFT, or RIGHT.\
\SQH\Remember that the direction of the
aA\T\CONTAINS\r\Correct. The fact that the arrow
points to the right indicates that
the ball is moving to the right.\
\SQA\T\CONTAINS\l\No, the fact that the arrow points to the right indicates that the ball is
moving to the right.\
\DQ\
\BQESC\
\Does the lower vector represent the
velocity of the ball moving to the
left or to the right?\
\SQH\Enter left or right.\
\SQANS\right\
\SQLAST\No, the fact that the arrow points to the right indicates that the ball is
moving to the right.\
\SQat the arrow
points to the left indicates that the ball is moving to the left.\
\SQA\T\CONTAINS\r\No, the fact that the arrow points to the left indicates that the ball is
moving to the left.\
\DQ\
\BQESC\
\DCW\
\SQ\1\N\
\SQM\2\
\SQF\S\5\
\SQT\y of the ball moving to the
left, or moving to the right?\
\SQH\Enter left or right.\
\SQANS\L\left\
\SQLAST\No, the fact that the arrow points to the left indicates that the ball is
moving to the left.\
\SQA\T\CONTAINS\l\Correct. The fact thf the object.\
\DTN\ The direction in which the arrow is
pointing gives the direction of the
vector and indicates the direction in
which the object was moving.\
\DCW\
\SQ\1\N\
\SQM\2\
\SQF\S\5\
\SQT\Does the upper vector represent the
velocit A vector consists of a magnitude and a direction.\
\SWS\1\
\DF\VECTORS1\1\1\1\0\
\SWT\\\R2\\1\
\DTN\ A velocity vector is represented by
an arrow.
The length of the arrow gives the
magnitude of the vector and represents
the speed oistance it moves
each second is constant, its direction
is constantly changing.\
\DQ\
\DCW\
\SWT\\\8\\
\DTN\ Whereas a speed can be represented
simply by a number, a velocity cannot.\
\DTN\ Physicists represent velocities by
vectors.
the distance it moves
each second is constant, its direction
is constantly changing.\
\SQA\T\CONTAINS\y\No! Although the distance it moves
each second is constant, its direction
is constantly changing.\
\SQA\T\CONTAINS\n\Right! Although the d0\2\1\-10\1\
\SAL\10\11\-1\1\1\
\SAL\10\-2\-1\10\-1\
\SAL\10\-11\1\-1\-1\
\SAL\10\2\1\-10\1\
\DA\
\DCW\
\SWT\\\15\\
\SQ\1\N\
\SQM\2\
\SQF\S\3\\QH\
\SQT\Was the ball's velocity constant?\
\SQH\Enter Y or N\
\SQANS\no\
\SQLAST\No! Although circle at a
constant speed.
But its velocity is constantly
changing.\
\DTN\ While the magnitude of the velocity
remains constant, its direction changes constantly.\
\DTN\ The instantaneous velocity of the
ball at a given moment is its
instantaneous speed and the direction
it is moving at that moment.\
\SFF\0\1\
\DCW\
\DFB\V5OLD\56\0\
\SWT\\\17\\
\DTN\ These arrows indicate the
instantaneous velocity of the ball at
various moments as it traveled around
the circle.\
\
\DFA\BALL1\8\1\
\SAS\40\0\1\-1\
\SAL\18\0\0\0\1\
\DA\
\GCR\
\DCW\
\DTN\ You can see the ball
speeding up more clearly if we
take stroboscopic pictures of
it falling.\
\XERASE\10\152\25\164\
\SA\
\DFA\BALL1\8\0\
\SAS\40\0\0\-1\
\SAL\18\0
speed gradually increases.\
\SFF\3\\
\SA\
\DFA\BALL1\8\1\
\SAS\40\0\1\-1\
\SAL\18\0\0\0\1\
\DA\
\DTNNW\ You might imagine that this
is a picture of a ball falling
from a height of 50 m to the
ground below in 3 s.\
\XERASE\10\152\25\164\
\SA The speed or velocity of an object
tells you how rapidly its position is
changing.\
\DTN\ The acceleration of an object tells
you how rapidly its speed or velocity
is changing.\
\DCW\
\SWT\8\\\\
\DTNNW\ When this ball falls, its \
\BMC\2\ACCEL CONSTANT\
\BMC\3\ACCEL NEWT1\
\BMC\4\ACCEL NEWT2\
\BMC\5\ACCEL VOCAB\
\BMC\6\ACCEL REVIEW\
\BL\FORCE\
\L\ACCEL ACCEL\
\SCH\2\2\
\DTCNW\Acceleration\
\SCH\1\1\
\DTN\ The position of an object tells you
where it is located.
ERATION\
\SM\
\SMT\Acceleration and Force\
\SMO\Acceleration\
\SMO\Constant Acceleration\
\SMO\Newton's First Law\
\SMO\Newton's Second Law\
\SMO\Vocabulary Review\
\SMO\Review Questions\
\SMO\Return to Main Menu\
\DM\
\BMC\1\ACCEL ACCELt's moving upward, as indicated by the arrow at point D.\
\SQA\T\CONTAINS\up\Correct! The arrow at Point D shows
the direction of this vector.\
\DQ\
\SQSEND\
\BESC\
\L\VELOCITY VOCAB\
\RVR\1\
\BESC\
\L\VELOCITY REVIEW\
\RRQ\1\
\BESC\
\L\ACCEL direction of the velocity
vector at point D? \
\SQH\No, remember that the velocity vector
is represented by the arrow with its
end at point D.\
\SQH\No, think about the direction that the
arrow at point D is pointing.\
\SQANS\up\
\SQLAST\No, iNS\left\
\SQLAST\No, it's moving to the left, as
indicated by the arrow at point C.\
\SQA\T\CONTAINS\left\Correct! The arrow at Point C shows
the direction of this vector.\
\DQ\
\BQESC\
\DCW\
\SQ\1\Y\
\SQM\3\
\SQF\S\6\\NQAH\
\SQT\What is theAH\
\SQT\What is the direction of the velocity
vector at point C? \
\SQH\No, remember that the velocity vector
is represented by the arrow with its
end at point C.\
\SQH\No, think about the direction that the
arrow at point C is pointing.\
\SQAt point B is pointing.\
\SQANS\down\
\SQLAST\No, it's moving down, as indicated by
the arrow at point B.\
\SQA\T\CONTAINS\down\Correct! The arrow at Point B shows
the direction of this vector.\
\DQ\
\BQESC\
\DCW\
\SQ\1\Y\
\SQM\3\
\SQF\S\6\\NQ
\SQT\What is the direction of the ball's
velocity vector at point B? \
\SQM\3\
\SQF\S\6\\NQAH\
\SQH\No, remember that the velocity vector
is represented by the arrow with its
end at point B.\
\SQH\No, think about the direction that the
arrow aion that the
arrow at point A is pointing.\
\SQANS\right\
\SQLAST\No, it's moving to the right, as
indicated by the arrow at point A.\
\SQA\T\CONTAINS\right\Correct! The arrow at Point A shows
the direction of this vector.\
\DQ\
\DCW\
\SQ\1\Y\
DCW\
\SQ\1\Y\
\SQM\3\
\SQF\S\6\\NQAH\
\SQT\What is the direction of the velocity
vector for the ball at point A? \
\SQH\No, remember that the velocity vector
is represented by the arrow with its
end at point A.\
\SQH\No, think about the direct\DTN\ In each case, the tail end of the
vector arrow is placed at the position
of the ball where the velocity is being measured.\
\SQS\1\Y\
\SWS\1\
\DCW\
\SWT\M\M\17\M\
\SWS\1\
\DCW\
\SWT\M\M\17\M\
\SFF\0\\
\DFB\VECTORS5\56\0\
\DFL\1\4\
\\0\0\1\
\DA\
\DTN\ The distance between frozen
pictures is an indication of
the speed of the ball at each
point.\
\DTN\ If it moves 50 m in 3 s, its average speed is 16.7 m/s.
This can be seen from a
graph of position versus time.\
\SWS\1\
\SFF\0\\
\DF\ACCEL1\1\1\1\0\
\SWT\M\M\17\M\
\DT\ As time passes, the curve becomes
steeper and steeper.
This increase in slope reflects an
increase in velocity.\
\XPLOT\60\4\
\XDRAW\176\100\
\DTN\ The line connecting the initial d\Right, it increased from 10 to 20 m/s.\
\SQA\F\=\e\No, that's the total velocity at two
seconds. You need the increase in
velocity from t = 1 to t = 2.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\2\
\SQF\S\1\\QH\
\SQT\What is the velocity of the ball ang the second second?
a. 0 m/s c. 5 m/s e. 20 m/s
b. 1 m/s d. 10 m/s f. 30 m/s\
\SQH\The velocity was 10 m/s at t = 1 s and
20 m/s at t = 2 s.\
\SQANS\d\
\SQLAST\No, the speed increased by 10 m/s, from 10 m/s to 20 m/s.\
\SQA\T\=\ity graph, for t = 2 s\
\SQANS\e\
\SQLAST\No, the velocity was 20 m/s.\
\SQA\T\=\e\Right. The graph shows the velocity
was 20 m/s at 2 s.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\2\
\SQF\S\1\\QH\
\SQT\How much does the velocity of the ball
change duri\
\SQA\T\=\d\Right, it increased from 0 to 10 m/s.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\2\
\SQF\S\1\\QH\
\SQT\What is the velocity of the ball at 2
s?
a. 0 m/s c. 5 m/s e. 20 m/s
b. 1 m/s d. 10 m/s f. 30 m/s\
\SQH\Look at the velocl
change during the first second?
a. 0 m/s c. 5 m/s e. 20 m/s
b. 1 m/s d. 10 m/s f. 30 m/s\
\SQH\The velocity was 0 m/s at t = 0 s, and
10 m/s at t = 1 s.\
\SQANS\d\
\SQLAST\No, the speed increased by 10 m/s, from 0 m/s to 10 m/s. at the velocity graph for t = 1 s\
\SQANS\d\
\SQLAST\No, the velocity was 10 m/s.\
\SQA\T\=\d\Right. The graph shows that the speed
was 10 m/s at 1 s.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\2\
\SQF\S\1\\QH\
\SQT\How much does the velocity of the balng at all.\
\SQA\T\=\a\Right. It wasn't moving at all.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\2\
\SQF\S\1\\QH\
\SQT\What is the velocity of the falling
ball at 1 s?
a. 0 m/s c. 5 m/s e. 20 m/s
b. 1 m/s d. 10 m/s f. 30 m/s\
\SQH\LookQM\2\
\SQF\S\1\\QH\
\SQT\What is the velocity of the ball at
time zero?
a. 0 m/s c. 5 m/s e. 20 m/s
b. 1 m/s d. 10 m/s f. 30 m/s\
\SQH\Look at the velocity graph for t = 0 s\
\SQANS\a\
\SQLAST\No, the velocity was 0 m/s; it wasn't
movi constant rate.\
\DCW\
\DTN\ Since the height of the curve
increases by 10 m/s every second, the
slope of the line is 10 m/s per second, or 10 meters per second per second.\
\DCW\
\SQS\2\Y\
\SFF\0\\
\DF\ACCEL3\1\0\1\0\
\SWT\\\R2\\1\
\SQ\
\Sn of time on this same graph.\
\SFF\0\\
\DF\ACCEL3\1\1\1\0\
\SWT\\\R2\\
\DTN\ The curve on the left shows the
velocity of the ball as a function of
time.\
\DTN\ The curve forms a straight line,
indicating that the velocity is
increasing at a0\1\-16\0\
\GCR\
\SWT\M\M\17\M\
\DCW\
\SWT\M\M\18\M\
\SCH\1\1\
\DTN\ This acceleration is in units of
meters per second per second, or meters per second squared.\
\DTN\ We can show this by plotting the
instantaneous velocity as a functioSWT\M\M\17\M\
\SCH\2\3\
\DCW\
\SWT\M\M\17\M\
\DTNNW\ 30 m/s 10 m
\
\DTNW\ 3 s s2\
\XPLOT\27\160\
\XDRAW\84\\0\1\-84\0\
\XPLOT\165\160\
\XDRAW\63\0\0\1\-63\0\
\XPLOT\129\156\
\XDRAW\16\0\0\-1\-16\0\
\XPLOT\129\164\
\XDRAW\16\0\0 and 30 m/s, respectively.\
\DTN\ At time zero, the instantaneous
velocity is actually zero.\
\DCW\
\DTN\ To calculate the average
acceleration, take the change in
velocity over the entire period, 30
m/s, and divide it by the time, 3 s.\
\and
final position of the curve indicates
an average velocity of 16.7 m/s.\
\SCH\1\3\
\XPLOT\93\4\
\XDRAW\57\19\
\XPLOT\212\62\
\XDRAW\28\33\
\SCH\1\1\
\DTN\ The two tangent lines show that the
instantaneous velocities at 1 s and 3 s are 1t 3
s?
a. 0 m/s c. 5 m/s e. 20 m/s
b. 1 m/s d. 10 m/s f. 30 m/s\
\SQANS\f\
\SQLAST\No, the velocity was 30 m/s.\
\SQH\Look at the graph at t = 3 s.\
\SQA\T\=\f\Right. The graph shows that the
velocity was 30 m/s at 3 s.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\2\
\SQF\S\1\\QH\
\SQT\How much does the velocity of the ball
change during the third second?
a. 0 m/s c. 5 m/s e. 20 m/s
b. 1 m/s d. 10 m/s f. 30 m/s\
\SQH\The velocity was 20 m/s at t = 2 s and
30 m/s ue as a function of time as well.\
\DCW\
\XERASE\0\0\279\47\
\DFB\CRAZYV\14\0\
\DFB\CRAZYP\0\48\
\SWT\\\14\\
\DCW\
\DTN\ Notice that the velocity is positive when the ball is moving toward the
right and negative when it's moving
toward the of the ball
during its movement.\
\DTN\ The curve goes up and down as the
ball moves to the right and then back
to the left.\
\DTN\ At every point the slope of the
curve gives the instantaneous velocity
of the ball.
We can plot this valA\BALL1\4\17\
\SAS\50\\1\3\
\SAL\10\0\1\0\0\
\SAL\10\10\0\0\0\
\SAL\10\10\-3\0\0\
\SAL\5\-20\8\0\0\
\SAL\12\20\0\0\0\
\DA\
\SWT\\\10\\
\DTN\ Got that?\
\SWS\1\
\DCW\
\SFF\0\\
\DFB\CRAZYP\0\0\
\SWT\M\M\8\M\
\DTN\ This shows the positionEXT\30\180\33\
\XTEXT\40\236\33\
\SA\
\DFA\BALL1\4\17\
\SAS\50\\1\3\
\SAL\10\0\1\0\0\
\SAL\10\10\0\0\0\
\SAL\10\10\-3\0\0\
\SAL\5\-20\8\0\0\
\SAL\12\20\0\0\0\
\DA\
\SWT\\\10\\
\DTNNW\ Let's see that again.\
\XERASE\237\17\250\27\
\SA\
\DF very smoothly.\
\DTN\ This need not be the case.\
\SWS\1\
\DCW\
\SWT\M\M\17\M\
\SWT\M\M\17\M\
\XTEXT\........................................\0\24\
\XPLOT\0\28\
\XDRAW\279\0\0\1\-279\0\
\XTEXT\0\13\33\
\XTEXT\10\68\33\
\XTEXT\20\124\33\
\XTT\\\R2\\1\
\DCW\
\DTN\ Interestingly, the acceleration is
greatest at the two ends, where the
velocity is smallest. The acceleration is smallest where the velocity is
greatest.\
\DTN\ In this example, the acceleration is still changingo at each end.\
\DTN\ Notice that the slope of the
velocity curve is far from constant.\
\DTN\ This means that the velocity is not
changing at a constant rate.
The acceleration of the ball is
constantly changing.\
\DF\HARMONY3\1\1\1\0\
\SWCW\
\SFF\0\\
\DF\HARMONY1\1\1\1\0\
\SWT\\\R2\\1\
\DTN\ This graph shows the position as a
function of time.\
\DF\HARMONY2\1\1\1\0\
\SWT\\\R2\\1\
\DTN\ The velocity of the ball is greatest in the middle of its path. Of course,
it must be zerare
constantly changing.\
\SA\
\DFA\BALL1\80\12\
\SAS\90\\1\\
\SAL\4\1\3\0\0\
\SAL\4\18\0\0\0\
\SAL\4\10\-3\0\0\
\SAL\4\-1\-3\0\0\
\SAL\4\-18\0\0\0\
\SAL\4\-10\3\0\0\
\SAL\4\1\3\0\0\
\SAL\4\18\0\0\0\
\SAL\4\10\-3\0\0\
\DA\
\GCR\
\SWS\1\
\DPLOT\148\25\
\XDRAW\0\2\1\0\0\-2\
\XPLOT\86\25\
\XDRAW\0\2\1\0\0\-2\
\XPLOT\208\25\
\XDRAW\0\2\1\0\0\-2\
\XTEXT\0\84\29\
\XTEXT\1\146\29\
\XTEXT\2\206\29\
\SWT\\\7\\
\DTNNW\ In this example, both the velocity
and the acceleration of the ball \No, this is its velocity at 3 s.\
\SQA\F\=\f\No, at this rate it would reach a
velocity of 30 m/s in just 1 s.\
\DQ\
\SQSEND\
\DCW\
\DTN\ Acceleration need not always be
constant.\
\SWS\1\
\DCW\
\XPLOT\83\23\
\XDRAW\130\0\0\1\-130\0\
\Xond, for a rate of 10 m/s/s (or 10
m/s2).\
\SQA\T\=\e\Right. It accelerated 10 m/s each
second, a rate of 10 m/s/s (10 m/s2).\
\SQA\F\IS CONTAINED IN\ab\No, this is a distance!\
\SQA\F\=\c\No, this is its velocity at 1 s.\
\SQA\F\=\dhat is the acceleration of the ball
during these three seconds?
a. 10 m c. 10 m/s e. 10 m/s2
b. 50 m d. 30 m/s f. 30 m/s2\
\SQH\How much did the ball speed up each
second?\
\SQANS\e\
\SQLAST\No, it accelerated by 10 m/s every
secat t = 3 s.\
\SQANS\d\
\SQLAST\No, the speed increased by 10 m/s, from 20 m/s to 30 m/s.\
\SQA\T\=\d\Right, it increased from 20 to 30 m/s.\
\SQA\F\=\f\No, that's the velocity at t = 3 s.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\4\
\SQF\S\1\\QH\
\SQT\Wleft.\
\DTN\ When the velocity is changing, the
ball is accelerating.\
\SWS\1\
\DCW\
\DFB\CRAZYA\0\0\
\DFB\CRAZYV\14\42\
\DFB\CRAZYP\0\90\
\SWT\M\M\19\M\
\DCW\
\DTN\ The faster the |C velocity changes|C, the more the ball ~
is |C accelerating|C.\
\DTN\ The slower the velocity changes, the less the ball is accelerating.\
\DTN\ When the velocity is constant,
acceleration is zero.\
\DTN\ Notice that when the ball is
speeding up, the acceleration is
greater than zero. When the acceleration is positive,
the velocity curve moves upward.\
\DTN\ When the acceleration is negative,
the ball moves less rapidly toward the
right, or more rapidly toward the left.\
\DTN\ When the acceleration is negative,
the veloc\DTN\ When the velocity is negative, the
ball moves toward the left and the
position curve moves downward.\
\DCW\
\DTN\ When the acceleration is positive,
the ball moves more rapidly toward the
right or less rapidly to the left.\
\DTN\ me zero.\
\DQ\
\SQSEND\
\SWS\1\
\DCW\
\DTN\ Notice that both the velocity and
acceleration can be either positive or
negative.\
\DTN\ When the velocity is positive, the
ball moves toward the right and the
position curve moves upward.\
= 0 s) is ____ m/s2. \
\SQH\Look at the acceleration curve for t =
0 s.\
\SQANS\50\
\SQLAST\No, the acceleration curve shows a
value of 50 m/s2 at time zero.\
\SQA\T\=\50\Right. The acceleration curve shows a
value of 50 m/s2 at tilocity graph for t = 0 s.\
\SQANS\0\
\SQLAST\No, it isn't moving at all at time
zero.\
\SQA\T\=\0\Right. It isn't moving at all at t =
0.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\3\
\SQF\S\3\\NQAH\
\SQT\The acceleration of the ball at this
time (ts
lowest point at t = 0 s.\
\SQA\T\=\0 s\Right. The position graph is at its
lowest point at t = 0 s.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\3\
\SQF\S\3\\NQAH\
\SQT\The speed of the ball at this time (t = 0 s) is ____ m/s. \
\SQH\Look at the veT\The ball is furthest to the left at
t = ____ s. \
\SQH\At what time is the position graph at
its lowest level?\
\SQANS\0\
\SQLAST\No, the position graph is at its lowest level at time zero.\
\SQA\T\=\0\Right. The position graph is at it fast
is it changing?!?\
\SQANS\0\
\SQLAST\No, its acceleration is 0 m/s2. It
isn't accelerating at all if its
velocity stays constant!\
\SQA\T\=\0\Right. It isn't accelerating at all.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\3\
\SQF\S\3\\NQAH\
\SQ=\0\Right. The acceleration curve at t =
2.5 s shows a value of 0 m/s2.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\3\
\SQF\S\3\\NQAH\
\SQT\The acceleration at times when the
velocity is constant is ____ m/s2. \
\SQH\If the velocity is constant, how3\\NQAH\
\SQT\The acceleration of the ball at this
time (t = 2.5 s) is ____ m/s2. \
\SQH\Look at the acceleration curve
at t = 2.5 s.\
\SQANS\0\
\SQLAST\No, the curve shows that the rate of
acceleration at t = 2.5 s is 0 m/s/s.\
\SQA\T\\
\SQH\Look at the graph of velocity versus
time, at t = 2.5 s.\
\SQANS\50\
\SQLAST\No, at t = 2.5 s the velocity is 50
m/s.\
\SQA\T\CONTAINS\50\Right. The velocity graph is at 50 m/s at 2.5 s.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\3\
\SQF\S\SQA\T\=\2.5\Right. At 2.5 s the graph is at its
highest point.\
\SQA\F\IS CONTAINED IN\2.\No, it's a little later than that!\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\3\
\SQF\S\3\\NQAH\
\SQT\The speed of the ball at this time
(t = 2.5 s) is ____ m/s. the right at
t = ____ s. \
\SQH\Look at the graph of position versus
time.\
\SQH\At what time is the position highest?\
\SQANS\2.5\
\SQLAST\No, the ball is furthest to the right
at 2.5 s, when the graph of position
versus time is highest.\
\\
\DTN\ When the ball is slowing down, the
acceleration is less than zero.\
\SWS\1\
\DCW\
\SQS\2\Y\
\SFF\0\\
\DFB\CRAZYA\0\0\
\DFB\CRAZYV\14\42\
\DFB\CRAZYP\0\92\
\SWT\M\M\18\M\
\SQ\
\SQM\3\
\SQF\S\3\\NQAH\
\SQT\The ball is furthest to ity curve moves downward.\
\BESC\
\L\ACCEL CONSTANT\
\SCH\2\2\
\DTCNW\Constant Acceleration\
\SCH\1\1\
\DTN\ When an object falls without
hindrance, it accelerates at a constant rate of 9.8 m/s2, or approximately 10
m/s2.\
\DCW\
\SWT\8\M\M\M\
\DTN\ When this ball falls, its
speed gradually increases.\
\SA\
\DFA\BALL1\8\0\
\SAS\40\0\1\-1\
\SAL\18\0\0\0\1\
\DA\
\DTN\ You might imagine that this
is a picture of a ball falling
from a height of 50 m to the
ground below in 3 s.\QF\S\5\
\SQT\Its velocity at t = 1 s will be ____
m/s.\
\SQH\Remember that the final speed is the
initial speed plus the amount it sped up.\
\SQH\Remember that its initial speed was
10 m/s and it sped up 9.8 m/s each
second for 1 second.\
\SQANSnitial velocity.\
\DTN\ This equation is correct in all
instances of constant, or uniform,
acceleration.\
\SQS\2\Y\
\DCW\
\DTNNW\A ball with an initial velocity of 10
m/s, accelerating at a rate of 9.8
m/s2.\
\SWT\\\R2\\
\SQ\
\SQM\3\
\S(v0).\
\DCW\
\DTNNW\ This gives us the equation:\
\SCH\2\3\
\DTC\v(t) = at + v\
\SCH\1\1\
\SWT\M\M\6\M\
\DTN\ In other words, the velocity at
time, t, is equal to the rate of
acceleration, a, multiplied by the
time, t, plus the id by the time, t.\
\DTN\ If the ball is moving at time t0,
this equation will not be correct.\
\DTN\ While the change in velocity can
still be calculated as acceleration
multiplied by time, this change must be added to the initial velocity r a body moving with constant
acceleration, with a velocity (v) equal to 0 m/s when the time (t) is 0, we can say:\
\SCH\2\3\
\DTC\v(t) = at\
\SCH\1\1\
\DTN\ In other words, the velocity at time t is equal to the rate of acceleration, a, multiplie29.4 m/s\
\SQH\Remember, it has been speeding up by
9.8 m/s each second for 1.5 seconds.\
\SQANS\d\
\SQLAST\No, (9.8 m/s2) x (1.5 s) = 14.7 m/s\
\SQA\T\=\d\Right. (9.8 m/s2) x (1.5 s) = 14.7
m/s\
\DQ\
\SQSEND\
\SWS\1\
\DCW\
\DTNNW\ Fom/s\
\SQA\F\=\c\No, this is its speed at 1 s.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\3\
\SQF\S\1\
\SQT\What is the velocity of the falling
ball at 1.5 s?
a. 0 m/s d. 14.7 m/s
b. 2 m/s e. 19.6 m/s
c. 9.8 m/s f. s e. 19.6 m/s
c. 9.8 m/s f. 29.4 m/s\
\SQH\Remember that it has been
accelerating at the rate of 9.8 m/s
for two seconds.\
\SQANS\e\
\SQLAST\No, (9.8 m/s2) x (2 s) = 19.6 m/s\
\SQA\T\=\e\Right. (9.8 m/s2) x (2 s) = 19.6 during the one second.\
\SQA\F\IS CONTAINED IN\abc\No, this a rate of acceleration, not
a speed.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\3\
\SQF\S\1\
\SQT\What is the velocity of the falling
ball at 2 s?
a. 0 m/s d. 14.7 m/s
b. 2 m/ 1 m/s e. 1 m/s2
c. 9.8 m/s f. 9.8 m/s2\
\SQH\Remember that its speed increases by
9.8 m/s every second.\
\SQANS\c\
\SQLAST\No, the speed at one second will be
just 9.8 m/s.\
\SQA\T\=\c\Right, it speeds up by 9.8 m/s t.\
\SWS\1\
\DCW\
\SQS\2\Y\
\SQ\
\SQM\3\
\SQF\S\1\
\SQT\Given that the ball is not moving at
time zero (t = 0) and that it falls
with an acceleration of 9.8 m/s2,
what is its velocity at 1 s?
a. 0 m/s d. 0 m/s2
b.e ball
at each moment during its decent.\
\DCW\
\DTN\ As was mentioned earlier, the
acceleration is constant, at:\
\SCH\2\3\
\DTC\9.8 m/s\
\SCH\1\1\
\DTN\ Knowing this, you can calculate the
velocity of the ball at any time after
the star\-1\
\SAL\18\0\0\0\1\
\DA\
\DTN\ The distance between frozen
pictures is an indication of
the speed of the ball at that
point.\
\SFF\0\\
\DF\ACCEL2\1\1\1\0\
\SWT\M\\R2\\
\DTN\ This graph shows the position,
velocity, and acceleration of th
\XERASE\10\152\25\165\
\SA\
\DFA\BALL1\8\0\
\SAS\40\0\1\-1\
\SAL\18\0\0\0\1\
\DA\
\DTN\ You can see the ball
speeding up more clearly if we
take stroboscopic pictures of
it falling.\
\XERASE\10\152\25\165\
\SA\
\DFA\BALL1\8\0\
\SAS\40\0\0\19.8\
\SQLAST\No, (10 m/s) + (9.8 m/s2) x (1 s) =
19.8 m/s\
\SQA\T\=\19.8\Right. (10 m/s) + (9.8 m/s2) x (1 s) = 19.8 m/s\
\SQA\T\=\19.80\Right. (10 m/s) + (9.8 m/s2) x (1 s) = 19.8 m/s\
\SQA\F\CONTAINS\20\Please be more accurate.\
\SQA\F\CONTAINS\9.8\Don't forget to include the initial
speed.\
\DQ\
\DCW\
\SQ\
\SQM\3\
\SQF\S\5\
\SQT\Its velocity at t = 2.5 s will be
____ m/s.\
\SQH\Remember that its initial speed was
10 m/s and it sped up 9.8 m/s each
second for 2.5 seconds. on it?\
\SQANS\yes\
\SQH\Did its velocity change as it rolled
to a stop?\
\SQANS\Yes\
\SQLAST\Yes, a force did act. Otherwise, it
wouldn't have changed its velocity!\
\SQA\T\CONTAINS\Y\Right! Otherwise it wouldn't have
changed its velocity.\
\
\DCW\
\SQS\2\Y\
\SWT\M\M\6\M\
\DTNNW\ Watch this animation.\
\DCW\
\SFF\3\\
\SA\
\DFA\BALL1\0\8\
\SAS\10\\1\3\
\SAL\22\22\-1\0\0\
\DA\
\SWT\\\6\\
\DCW\
\SQ\
\SQM\3\
\SQF\S\7\
\SQT\As this ball rolled to a stop, was
any force actingn.\
\SWS\1\
\DCW\
\DTNNW\ Newton's First Law of Motion states
this fact:\
\DTN\ If no force is applied to an object, its velocity cannot change. As a
result, an object at rest will stay at
rest and an object in motion will stay
in motion.0\16\253\26\
\SFF\3\\
\SA\
\DFA\BALL1\4\16\
\SAS\10\\1\3\
\SAL\13\0\1\0\0\
\SAL\10\13\0\0\0\
\SAL\13\13\-3\0\0\
\SAL\5\-26\8\0\0\
\SAL\14\14\0\0\0\
\DA\
\DTN\ If there were no forces acting on
it, the ball would neither speed up nor slow dow\1\0\0\
\SAL\10\13\0\0\0\
\SAL\13\13\-3\0\0\
\SAL\5\-26\8\0\0\
\SAL\14\14\0\0\0\
\DA\
\DCW\
\DTN\ If this were a real ball rolling
along a surface, there would have to be some force acting on it each time it
sped up or slowed down.\
\XERASE\24 ball as it moves.\
\XTEXT\........................................\0\24\
\XPLOT\0\28\
\XDRAW\279\0\0\1\-279\0\
\XTEXT\0\13\33\
\XTEXT\10\68\33\
\XTEXT\20\124\33\
\XTEXT\30\180\33\
\XTEXT\40\236\33\
\SA\
\DFA\BALL1\4\16\
\SAS\10\\1\3\
\SAL\13\0SQSEND\
\BESC\
\L\ACCEL NEWT1\
\SCH\2\2\
\DTCNW\Newton's First Law\
\SCH\1\1\
\DTN\ A force is anything that can change
the velocity of an object.\
\DTN\ A push or pull is a force.\
\SWS\1\
\DCW\
\SWT\M\M\9\M\
\DTNNW\ Watch this_ m.\
\SQANS\44.1\
\SQLAST\No, (1/2) x (9.8 m/s2) x (3 s)2 =
44.1 m\
\SQA\T\=\44.1\Right. (1/2) x (9.8 m/s2) x (3 s)2 = 44.1 m\
\SQA\T\=\44.10\Right. (1/2) x (9.8 m/s2) x (3 s)2 = 44.1 m\
\SQA\F\=\44\More accuracy, please.\
\DQ\
\=\19.6\Right. (1/2) x (9.8 m/s2) x (2 s)2 = 19.6 m\
\SQA\T\=\19.60\Right. (1/2) x (9.8 m/s2) x (2 s)2 = 19.6 m\
\SQA\F\=\20\More accuracy, please.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\3\
\SQF\S\5\
\SQT\When t = 3 s, it will have fallen
___(1/2) x (9.8 m/s2) x (1 s)2 = 4.9 m\
\SQA\F\=\5\More accuracy, please.\
\DQ\
\BQESC\
\DCW\
\SQ\
\SQM\3\
\SQF\S\5\
\SQT\When t = 2 s, it will have fallen
____ m.\
\SQANS\19.6\
\SQLAST\No, (1/2) x (9.8 m/s2) x (2 s)2 =
19.6 m\
\SQA\T\
\SQM\3\
\SQF\S\5\
\SQT\At t = 1 s, it will have fallen ____
m.\
\SQH\d = (1/2)at2\
\SQANS\4.9\
\SQLAST\No, (1/2) x (9.8 m/s2) x (1 s)2 = 4.9 m\
\SQA\T\=\4.9\Right. (1/2) x (9.8 m/s2) x (1 s)2 = 4.9 m\
\SQA\T\CONTAINS\4.90\Right. \1\
\DTN\where d is the distance traveled, a is
the rate of acceleration and t is the
time since the ball was released.\
\DCW\
\SQS\2\Y\
\DTNNW\A ball is dropped at time zero, and
then accelerates at a rate of 9.8 m/s2.\
\SWT\\\R2\\
\SQ\
velocity.\
\DQ\
\SQSEND\
\SWS\1\
\DCW\
\DTNNW\In the case of a ball being dropped,
the position of the ball at some time,
t, can be calculated from the equation:\
\SCH\2\3\
\DTNNW\ 1
\
\DTNW\ d = -at
\
\DTNW\ 2\
\SCH\1\
\SQANS\34.5\
\SQLAST\No, (10 m/s) + (9.8 m/s2) x (2.5 s) = 34.5 m/s\
\SQA\T\CONTAINS\34.5\Right. (10 m/s) + (9.8 m/s2) x (2.5
s) = 34.5 m/s\
\SQA\F\CONTAINS\34\More accuracy, please.\
\SQA\F\CONTAINS\24\Don't forget to include the initial
\SQA\T\CONTAINS\N\Yes, a force did act! Otherwise it
wouldn't have changed its velocity.\
\DQ\
\SWS\1\
\DCW\
\SWT\8\\\\
\DTNNW\Watch this animation.\
\SA\
\DFA\BALL1\8\0\
\SAS\40\0\1\-1\
\SAL\18\0\0\0\1\
\DA\
\SWT\M\M\M\17\
\DCW\
\SWS\1\
\SQ\
\SQM\3\
\SQF\S\7\
\SQT\As this ball fell, was any force
acting on it?\
\SQANS\yes\
\SQH\Did its velocity change as it rolled
to a stop?\
\SQANS\Yes\
\SQLAST\Yes, a force did act. Otherwise, it
wouldn't have changed its velocity!\
\SQ.\
\DTN\ A gram is one-thousandth of a
kilogram.
The mass of a quarter teaspoon of
water is about 1 g.\
\DCW\
\DTNNW\ Newton's Second Law states that the
acceleration caused by the application
of a force is proportional to the
magnitude o the earth.\
\DCW\
\DTN\ Like length and time, mass is a
fundamental physical quantity.
Scientists measure the mass of an
object in kilograms.
The unit kilogram is normally
abbreviated kg.
The mass of 1 liter of water is 1
kgan object is
actually a measure of the magnitude of
the gravitational force acting on the
object.\
\DTN\ Whereas the mass of an object is a
property of that object, the weight of
an object is a measure of the
interaction between that object anderty of all matter.
It is a measure of how difficult it
is to accelerate the object when a
force is applied to it.\
\DTN\ At the surface of the earth, the
mass of an object is proportional to
the weight of the object.
But the weight of twice as massive
(and the force is not doubled) it will
accelerate at half the rate.\
\XERASE\74\101\109\125\
\SA\
\DFA\BIGBALL5\70\107\
\SAS\300\\1\3\
\SAL\15\0\1\0\-1\
\DA\
\XERASE\176\0\212\20\
\GCR\
\SWS\1\
\DCW\
\DTN\ Mass is a prop as hard, the
object will accelerate at twice the
rate.\
\SWT\M\M\19\M\
\SFF\3\\
\SA\
\DFA\BIGBALL\70\107\
\SAS\100\\1\3\
\SAL\10\4\2\-4\-2\
\DA\
\XERASE\160\18\195\37\
\GCR\
\DFB\BIGBALL4\70\100\
\SWT\\\19\\
\DCW\
\DT\ If the object isL\15\0\1\0\-1\
\DA\
\XERASE\176\0\212\20\
\GCR\
\DCW\
\DTN\ The rate at which it speeds up will
depend on how hard you push and how
massive the object is.\
\DCW\
\SFF\0\\
\SWT\\\19\\
\DFB\BIGBALL3\70\93\
\DCW\
\DTN\ If you push twiceWT\\\17\\
\DTN\ If you push on something, it will
start to move.
If there is no friction to slow it
down, the object will continue to speed up as long as you keep pushing.\
\XERASE\74\101\109\125\
\SA\
\DFA\BIGBALL\70\107\
\SAS\200\\1\3\
\SA object causes the
object to accelerate at a constant
rate, proportional to the magnitude of
the force and inversely proportional to the mass of the object.\
\DTN\ This law is rather easily understood intuitively.\
\DCW\
\DFB\BIGBALL2\70\100\
\SDTN\ The direction of the force is the
direction in which you try to push the
cylinder.
It is indicated by the direction in
which the arrow is pointing.\
\SWS\1\
\DCW\
\DTNNW\ Newton's Second Law states that:\
\DTN\A force acting on an To move this cylinder, you must
apply a force to it.\
\XERASE\72\107\106\125\
\DFB\BIGBALL2\70\100\
\DTN\ The magnitude of the force is a
measure of how hard you push on the
cylinder.
It is indicated here by the length
of the arrow.\
\ Force, like velocity, is a vector
quantity.
A force has both a magnitude and a
direction.\
\DCW\
\DFB\BIGBALL1\70\107\
\SWT\\\17\\
\DTN\ Imagine that you are looking down on a large steel cylinder sitting on a
flat floor.\
\DTN\ Law\
\SCH\1\1\
\DTN\ Newton's First Law states that the
velocity of an object can be altered
only as a result of a force acting on
the object.\
\DTN\ Newton's Second Law states how a
given force will affect the velocity of an object.\
\DTN\A\T\CONTAINS\Y\Right! Otherwise it wouldn't have
changed its velocity.\
\SQA\T\CONTAINS\N\Yes, a force did act! Otherwise it
wouldn't have changed its velocity.\
\DQ\
\SWS\1\
\SQSEND\
\BESC\
\L\ACCEL NEWT2\
\SCH\2\2\
\DTCNW\Newton's Secondf that force and is
inversely proportional to the mass of
the object. This can be expressed
mathematically as:\
\SCH\2\2\
\DTCNW\a = F/m\
\SCH\1\1\
\DTN\The acceleration equals the force
divided by the mass.\
\DCW\
\DTN\ Force is measured in units called
newtons.
A force of one newton is required to accelerate a one kilogram mass at the
rate of one meter per second per
second.
The unit newton is abbreviated N:\
\SCH\2\2\
\DTN\ 1 m/s = 1 N / 1 kg\
\SQS\2\Y\
\T\An object that accelerates at 2 m/s2
when exposed to a force of 4 N has a
mass of ____ kg.\
\SQH\Remember, m = F/a\
\SQANS\2\
\SQLAST\No, (4 N)/(2 m/s2) = 2 kg\
\SQA\T\=\2\Right. (4 N)/(2 m/s2) = 2 kg\
\SQA\F\=\8\No, m = F/a, not F x aSQA\F\=\1000\No, 50 g = 0.05 kg and 20 mm/s2 =
0.02 m/s2\
\SQA\F\=\1\No, 50 g = 0.05 kg and 20 mm/s2 =
0.02 m/s2\
\SQA\F\=\1.0\No, 50 g = 0.05 kg and 20 mm/s2 =
0.02 m/s2\
\DQ\
\SQSEND\
\SQS\2\Y\
\DCW\
\SQ\
\SQM\3\
\SQF\S\3\
\SQM\3\
\SQF\S\5\
\SQT\To accelerate a 50 g mass at the rate of 20 mm/s2 requires a force of ____
N.\
\SQH\F = ma\
\SQANS\.001\
\SQLAST\No, (0.05 kg) x (0.02 m/s2) = 0.001 N\
\SQA\T\CONTAINS\.001\Right. (0.05 kg) x (0.02 m/s2) =
0.001 N\
\5\
\SQT\To accelerate a 20 kg mass at the
rate of 10 m/s2 requires a force of
____ N.\
\SQANS\200\
\SQH\F = ma\
\SQANS\200\
\SQLAST\No, (20 kg) x (10 m/s2) = 200 N\
\SQA\T\=\200\Right. (20 kg) x (10 m/s2) = 200 N\
\DQ\
\DCW\
\SQ\
\SQs at the rate of 2 m/s2 requires a force of ____ N.\
\SQH\F = ma\
\SQANS\4\
\SQLAST\No, (2 kg) x (2 m/s2) = 4 N\
\SQA\T\=\4\Right. (2 kg) x (2 m/s2) = 4 N\
\SQA\T\=\4.0\Right. (2 kg) x (2 m/s2) = 4 N\
\DQ\
\DCW\
\SQ\
\SQM\3\
\SQF\S\ requires a force of ____ N.\
\SQH\F = ma\
\SQANS\2\
\SQLAST\No, (2 kg) x (1 m/s2) = 2 N\
\SQA\T\=\2\Right. (2 kg) x (1 m/s2) = 2 N\
\SQA\T\=\2.0\Right. (2 kg) x (1 m/s2) = 2 N\
\DQ\
\DCW\
\SQ\
\SQF\S\5\
\SQT\To accelerate a 2 kg masTCNW\F = ma\
\SCH\1\1\
\DTN\ The force, F, required to accelerate an object of mass, m, at the rate a, is equal to the mass times the
acceleration.\
\SQS\2\Y\
\DCW\
\SQ\
\SQM\3\
\SQF\S\5\
\SQT\To accelerate a 2 kg mass at the rate of 1 m/s2No, A = F/m = (3 N)/(3 kg) = 1 m/s2\
\SQA\T\=\c\Right. A = F/m = (3 N)/(3 kg) = 1
m/s2 of acceleration.\
\DQ\
\SQSEND\
\DCW\
\DTNNW\ The formula a = F/m, can be
rearranged to give the more common form of Newton's Second Law:\
\SCH\2\3\
\DSQF\S\1\
\SQT\How rapidly will a 3 kg object
accelerate when acted on by a force
of 3 N?
a. 0 m/s2 d. 3 m/s2
b. 0.3 m/s2 e. 9 m/s2
c. 1 m/s2 f. 27 m/s2\
\SQH\Remember, a = F/m\
\SQANS\c\
\SQLAST\ c. 1 m/s2 f. 50 m/s2\
\SQH\Remember, a = F/m\
\SQANS\b\
\SQLAST\No, a = F/m = (2 N)/(10 kg) = 0.2
m/s2\
\SQA\T\=\b\Right. a = F/m = (2 N)/(10 kg) = 0.2
m/s2\
\SQA\F\=\e\No, a = F/m, not m/F.\
\DQ\
\DCW\
\SQ\
\SQM\3\
\kg) = 0.5
m/s2\
\SQA\F\=\d\No, a = F/m, not m/F.\
\DQ\
\DCW\
\SQ\
\SQM\3\
\SQF\S\1\
\SQT\How rapidly will a 10 kg object
accelerate when acted on by a force
of 2 N?
a. 0 m/s2 d. 2 m/s2
b. 0.2 m/s2 e. 5 m/s2
ce
of 2 N?
a. 0 m/s2 d. 2 m/s2
b. 0.5 m/s2 e. 5 m/s2
c. 1 m/s2 f. 80 m/s2\
\SQH\Remember, a = F/m\
\SQANS\b\
\SQLAST\No, A = F/m = (2 N)/(4 kg) = 0.5 m/s2\
\SQA\T\=\b\Right. A = F/m = (2 N)/(4
\SQANS\e\
\SQLAST\No, (5 N)/(1 kg) = 5 m/s2\
\SQA\T\=\e\Right. a = F/m = (5 N)/(1 kg) = 5
m/s2\
\SQA\F\=\b\No, a = F/m, not m/F.\
\DQ\
\DCW\
\SQ\
\SQM\3\
\SQF\S\1\
\SQT\How rapidly will a 4 kg object
accelerate when acted on by a forDCW\
\SQ\
\SQM\3\
\SQF\S\1\
\SQT\How rapidly will a 1 kg object
accelerate when acted on by a force
of 5 N?
a. 0 m/s2 d. 2 m/s2
b. 0.2 m/s2 e. 5 m/s2
c. 1 m/s2 f. 50 m/s2\
\SQH\Remember, a = F/m\\
\SQA\F\=\.5\No, m = F/a, not a/F\
\DQ\d
\DCW\
\SQ\
\SQM\3\
\SQF\S\3\
\SQT\If the force of gravity accelerates a falling 1 kg mass at the rate of 9.8
m/s2, the force is ____ N.\
\SQH\F = ma\
\SQANS\9.8\
\SQLAST\No, F = ma = (1 kg) x (9.8 m/s2) =
9.8 N\
\SQA\T\=\9.8\Right. F = ma = (1 kg) x (9.8 m/s2)
= 9.8 N\
\SQA\T\=\10\More accuracy, please.
\
\DCW\
\SQ\
\SQM\3\
\SQF\S\3\
\SQT\If the force of gravity accelerates a falling 5 kg mass at the rate of 9.8
m/s2, the forcSFF\0\\
\DF\ADDV5\1\1\1\0\
\SWT\\\12\\
\DTN\ Two forces were applied to the ball
at the same time.
Forces applied at the same time are
called concurrent forces
To add them together, you must place the tail of one vector at the head of
t\SWT\\\M\16\
\XERASE\84\55\91\89\
\XERASE\94\92\127\98\
\SA\
\DFA\BALL1\77\90\
\SAS\75\\1\\
\SAL\16\2\2\-2\-2\
\DA\
\XERASE\170\0\183\10\
\SWT\M\M\17\M\
\DCW\
\DTN\ The ball moves up and to the right.
How could we have predicted this?\
\LOT\87\95\
\XDRAW\40\0\-3\-3\3\3\-3\3\
\XPLOT\87\95\
\XDRAW\0\-40\-3\3\3\-3\3\3\
\SWT\\\17\\
\DTN\ Here you see a ball with two equal
forces being applied to it: one from
the left and one from below.\
\DTNNW\ Let's see which way it moves.\
net force is
applied.
The object does not move at all.\
\DTN\ The net force can be calculated by
adding up all the forces being applied.
This is not as easy as it might
sound, because each force is a vector.\
\DCW\
\DFB\B2\81\90\
\XPther\
\SQA\F\CONTAINS\left\No, the force isn't greater toward
the left.\
\SQA\F\CONTAINS\right\No, the force isn't greater toward
the right.\
\DQ\
\SFF\0\\
\DF\ADDV3\1\1\1\0\
\DTN\ Since two equal and opposite forces
are being applied, no ch
direction will it move - left, right, or neither?\
\SQH\Figure out the direction in which the force is greater.\
\SQANS\neither\
\SQLAST\No, the ball won't move either way.\
\SQA\T\=\neither\Right, it won't move either way.\
\SQA\T\CONTAINS\nei\SA\
\DFA\BALL1\245\10\
\SAS\75\\1\\
\SAL\14\-2\-2\\\
\SAL\1\-29\\\\
\DA\
\XERASE\9\10\22\20\
\GCR\
\SWS\1\
\DCW\
\SQ\3\Y\
\SQM\3\
\SQF\S\7\
\SQT\If the same force is applied
simultaneously to both the left and
right sides of a ball, in whi0\
\SAS\75\\1\\
\SAL\16\2\2\\\
\DA\
\XERASE\243\10\256\20\
\GCR\
\SWT\\\6\\
\DCW\
\DTNNW\ When the same force is applied to
the right side, the ball accelerates
toward the left.\
\DFB\B2\245\10\
\XPLOT\250\15\
\XDRAW\-40\0\3\3\-3\-3\3\-3\
acting on the
refrigerator simultaneously.\
\SWS\1\
\DCW\
\SWT\\\6\\
\DTNNW\ When a force is applied to the left
side of this ball, it accelerates
toward the right.\
\DFB\B2\0\10\
\XPLOT\5\15\
\XDRAW\40\0\-3\-3\3\3\-3\3\
\SA\
\DFA\BALL1\0\1 If you push gently against the side
of a refrigerator, it probably won't
move at all.
Yet Newton's second law states that
when a force is applied to an object,
it accelerates.\
\DTN\ The problem here is that there are
actually two forcesrn to Main Menu\
\DM\
\BMC\1\FRICTION ADD\
\BMC\2\FRICTION FRICTION\
\BMC\3\FRICTION RESOLVE\
\BMC\4\FRICTION INCLINE\
\BMC\5\FRICTION VOCAB\
\BMC\6\FRICTION REVIEW\
\BL\FORCE\
\L\FRICTION ADD\
\SCH\2\2\
\DTCNW\Adding Vectors\
\SCH\1\1\
\DTN\RVR\2\
\BESC\
\L\ACCEL REVIEW\
\RRQ\2\
\BESC\
\L\FRICTION\
\SM\
\SMT\Friction & Vectors\
\SMO\Adding Vectors\
\SMO\Friction\
\SMO\Resolving Vectors\
\SMO\Friction on an Incline\
\SMO\Vocabulary Review\
\SMO\Review Questions\
\SMO\Retu falling 10 kg mass at the rate of 9.8 m/s2, the force is ____ N.\
\SQANS\98\
\SQLAST\No, F = ma = (10 kg) x (9.8 m/s2) =
98 N\
\SQH\F = ma\
\SQA\T\=\98\Right. F = ma = (10 kg) x (9.8 m/s2) = 98 N\
\DQ\
\SQSEND\
\BESC\
\L\ACCEL VOCAB\
\e is ____ N.\
\SQH\F = ma\
\SQANS\49\
\SQLAST\No, F = ma = (5 kg) x (9.8 m/s2)
= 49 N\
\SQA\T\=\49\Right. F = ma = (5 kg) x (9.8 m/s2)
= 49 N\
\DQ\
\DCW\
\SQ\
\SQM\3\
\SQF\S\3\
\SQT\If the force of gravity accelerates ahe other.\
\DCW\
\DF\ADDV6\1\0\1\0\
\SWT\\\R2\\
\DTN\ Now, draw a new vector from the tail of the first one to the head of the
second one.\
\DF\ADDV7\1\0\1\0\
\SWT\\\R2\\
\DCW\
\DTN\ This resultant vector is the sum of
the two concurrent vectors.
It has a length that indicates the
magnitude of the resulting vector.
It has a direction that indicates
the direction of the resulting vector.\
\DCW\
\DTN\ The vector points in the direction
the ball moves when pushed by both of\DQ\
\DCW\
\SQ\3\Y\
\SQM\3\
\SQF\S\3\
\SQT\Two vectors of magnitudes 7 N and 10
N are applied to an object
concurrently. The smallest possible
magnitude for the resultant vector is ____ N.\
\SQANS\3\
\SQLAST\No, the smallest resultant vectortors are
parallel and pointing in the same
direction.\
\SQA\T\=\17\Right. The largest is 17 N (10 N + 7
N), obtained when the vectors are
parallel and pointing in the same
direction.\
\SQA\F\=\3\No, 3 N is actually the smallest
force possible!\
The largest possible
magnitude for the resultant vector is ____ N.\
\SQANS\17\
\SQH\The resultant vector will be greatest if they are pointing in the same
direction.\
\SQANS\17\
\SQLAST\No, the largest is 17 N (10 N + 7 N), obtained when the vec8\0\
\XDEL\20\
\DFB\ADDV12D\48\0\
\DTN\ It can be as small as the difference between the two magnitudes.\
\SWS\1\
\DCW\
\SQS\3\Y\
\SQ\3\Y\
\SQM\3\
\SQF\S\3\
\SQT\Two vectors of magnitudes 7 N and 10
N are applied to an object
concurrently. e added, both the magnitude and the direction of the
resultant vector are dependent on the
directions of the two vectors.\
\DFB\ADDV12B\48\0\
\DTN\ The magnitude can be as great as the sum of the two magnitudes.\
\DCW\
\XDEL\5\
\DFB\ADDV12C\4hose
pushing toward the left.\
\SQA\T\CONTAINS\left\No! The total forces pushing toward
the right are greater than those
pushing toward the left.\
\DQ\
\SQSEND\
\SWS\1\
\DCW\
\SFF\0\\
\DFB\ADDV12\48\0\
\SWT\\\13\\
\DTN\ When two vectors arer (LEFT or RIGHT)?\
\SQANS\Right\
\SQLAST\No, more force is pushing toward the
right (10 N) than toward the left (7
N), so the net force pushes to the
right.\
\SQA\T\CONTAINS\right\Yes! The total forces pushing toward the right are greater than tthe right and a total of 7 N
pushing to the left, so the result is 10 - 7 = 3 N.\
\DQ\
\DCW\
\SQ\3\Y\
\SQM\2\
\SQF\S\7\
\SQT\Is the force pushing the ball to the
left or the right?\
\SQANS\right\
\SQH\In which direction is the total force great one way and
subtract the forces going the other.\
\SQANS\3\
\SQLAST\No, there is a 10 N force pushing
toward the right and a total of 7 N
pushing to the left, so the result is 10 - 7 = 3 N.\
\SQA\T\=\3\Right. There is a 10 N force pushing toward des of the
forces working in the opposite
direction.\
\SQS\3\Y\
\SFF\0\\
\DF\ADDV11\1\0\1\0\
\SWT\\\R2\\
\DCW\
\SQ\3\Y\
\SQM\3\
\SQF\S\3\
\SQT\The magnitude of the net force acting on this ball is ____ N.\
\SQANS\3\
\SQH\Add the forces goingt forces are
applied, the direction of the resultant vector will be along the same line as
the forces.\
\DTN\ The magnitude of the force is
calculated by adding the magnitudes of
all the forces working in one direction and subtracting the magnituesultant vector is drawn.\
\SWS\1\
\DCW\
\DFB\B2\4\10\
\XPLOT\9\15\
\XDRAW\30\30\-3\0\3\0\0\-3\
\SA\
\DFA\BALL1\0\10\
\SAS\75\\1\\
\SAL\16\2\2\2\2\
\DA\
\XERASE\159\166\172\176\
\DF\ADDV11\1\1\1\0\
\SWT\\\R2\\
\DTN\ When parallel concurrenXERASE\170\0\183\10\
\GCR\
\DF\ADDV8\1\1\1\0\
\SWT\\\12\\
\DTN\ Here are four concurrent forces.
\
\DF\ADDV9\1\1\1\0\
\SWT\\\12\\
\DTN\ Here they have been combined, head
to tail.\
\DCW\
\DF\ADDV10\1\0\1\0\
\SWT\\\12\\
\DTN\ Now the r
these forces.\
\SWS\1\
\DCW\
\SWT\M\M\17\M\
\DTNNW\ Any number of concurrent forces can
be combined in this manner:\
\DFB\B2\81\90\
\XPLOT\87\95\
\XDRAW\30\-30\-3\0\3\0\0\3\
\SA\
\DFA\BALL1\77\90\
\SAS\75\\1\\
\SAL\16\2\2\-2\-2\
\DA\
\ is
seen when they are parallel but
pointing in opposite directions: 10 N - 7 N = 3 N\
\SQH\The resultant vector will be smallest if they are parallel but pointing in
opposite directions.\
\SQA\F\CONTAINS\-\No, the magnitude of a vector can
never be less than zero!\
\SQA\T\=\3\Right. With the two forces parallel
but pointing in opposite directions,
the magnitude is 10 N - 7 N = 3 N.\
\DQ\
\DCW\
\SQ\3\Y\
\SQM\3\
\SQF\S\3\
\SQT\Three vectors of magnitudes 4 N, 5 N, and 10 N are applied tMon Aug 13 14:47:00 1990
\L\acceleration\
\2\the rate of change of velocity.\
\L\average speed\
\1\the total displacement (the
straight-line distance from the
starting point to the final position) divided by elapsed time.\
\L\average acceleration\ nature of the two surfaces (how
"slippery" they are) and the magnitude
of the force holding the object to the
surface (normally its weight).\
\DCW\
\DTNNW\ In general, the starting friction,
F, is proportional to the force
holding the object
way.
When the applied force becomes great enough, the object will begin to move.
The applied force is now greater
than the starting friction.\
\SWS\1\
\DCW\
\DF\FRICTN2\1\1\1\0\
\SWT\\\R2\\1\
\DTN\ This starting friction depends on
theen the two objects and
opposite to the direction of the
applied force.\
\DTN\ The magnitude of the frictional
force is equal to the applied force
(Fa).\
\DCW\
\DTN\ There is a limit to the amount of
frictional force that can develop this , which
opposes the movement.\
\DTN\ Like all forces, frictional force
has both a magnitude and a direction.\
\DTN\ For an object at rest with a force
applied to one side, the direction of
the frictional force is parallel to the surface betwehen one is sliding over the
other, or when a force is acting which
tends to slide one over the other.\
\DTN\ When an object is at rest on a
surface and a small force (Fa) is
applied to one side of the object, a
frictional force (Ff) developsave disappeared.\
\DTN\ Very strange!\
\DCW\
\SFF\0\\
\DF\FRICTN1\1\1\1\0\
\SWT\\\R2\\1\
\DTN\ This mysterious force is friction.\
\DTN\ Friction is a force that acts to
resist movement.
It acts at the surface between two
objects, w Therefore, there must be an equal
and opposite force acting on it
simultaneously, so that the resultant
force is zero.\
\DTN\ When you stop pushing, the
refrigerator continues to sit there.
Therefore, that equal and opposite
force must hther:
F = (10 N) - (4 N) - (5 N) = 1 N\
\DQ\
\SQSEND\
\BESC\
\L\FRICTION FRICTION\
\SCH\2\2\
\DTCNW\Friction\
\SCH\1\1\
\DTN\ When you push on the side of a
refrigerator, you are applying a force
to it.
But it doesn't move.\
\DTN\ ng the two smaller vectors
in one direction and the largest one
in the other:
F = (10 N) - (4 N) - (5 N) = 1 N\
\SQA\T\=\1\Right. The smallest result is
obtained by pointing the two smaller
vectors in one direction and the
largest one in the opossible
magnitude for the resultant vector is ____ N.\
\SQH\The smallest result is obtained by
pointing the two smaller vectors in
one direction and the largest one in
the other.\
\SQANS\1\
\SQLAST\No, the smallest result is obtained
by pointi result is 19 N,
obtained when all three vectors point in the same direction; 10 N + 5 N + 4 N = 19 N.\
\DQ\
\DCW\
\SQ\3\Y\
\SQM\3\
\SQF\S\3\
\SQT\Three vectors of magnitudes 4 N, 5 N, and 10 N are applied to an object
concurrently. The smallest ude of the
resultant vector is equal to the sums of the individual magnitudes.\
\SQANS\19\
\SQLAST\No, the largest result is 19 N,
obtained when all three vectors point in the same direction; 10 N + 5 N + 4 N = 19 N.\
\SQA\T\=\19\Right. The largesto an object
concurrently. The largest possible
magnitude for the resultant vector is ___ N.\
\SQANS\19\
\SQH\The largest result will be obtained
if they all point in the same
direction.\
\SQH\If they all point in the same
direction, the magnit
\2\the total change of velocity divided
by the elapsed time.\
\L\coefficient of starting friction\
\3\the ratio of the force required to
start sliding motion to the normal
(perpendicular) force between the two surfaces.\
\L\concurrent forces\
\3to the surface (F):\
\SCH\2\3\
\DTC\F = F\
\SCH\1\1\
\DTN\ The proportionality factor, , (the
Greek letter, mu) is called the
coefficient of starting friction.
It is a function of the two surfaces between which sliding might occur.\-motion\Newton's first law of motionnewtonsecond%slopespeed starting friction tangent time'
unit
vector
velocity+accelerationdaverage speedaverage accelerationhcoefficient of starting frictionconcurrent forces;directiondistanceforceinstantaneous velocityXkilogrammagnitudemassfmetermilliular direction.\
exists or something occurs.\
\L\unit\
\1\the basic measure of a property, such as 1 meter, 1 sec, or meter per
second per second.\
\L\vector\
\1\a measurement that includes a
direction as well as a magnitude.\
\L\velocity\
\1\a speed in a particing friction\
\3\the maximum frictional force
resisting motion between two objects
in contact with one another.\
\L\tangent\
\1\a line drawn so as to touch a curve
at only a single point.\
\L\time\
\1\a measured period during which a
condition ne sixtieth of a minute.\
\L\slope\
\1\an indication of how steep a line is
- the ratio of the vertical
displacement to the horizontal
displacement along the line.\
\L\speed\
\1\the rate of movement, measured as
distance per unit time.\
\L\startton\
\2\the basic unit of force - the force
needed to accelerate a 1 kilogram
mass at a rate of 1 meter per second
per second.\
\3\a vector obtained through the
addition or subtraction of other
vectors.\
\L\second\
\1\the basic unit of time - oin
relation to objects that are
considered stationary.\
\L\Newton's first law of motion\
\2\an object at rest will remain at rest and an object in motion will remain
in motion with constant velocity
unless an unbalanced force acts upon
it.\
\L\newthe amount of material
in an object, defined by its inherent resistance to acceleration.\
\L\meter\
\1\the basic metric measure of length
(39.37 inches).\
\L\milli-\
\1\a prefix meaning "thousandth."\
\L\motion\
\1\the displacement of an object which it is measured approaches zero.\
\1\a prefix meaning "thousand."\
\L\kilogram\
\2\the basic metric unit of mass, equal
to 1,000 grams (2.2046 pounds).\
\L\magnitude\
\13\a numerical measure of size or
quantity.\
\L\mass\
\2\a measure of t resists the relative
motion of objects in contact with one another.\
\1\the magnitude component (size) of the instantaneous velocity vector.\
\L\instantaneous velocity\
\1\the limit which the average velocity
approaches as the time interval over
\forces simultaneously acting on a
body.\
\L\direction\
\1\the line or course along which an
object moves.\
\L\distance\
\1\the length of a line segment between
two points.\
\L\force\
\2\a push or pull exerted on an object.\
\3\the force tha
\SCH\2\3\
\DCW\
\DTNNW\
F
\
\DTNW\ = --
\
\DT\ F\
\SCH\1\1\
\DTN\ Notice that is the ratio of two
forces.
As a result, it has no units.\
\DTN\
If the starting force for a 5 kg
wooden cube sitting on a plank of wood
is 24.5 N,
= (24.5 N)/(5 kg)(9.8 m/s2)
= (24.5 N)/(49.0 N)
= 0.5
The coefficient of starting friction is one-half.\
\SWS\1\
\DCW\
\SCH\1\3\
\DTNNW\ F
= --
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