>> AND SO RHONI BARTON
TAKES IT ALL THE WAY HOME,
HERE ON THE SPORTSFIGURES 
CLASSIC SLALOMING 
CHAMPIONSHIP. 
>> OK, RHONI, IT SEEMED LIKE
YOU RE HAVING
A LITTLE TUBLE
OUT THERE TODAY.
YOU KEPT ZIGZAGGING ALL OVER
THE PLACE, BUT YOU MAGED
TO STAY UP.
>> WELL, IT'S SLALOM.
YOU'RE SUPPOSED TO ZIGZAG.
>> OH.
IS THAT WHAT ALL THOSE
LITTLE BALLS ARE FOR,
FOR YOU TO GO AROUND?
>> UH, YEAH.
>> NOW, THERE SEEMED TO BE
A TROUBLE WITH THE ROPE.
IT SEEMED TO GET SHORTER
EVERY ROUND.
DID IT KEEP BREAKING?
>> NO.FTER YOU SUCCESSFULLY
COMPLETE A PASS,
YOU STOP ON THE END
AND THE JUDGE ORTENS THE ROPE
TO MAKE IT MORE DIFFICULT.
>> OH, WOW.
>> HAVE YOU EVER COVERED
WATERSKIINBEFORE?
>> WELL, ACTUALLY,
SCNCE IS MORE MY THING.
>> SPORTSFIGURES. 
PUT YOUR BRAIN IN THE GAME.
>> IN THE OLD DAYS,
PEOPLE USED TO WATER-SKI
LIKE THIS.
 NOWADAYS, THEY WATER-SKI
LIKE THIS.
>> IF YOU'VE EVER WATER-SKIED,
OR WATCHED SOMEONE WATER-SKI,
YOU MIGHT HAVE NOTICEDHIS.
>>EE?
RHONI CAUGHT UP WITH THE BOAT.
AT SHOULD BE
IMPOSSIBLE, RIGHT?
I MEAN, THE BOAT IS WHAT'S
PULLING HER.
TO CATCH UP LIKE THAT,
SHE'D HAVE TO BE GOING
FASTER THAN THE BOAT.
>> WEI, RIGHT?
HOW CAN A WATER-SKIER BE GOING
FASTER THAN THE BOAT THAS
TOWING THEM?
AND HOW MUCH FASTER CAN SHE GO?
>> HEY, RHONI, COULD YOU GIVE US
SOME OF THE SICS OF SLALOM?
HOW DOES IT WORK?
>> OK, WELL, THE COURSE IS
MADE UP OF ENTRANCE GATES,
EXIT GATES, AND 6 BUOYS.
THE SKIER HAS TO GO THROUGH
THE GATES,ROUND ALL 6 BUOYS,
AND OUT THE EXIT GATES.
>>T SOMETIMES SEEMS LIKE
YOU'RE GOING FASTER THAN
THE BOAT.
IS THAT AN OPTICAL ILLUSION?
>> NO.
IN FACT, WHEN WE'RE CUTTING
THROUGH THE WAKES,
WE CAN ALMOST DOUBLE O SPEED,
AND THE SPEED THAT WE GENERATE
GOING THROH THE WAKES
GIVES US ENOUGH SPEED
UP ON THE BUOY THAT WE'RE
ACTUALLY FREE FROM
THE PULLF THE BOAT.
>> THE BOAT IS TRAVELING AT
A CONSTANT VELOCITY.
WE KNOW THAT WHEN RHONI IS
DIRECTLY BEHIND THE BOAT,
AND NOT SWINGING OUT
TO EITHER SIDE,
SHE'S  A STATE OF EQUILIBRIUM.
THAT MEANS THAT ALL THE FORCES
AND TORQUES ACTING ON R
D UP TO ZERO.
THERE'S NO NET FORCE,
AND WE KW THIS BECAUSE
THERE'S NO ACCELERATION.
>> OK, GUYS, SO WHAT DOES IT
MEAN THAT THERE'S
NO ACCELERATION?
A SKIER  MOVING.
>> SOMETHING CAN BE MOVING
WITHOUT ACCELERATING.
>> ACCELATION MEANS
A CHANGE IN SPEED OR DIRECTION.
>> AT A CONSTANT SPEED,
AND JUST GOING STRAIGHT
BEHIND THE BOAT,
THE SKIER'S NOT ACCELERATING.
>> THE ONLY TIME THERE'S
ACCELERATION IS WHEN THE FORCES
DON'T ADD UP TO ZERO.
THEY'RE OUT OFALANCE.
>> LET'S LOOK AT OUR FORCE
PROBLEM FROM DIRECTLY ABOVE.
WE'LL PUT IT ON THE i-j AXES,
BECAUSE WE'RE WORKING WITH
FORCE VECTORS.
EVERYTHING'S IN EQUILIBRIUM,
OR BALANCE.
T'S SAY THE TOW ROPE
EXERTS A FORCE OF ZERO i PLUS
50j ON THE SKIER,
AND THE WATER EXERTS
A FORCE OF ZERO i
PLUS -50j.
50 PLUS -50 EQUALS ZERO.
WITH THE BOAT TRAVELING AT
A CONSTANT VELOCITY,
THE FORCES CANCEL OUT,
AND THE SKIER
DOESN'T ACCELERATE.
BY LEANING BACK AND TURNING,
THE SKIER DIGS THE SKI DEEPER
INTO THE WATER,
AND THE WATER CREATES A GREATER
FORCE AGAINST THE SKI.
NOW SHE'S TRAVELING SIDEWAYS
IN RELATION TO THE BOAT.
NOW THERE'S A GREATER FORCE
FROM THE WATER AND A GREATER
FORCE FROM THE ROPE.
LES SAY 0i PLUS 80j.
THE WATER'S FORCE HAS TO
EQUAL THAT, BUT THE SKI IS
AT AN ANGLE TO THE TOW ROPE
FORCE, RIGHT?
SO, TO FIGURE IT OUT, WE
HAVE TO BREAK THE WATER'S VECTOR
DOWNNTO ITS VERTICAL
AND HORIZONTAL COMPONENTS.
THE WATER'S FORCE VECTOR
WILL BE 40i, -70j.
>> I THOUGHT YOU SAID THAT
IT HAD TO EQUAL THE FORCE
FROM THE TOW ROPE.
>> BUT YOU CAN'T ADD TM
THAT WAY.
BECAUSE YOU'RE MAKING
A TRIANGLE, YOU HAVE TO USE
THE PYTHAGOREAN EOREM.
>> OUR ANGLED FORCE IS
THE RESULTANT VECT OF OUR
HORIZONTAL AND VERTICAL FORCES.
ALL GETHER, THEY FORM
A RIGHT TRIANGLE, WITH
THE RESULTANT AS A HYPOTENUSE.
SO WE HAVE TO USE "c" SQUARED
EQUALS "a" SQUARED
PLUS "b" SQUARED.
FOR OUR "a" SIDE, WE GET
40i, AND FOR OUR "b" SIDE WE GET
-7, AND SOLVING FOR THEM,
WE CAN GET ABOUT 80
AS OUR RESULTANT.
NOW WE CAN ADD THE i AND j
COMPONENTS OF OUR VECTORS.
>> OK, WE HAVE THE ROPE
PULLING WITH THE VECTOR OF
0i PLUS 80j, AND THE WATER'S
EXERTING A FORCE ON THE SKI
OF 40i PLUS -70j.
THEY'RE DIFFERENT, RIGHT?
>> WHILE THE WATER'S EXERTING
A FORCE TOWARDHE OUTSIDE,
THE ROPE DOESN'T EXERT
ANY FORCE AGAINST THE i AXIS
TO BALANCE IT OUT.
>> AND THERE'S AN IMBALANCE
 THE j AXIS OF NET FORCE
TOWARD THE BOAT,
AND WHEN YOU HAVE FORCES
OUT OF BALANCE?
>> YOU GET AN ACCELERATION.
>> BEFORE, WN WE ADDED
OUR VECTORS,E GOT A SUM OF
ZERO, RIGHT?
EVERYTHING WAS IN BALANCE.
NOW WHEN WE D OUR VECTORS,
THE SUM ISN'T ZERO.
THEY DON'T BALANCE OUT.
WE GET 40i PS 10j.
THERE'S A NET FORCE ON
THE i AND j AXES.
THAT MEANS TOWARD THE BOAT
AND TO THE OUTSIDE.
THE FORCE VECTOR CONSTANTLY
CHANGES AS A SKIER COMES UP
EVEN WITH THE BOAT.
WATCH THE ROPE'S VECTOR.
EVENTUALLY, ERE'S NO NET FORCE
IN TOWARD THE BOAT.
>> WITNO FORCE FROM THE BOAT,
THE SKI CAN'T PUSH BACK
AGAINST THE WATER WITH
AN OPPOSITE FOE.
NOT ONLY IS THAT FORCE WHAT
ACCELERATES HER, IT'S WHAT
KEEPS HER UP.
SOY, RHONI.
SO, NOW WE KNOW WHY
A WATER-SKIER ACLERATES.
BUT HOW MUCH
CATHEY ACCELERATE?
>> WELL, SPEED IS A MEASURE OF
DISTANCE OVER TIME.
>> SO WE NEED TO FIGURE OUT
HOW FAR A SKIER IS GOING
IN HOW LONG A ME.
>> OK, SO FIRST WE NEED TO
FIGURE OUT A WATER-SKIER PATH,
AND THEN HOW LONG IT IS.
>> A WATER-SKIER IS JUST LIKE
THE BALL AT THE END OF A STRING.
RHONI IS DESCRIBING THE ARC OF
A CIRCLE IN THE WATER, RIGHT?
AND JUST LIKE THE BALL
AT THE END OF THIS STRING,
THE ROPE LENH STAYS CONSTANT
AND FORMS THE RADIUS
OF THE CIRCLE.
TO FIND HER SPEED,
WE NEED DISTANCE AND TIME.
WE NEED TO FIND OUT
THE DIMENSIONS OF THE CIRCLE
IN ORDER TO FIND OUT
THE DISTANCE RHONI'S TRAVELED.
LET'S E THE FIRST ROUND
ROPE LENGTH OF 75 FEET.
>> THIS IS THE TOTAL CIRCLE
THAT OUR SKIER COULD MAKE.
AND, WE KNOW THE TOT DISTANCE
AROUND A CIRCLE IS CALLED
THE CIRCUMFERENCE.
NOW, WE'RE OY INTERESTED IN
THIS PART OF THE CIRCLE HERE,
WHERE THE SKIER ACCELERATES.
SO, TO FIND THAT DISTANCE IS
REALLY SIMPLE.
WE JUST TAKE OUR FORMULA
FOR CIRCUMFERENCE, 2 TIMES PI
TIMES RADIUS,
AND THEN DIVIDE BY 4.
>> THE RADIUS OF OUR CIRCLE
IS THE LENGTH OF THE ROPE,
75 FEET.
SO WE JUST MULTIPLY 2 TIMES PI,
3.14, TIMES 75, AND WE GET
A TOTAL CIRCUMFERENCE OF
471 FEET.
NOW WE TAKE 1/4 OF
OUR CIRCLE, DIVIDE BY 4,
AND GET ABOUT 118 FEET THAT
THE SKIER TRAVELS.
>> BECAUSE SPEED IS ABOUT
DISTANCE AND TIME, WE NEED TO
FIND OUT HOW LONG IT TAKES
RHONI TO GO THIS DISTANCE.
AND WE CAN DO THAT REALLY EASILY
BY COUNTING VIDEO FRAMES.
>> VIDEO RECORDS AT 30 FRAMES
A SECOND, SO ALL WE HAVE TO DO
IS COUNT THE NUMBER OF FRAMES
AND DIVIDE BY .
IT ONLY TAKES 53 FRAMES
FOR RHONI TO REACH THE CENTER
OF THE WAKE.
LET'S CALL THAT 1.75 SECONDS.
>> OK, SO NOW WE HAVE
THE DISTANCE AND THE TIME.
WHAT'S NEX
>> WELL, NOW WE CAN FIND
THE SPEED, THE VELOCITY.
>> THE VELOCITY EQUALS
DISTANCE DIVED BY TIME.
>> HOW LONG IT TAKES YOU
TO GO A DISTANCE.
>> OK, THAT MAKES SENSE.
TO GET THE SKIER'S
AVERAGE VELOCITY, WE JUS
DIVIDE DISTANCE BY TIME.
118 FEET DIVIDED BY 1.75ECONDS
EQUALS 67 FEET PER SOND.
67 FEET PER SECOND?
THAT'S PRETTY FAST.
THAT'S 45 MILES PER HOUR.
HEY, THAT WAS
A STOP SIGN, BUDDY.
67 FEET PER SECOND IS
RHONI'S AVERAGE SPEED
OVER THIS DISTANCE.
SHLL BE GOING SLOWER
AT THE START AND FASTER WHEN
SHE HITS THE WAKE.
THE TOW BOAT STAYS AT
A CONSTANT SPEED OF 34 MILES
PER HOUR.
THAT'S ABO 50 FEET PER SECOND.
RHONI HAS AN AVERAGE SPEED OF
45 MILES PER HOUR,
67 FEET PER SECOND.
 CAN FIGURE OUT
WHAT PERCENTAGE OF
THE BOAT'S SPEED THAT IS
BY JUST DIVIDING,
AND WE FIND THAT RHONI
GETS GOING AROUND 75% FASTER
TH THE BOAT.
>> BUT SOMETIMES I GET GOING
WAY FASTERHAN 45 MILES
AN HOUR.
>> YEAH, THAT'S JUST
YOUR AVERAGE SPEED
OVERHAT DISTANCE.
WHEN YOU HIT THE WAKE,
YOU COULD BE GOING 55 TO 60
MILES PER HOUR.
[ELEPHANT TRUMPETS]
SO WE HAVE WHY A SKIER
ACCELERATES, HOW FAST THEY
GET GOING, AND HOW MUCH
THEY ACCELERATE.
WHOO!
WHO THOUGHT WATERSKIING
WOULD BE SO MPLICATED?
OR SO T?
[ELEPHANT TRUMPETS]
OK, GUYS, THAT WAS A LOT.
SO, WHAT HAVE WE LEARNED?
>> THAT A WATER-SKIER
ACCELERATES BECAUSE THFORCES
ARE OUT OF BALANCE,
BECAUSE THEY DIG THEIR SKI
INTO THE WAT.
>> YEAH, AND WHENEVER FORCES
ARE OUT OF BALANCE,
SOMETHING ACCELERATES.
 AND ACCELERATION CAN CHANGE
IN SPEED OR DIRECTION.
>>ND YOU CAN FIGURE IT OUT
WITH THE FORCE VECTORS.
>> WE LEARNED THAT YOU CAN FIND
AN OBJECT'S VELOCITY BY DIVIDING
DISTANCE BY TIME.
>> RIGHT.
WELL, THAT'S IT.
WE'D LIKE TO THANK RHONI BARTON
AND OUR STUDENTS,
KARA, MATT, CHRISTINA,
AND HEIDI FOR HELPING US OUT
HERE TODAY ON
ESPN SPORTSFIGURES: 
WATERSKIING IN CIRCLES. 
WHOO!
