- NOW WE'LL LOOK AT AN EXAMPLE INVOLVING THE VELOCITY
AND ACCELERATION FUNCTIONS GIVEN THE POSITION FUNCTION.
WE'RE GIVEN THAT S OF T IS THE EQUATION OF MOTION
FOR A PARTICLE.
WE WANT TO FIND THE VELOCITY FUNCTION, V OF T.
WE WANT TO DETERMINE WHERE THE VELOCITY EQUALS 0
AND ALSO FIND THE ACCELERATION FUNCTION "A" OF T.
WELL THE VELOCITY FUNCTION MEASURES THE CHANGE
IN THE POSITION WITH RESPECT TO TIME SO THE VELOCITY FUNCTION,
B OF T IS EQUAL TO THE DERIVATIVE
OF THE POSITION FUNCTION.
AND SINCE THE POSITION FUNCTION IS A POLYNOMIAL FUNCTION,
WE CAN FIND THE DERIVATIVE USING THE POWER RULE
OF DIFFERENTIATION GIVEN HERE IN RED.
SO V OF T IS GOING TO BE EQUAL TO 6 x THE DERIVATIVE OF T,
THE THIRD, WHICH WOULD BE 3 T TO THE SECOND
- 45 x THE DERIVATIVE OF T TO THE SECOND
WHICH WOULD BE 2T TO THE FIRST + 108 x THE DERIVATIVE OF T
WHICH IS EQUAL TO 1 WHICH MEANS OUR VELOCITY FUNCTION, V OF T,
IS EQUAL TO 18, T TO THE SECOND - 90T + 108.
NEXT WE'RE GOING TO FIND WHEN THE VELOCITY EQUALS 0,
SO USING OUR VELOCITY FUNCTION,
WE'LL SET V OF T EQUAL TO 0 AND THEN SOLVE FOR T.
SO WE'D HAVE 0 = 18T SQUARED - 90T + 108.
LOOKS LIKE THIS IS GOING TO BE FACTORABLE.
WE'LL START BY FACTORING OUT THE GREATEST COMMON FACTOR,
WHICH IS 18.
SO WE'LL HAVE 0 = 18 x THE QUANTITY T SQUARED.
THERE'S 5 18'S IN 90 SO WE HAVE - 5T
AND THERE ARE 6 18'S IN 108, SO WE HAVE + 6.
THEN WE'LL FACTOR AGAIN.
SO FACTOR INTO 2 BINOMIAL FACTORS. WE HAVE T AND T.
THE FACTORS OF 6 THAT ADD THE -5 ARE -2 AND -3.
SO THIS PRODUCT HERE IS EQUAL TO 0 WHEN T = 2 OR WHEN T = 3.
WHICH IS WHEN THE VELOCITY IS EQUAL TO 0.
AND BEFORE WE TALK ABOUT THE ACCELERATION FUNCTION,
I'D LIKE TO TAKE A LOOK AT GRAPHICALLY WHAT WE JUST FOUND.
WELL HERE'S A GRAPH OF OUR POSITION FUNCTION.
NOTICE HOW IT'S A DEGREE 3 POLYNOMIAL
AND THEN WE FOUND THE VELOCITY FUNCTION
WHICH WAS A DEGREE 2 POLYNOMIAL OR A QUADRATIC FUNCTION
AS WE SEE HERE.
AND IF WE GRAPH BOTH OF THESE ON THE SAME COORDINATE PLANE,
NOTICE HOW WHERE THE DERIVATIVE FUNCTION IS EQUAL TO 0,
HERE AND HERE,
THE POSITION FUNCTION HAS EITHER A HIGH POINT OR A LOW POINT
WHICH REPRESENTS A RELATIVE MAXIMUM OR RELATIVE MINIMUM
OF THE POSITION FUNCTION.
SO NOTICE WHEN T IS EQUAL TO 2, WE'RE AT THIS HIGH POINT HERE,
WHERE THE POSITION FUNCTION HAS A RELATIVE MAXIMUM
AND WHEN T IS EQUAL TO 3,
WE'RE AT THIS LOW POINT HERE WHERE THE POSITION FUNCTION
HAS A RELATIVE MINIMUM.
AND ALSO BECAUSE THIS IS WHERE THE VELOCITY IS EQUAL TO 0,
WE CAN SEE THIS REPRESENTS THE INSTANT
WHEN THE PARTICLE CHANGES DIRECTION
CREATING A HIGH POINT OR LOW POINT IN THE POSITION FUNCTION.
AND NOW GOING BACK TO OUR LAST QUESTION,
WE'RE ASKED TO FIND THE ACCELERATION FUNCTION.
WELL THE ACCELERATION FUNCTION MEASURES THE RATE OF CHANGE
OF THE VELOCITY WITH RESPECTS TO TIME
WHICH MEANS ACCELERATION FUNCTION
IS A DERIVATIVE OF THE VELOCITY FUNCTION.
SO -- ACCELERATION FUNCTION,
IS EQUAL TO THE DERIVATIVE OF THE VELOCITY FUNCTION.
REMEMBER THE VELOCITY FUNCTION WAS A DERIVATIVE
OF THE ACCELERATION FUNCTION
SO WE CAN ALSO SAY THIS IS EQUAL TO THE SECOND DERIVATIVE
OF A POSITION FUNCTION.
BUT SINCE WE ALREADY HAVE THE VELOCITY FUNCTION HERE IN BLACK,
LET'S GO AHEAD AND USE THIS FUNCTION
TO FIND OUR ACCELERATION FUNCTION.
SO IF ACCELERATION FUNCTION
IS GOING TO BE EQUAL TO 18 x THE DERIVATIVE OF T SQUARED
WHICH IS 2T - 90 x THE DERIVATIVE OF T,
WHICH IS JUST 1 + THE DERIVATIVE OF 108,
THE DERIVATIVE OF A CONSTANT IS EQUAL TO 0.
SO THIS IS OUR ACCELERATION FUNCTION.
LET'S GO AHEAD AND SIMPLIFY THIS.
"A" OF T IS EQUAL TO 36T - 90.
OKAY, THAT'S GOING TO DO IT FOR THIS EXAMPLE.
I HOPE YOU FOUND THIS HELPFUL.