# KINEMATICS The Study of Motion Copyright Sautter 2003.

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KINEMATICS The Study of Motion Copyright Sautter 2003

Measuring Motion The study kinematics requires the measurement of three properties of motion. (1) displacement – the straight line distance between two points (a vector quantity) (2) velocity – the change in displacement with respect to time (a vector quantity) (3) acceleration – the change in velocity with respect to time (a vector quantity) The term distance like displacement, refers to the change in position between two points, but not in a straight line. Distance is a scalar quantity. Speed refers to change in position with respect to time but unlike velocity, does not require straight line motion. Speed is a scalar quantity.

Speed ( a scalar ) & Velocity ( a vector )
Velocity = Displacement from A to B/ time Distance traveled from A to B x B Lake Tranquility Displacement from A to B A x Speed = Distance from A to B/ time

VELOCITY & ACCELERATION
OBJECTS IN MOTION MAY MOVE AT CONSTANT VELOCITY (COVERING EQUAL DISPLACEMENTS IN EQUAL TIMES) OR BE ACCELERATED (COVER INCREASING OR DECREASING DISPLACEMENTS IN EQUAL TIMES). VELOCITY MEASUREMENTS MAY BE OF TWO TYPES, AVERAGE VELOCITY (VELOCITY OVER A LARGE INTERVAL TIME) OR INSTANTANEOUS VELOCITY (VELOCITY OVER A VERY SHORT INTERVAL OF TIME). ACCELERATION MAY BE UNIFORM OR NON UNIFORM. UNIFORM OR CONSTANT ACCELERATION REQUIRES THAT THE VELOCITY INCREASE OR DECREASE AT A CONSTANT RATE WHILE NON UNIFORM ACCELERATION DISPLAYS NO REGULAR PATTERN OF CHANGE.

Uniformly Accelerated Motion
Constant Velocity 1 sec 2 sec 3sec 4sec 5 sec EQUAL DISPLACEMENTS IN EQUAL TIMES Uniformly Accelerated Motion 1 sec 2 sec 3sec 4sec CLICK HERE REGULARLY INCREASING DISPLACEMENTS IN EQUAL TIMES

Displacement vs Time for a Uniformily Accelerated Body S POSITIVE
ACCELERATION Equal time intervals result in increasingly larger displacements S S t t t time

Average Velocity for a Uniformily
Accelerated Body D I S P L A C E M N T Average velocity between t1 and t2 Is the slope of the Secant line = S/ t s2 S Secant line s1 t t1 t2 time

Instantaneous Velocity for a Uniformily Accelerated Body
Draw a tangent line at the point Finding velocity at point t1, s1 (instantaneous velocity) D I S P L A C E M N T S Find the slope of the tangent line s1 t Instantaneous velocity equals the slope of the tangent line t1 time

DISPLACEMENT, VELOCITY & CONSTANT ACCELERATION
The velocity of an object at an instant can be found by determining the slope of a tangent line drawn at a point to a graph of displacement versus time for the object. If several instantaneous velocities are found and plotted against time the graph of velocity versus time is a straight line if the object is experiencing constant acceleration. The slope of the straight line velocity versus time graph is constant and since acceleration can be determined by the slope of a velocity – time graph, the acceleration is constant. The graph acceleration versus time for a constant acceleration system is a horizontal line. (A slope of zero since constant acceleration means that acceleration is not changing with time!)

Finding Velocity & Acceleration from Displacement vs time
Slope of a tangent drawn to a point on a displacement vs time graph gives the instantaneous velocity at that point A C E L R T I O N Time PLOT OF INSTANTANEOUS VELOCITIES VS TIME V E L O C I T Y Time v Slope of a tangent drawn to a point on a velocity vs time graph gives the instantaneous acceleration at that point t

MEASURING VELOCITY & ACCELERATION
VELOCITY IS MEASURED AS DISPLACEMENT PER TIME. UNIT FOR THE MEASUREMENT OF VELOCITY DEPEND ON THE SYSTEM USED. IN THE MKS SYSTEM (METERS, KILOGRAMS, SECONDS) IT IS DESCRIBED IN METERS PER SECOND. IN THE CGS SYSTEM (CENTIMETERS, GRAMS, SECONDS - ALSO METRIC) IT IS MEASURED IN CENTIMETERS PER SECOND. IN THE ENGLISH SYSTEM IT IS MEASURED AS FEET PER SECOND. ACCELERATION IN THE MKS SYSTEM IS EXPRESSED AS METERS PER SECOND PER SECOND OR METERS PER SECOND SQUARED. IN CGS UNITS IT IS CENTIMETERS PER SECOND PER SECOND OR CENTIMETERS PER SECOND SQUARED. IN THE ENGLISH SYSTEM FEET PER SECOND PER SECOND OR FEET PER SECOND SQUARED ARE USED.

GRAVITY & CONSTANT ACCELERATION
Gravity is the most common constant acceleration system on earth. As object fall under the influence of gravity (free fall) they continually increase in velocity until a terminal velocity is reached. Terminal velocity refers to the limiting velocity caused by air resistance. In an airless environment the acceleration provided by gravity would allow a falling object to increase in velocity without limit until the object landed. In most problems in basic physics air resistance is ignored. In actuality, terminal velocity is related to air density, surface area, the velocity of the object and the aerodynamics of the object (the drag coefficient).

Acceleration Due to Gravity & Free Fall
g = 9.8 meters / second 2 g = 980 centimeters / second 2 g = 32 feet / second 2 CLICK HERE

Free Fall 78.4 m 44.1 m 19.6 m 19.6 m/s 2.0 sec 29.4 m/s 3.0 sec

CALCULATING AVERAGE VELOCITY
Average velocity for an object moving with uniform (constant) acceleration can be calculated in two ways. (1) average velocity = the change in displacement (displacement traveled, s) divided by the change in time ( t). (s is the symbol used for displacement) (2) average velocity = the sum of two velocities divided by two (an arithematic average). V = s t ave 2 1 v v V = ave

CALCULATING INSTANTANEOUS VELOCITY
Instantaneous velocity can be found by taking the slope of a tangent line at a point on a displacement vs. time graph (as previously discussed). Instantaneous velocity can also be determined from an acceleration vs. time graph by determining the area under the curve. For constant acceleration systems, the acceleration times the time (a x t) plus the original velocity (v0) also gives the instantaneous velocity. V = V + a t o i

GIVES THE INSTANTANEOUS
Area Under an Acceleration vs. Time Curve Gives the Instantaneous Velocity AREA UNDER THE CURVE (acceleration x time) GIVES THE INSTANTANEOUS VELOCITY AT TIME t1 A C E L R T I O N t1 Time

CALCULATING DISPLACEMENT
Displacement of a body in constant acceleration can be found in two ways. Displacement is given by the area under a velocity vs. time graph. Displacement can also be found using the follow equation where si = instantaneous displacement, vo = the original velocity of the object, a = the constant acceleration and t = elapsed time. s = v t a t o 1/2 2 i

GIVES THE INSTANTANEOUS
Area Under a Velocity vs. Time Curve Gives the Instantaneous Displacement AREA UNDER THE CURVE (velocity x time) GIVES THE INSTANTANEOUS DISPLACEMENT AT TIME t1 V E L O C I T Y Time t1

CALCULATING VELOCITY & ACCELERATION FROM DISPLACEMENT VS. TIME
The instantaneous velocity of an object can be found from a displacement versus time graph by measuring the slopes of tangent lines drawn to points on the graph. Since the derivate of an equation gives the formula for calculating slopes, the derivative of the displacement versus time equation will give the equation for velocity versus time. Additionally, the slope of an velocity versus time curve is the acceleration. Therefore, the derivative of the velocity versus time equation gives the acceleration versus time relationship.

Velocity from Displacement vs. time
s = v t a t o 1/2 2 i The first derivative of displacement versus time gives the instantaneous velocity in terms of time. 1 -1 = 0 s = v t a t o 1/2 2 i v = d / dt - 1 = 1 1 x 2 x v = v a t o i

Acceleration from velocity vs. time
v = v a t o i The first derivative of velocity versus time gives the instantaneous acceleration in terms of time. t 1 - 1 = 0 v = v a t o i a = d / d t 0 x 1 x a = a ( 1 ) acceleration is constant

CALCULATING VELOCITY & DISPLACEMENT FROM ACCELERATION
The instantaneous velocity of an object can be determined from the area under an acceleration versus time graph. Since the integration of an acceleration versus time equation gives the area under the curve, it also gives the velocity. The area under a velocity versus time graph gives the displacement. Therefore, the integral of the velocity versus time equation gives the displacement versus time equation.

Velocity from Acceleration vs. time
a = a x t V = a x t dt i The constant is the original velocity (V0) 0 + 1 v = a x t i + C 0 + 1 v = a t + v i

 Displacement from Velocity vs. time v = a t + v i s = a t + v i
s = a t v i ( t ) dt s = a t v t C 1 + 1 0 + 1 i s = v t a t o 1/2 2 i + C The constant C is the original displacement of the object If displacement is not measured from zero

Displacement, Velocity & Acceleration
vs time Acceleration vs. time slope Velocity vs time slope Area under curve Area under curve derivative derivative Acceleration vs. time Displacement vs time Velocity vs time integral integral

ACCELERATED MOTION SUMMARY
VAVERAGE = s/ t = (V2 + V1) / 2 VINST. = VORIGINAL + at SINST = V0 t + ½ at2 Instantaneous velocity at a point equals the slope of a tangent line drawn at that time point on a displacement vs. time graph Instantaneous acceleration at a point equals the slope of a tangent line drawn at that time point on a velocity vs. time graph. The derivative of a displacement vs. time equation gives the instantaneous velocity. The derivative of a velocity vs. time equation gives the instantaneous acceleration. The integral of an acceleration vs. time equation gives the instantaneous velocity. The integral of an velocity vs. time equation gives the instantaneous displacement.

PUTTING EQUATIONS TOGETHER
Often problems involving uniformly accelerated motion do not contain a time value. When this occurs these problems can be solved by combining equations which are already known. To simplify the algebra, V0 will assumed to be zero. Therefore, Vi = VO + at becomes Vi = at and Si= V0 t + ½ at2 becomes Si = ½ at2. Solving Vi = at for t we get t =Vi/a. Substituting into Si = ½ at2 gives Si = ½ a(Vi/a)2 or by simplifying the equation Si = ½ (Vi 2/a) If V0 is not equal to zero the equation becomes Si = ½ (Vi 2 – Vo2) /a (time is not required to solve this equation!)

In the next program the equations and relationships developed here will be used to solve one dimensional, uniform acceleration problems. Free fall problems will be included since they are the most common examples of constant acceleration . Problem involving variable acceleration and use to derivatives and integrals for their solution will be covered. Math concepts are required and if the program on Math for Physics has not yet been viewed, it may be a good idea to do so!

The End

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