Presentation on theme: "KINEMATICS The Study of Motion Copyright Sautter 2003."— Presentation transcript:
1KINEMATICSThe Study of MotionCopyright Sautter 2003
2Measuring MotionThe 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.
3Speed ( a scalar ) & Velocity ( a vector ) Velocity = Displacement from A to B/ timeDistance traveledfrom A to Bx BLake TranquilityDisplacementfrom A to BA xSpeed = Distance from A to B/ time
4VELOCITY & 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.
5Uniformly Accelerated Motion Constant Velocity1 sec2 sec3sec4sec5 secEQUAL DISPLACEMENTS IN EQUAL TIMESUniformly Accelerated Motion1 sec2 sec3sec4secCLICKHEREREGULARLY INCREASING DISPLACEMENTS IN EQUAL TIMES
6Displacement vs Time for a Uniformily Accelerated Body S POSITIVE ACCELERATIONEqual timeintervals resultin increasinglylarger displacementsSSttttime
7Average Velocity for a Uniformily Accelerated BodyDISPLACEMNTAverage velocitybetween t1 and t2Is the slope of theSecant line =S/ts2SSecantlines1tt1t2time
8Instantaneous Velocity for a Uniformily Accelerated Body Draw a tangent line at the pointFinding velocityat point t1, s1(instantaneous velocity)DISPLACEMNTSFind the slope ofthe tangent lines1tInstantaneous velocityequals the slope ofthe tangent linet1time
9DISPLACEMENT, 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!)
10Finding Velocity & Acceleration from Displacement vs time Slope of a tangent drawn to a point ona displacement vs time graph givesthe instantaneous velocity at that pointACELRTIONTimePLOT OF INSTANTANEOUSVELOCITIES VS TIMEVELOCITYTimevSlope of a tangent drawn to a point ona velocity vs time graph gives theinstantaneous acceleration at that pointt
11MEASURING 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.
12GRAVITY & 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).
13Acceleration Due to Gravity & Free Fall g = 9.8 meters / second2g = 980 centimeters / second2g = 32 feet / second2CLICKHERE
14Free Fall 78.4 m 44.1 m 19.6 m 19.6 m/s 2.0 sec 29.4 m/s 3.0 sec
15CALCULATING 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 tave21v vV =ave
16CALCULATING 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 toi
17GIVES THE INSTANTANEOUS Area Under an Acceleration vs. TimeCurve Gives theInstantaneous VelocityAREA UNDER THE CURVE(acceleration x time)GIVES THE INSTANTANEOUSVELOCITY AT TIME t1ACELRTIONt1Time
18CALCULATING 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 to1/22i
19GIVES THE INSTANTANEOUS Area Under a Velocity vs. TimeCurve Gives theInstantaneous DisplacementAREA UNDER THE CURVE(velocity x time)GIVES THE INSTANTANEOUSDISPLACEMENT AT TIME t1VELOCITYTimet1
20CALCULATING 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.
21Velocity from Displacement vs. time s = v t a to1/22iThe first derivative of displacement versus timegives the instantaneous velocity in terms of time.1 -1 = 0s = v t a to1/22iv = d/ dt- 1 = 11 x2 xv = v a toi
22Acceleration from velocity vs. time v = v a toiThe first derivative of velocity versus time givesthe instantaneous acceleration in terms of time.t1 - 1 = 0v = v a toia = d / d t0 x1 xa = a ( 1 ) acceleration is constant
23CALCULATING 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.
24Velocity from Acceleration vs. time a = a x tV = a x t dtiThe constantis the originalvelocity (V0)0 + 1v = a x ti+ C0 + 1v = a t + vi
25 Displacement from Velocity vs. time v = a t + v i s = a t + v i s = a t vi( t ) dts = a t v t C1 + 10 + 1is = v t a to1/22i+ CThe constant C is the original displacement of the objectIf displacement is not measured from zero
27ACCELERATED MOTION SUMMARY VAVERAGE = s/ t = (V2 + V1) / 2VINST. = VORIGINAL + atSINST = V0 t + ½ at2Instantaneous velocity at a point equals the slope of a tangent line drawn at that time point on a displacement vs. time graphInstantaneous 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.
28PUTTING 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 becomesSi = ½ (Vi 2 – Vo2) /a (time is not required to solve this equation!)
29In 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!