Lecture 5: Introduction to Physics PHY101Displacement and DistanceAverage Speed and VelocitySlide 4Instantaneous Velocity and SpeedConcept QuestionAccelerationSlide 8Slide 9Kinematics in One Dimension Constant AccelerationSlide 11Summary of ConceptsPhysics 101: Lecture 5, Pg 1Lecture 5: Introduction to Physics PHY101Chapter 2:•Distance and Displacement, Speed and Velocity (2.1,2.2)•Acceleration (2.3)•Equations of Kinematics for Constant Acceleration (2.4)Physics 101: Lecture 5, Pg 2Displacement and DistanceDisplacement and DistanceDisplacement is the vector that points from a body’s initial position, x0, to its final position, x. The length of the displacement vector is equal to the shortest distance between the two positions. x = x –x0Note:The length of x is (in general) not the same as distance traveled !Physics 101: Lecture 5, Pg 3Average Speed and VelocityAverage Speed and VelocityAverage speed is a measure of how fast an object moves on average: average speed = distance/elapsed timeAverage speed does not take into account the direction ofmotion from the initial and final position.Physics 101: Lecture 5, Pg 4Average Speed and VelocityAverage Speed and VelocityAverage velocity describes how the displacement of an object changes over time: average velocity = displacement/elapsed time vav = (x-x0) / (t-t0) = x / t Average velocity also takes into account the direction of motion.Note:The magnitude of vav is (in general) not the same as the average speed !Physics 101: Lecture 5, Pg 5Instantaneous Velocity and SpeedInstantaneous Velocity and SpeedAverage velocity and speed do not convey any information about how fast the object moves at a specific point in time.The velocity at an instant can be obtained from the average velocity by considering smaller and smaller time intervals, i.e. Instantaneous velocity: v = lim  t-> 0 x / t Instantaneous speed is the magnitude of v.Physics 101: Lecture 5, Pg 6Concept QuestionConcept QuestionIf the average velocity of a car during a trip along a straight road is positive, is it possible for the instantaneous velocity at some time during the trip to be negative? 1 - Yes 2 - No correctIf the driver has to put the car in reverse and back up some timeduring the trip, then the car has a negative velocity. However,since the car travels a distance from home in a certain amount oftime, the average velocity will be positive.Physics 101: Lecture 5, Pg 7AccelerationAccelerationAverage acceleration describes how the velocity of an object moving from the initial position to the final position changes on average over time: aav = (v-v0) / (t-t0) = v / tThe acceleration at an instant can be obtained from the average acceleration by considering smaller and smaller time intervals, i.e. Instantaneous acceleration: a = lim  t-> 0 v / tPhysics 101: Lecture 5, Pg 8Concept QuestionConcept QuestionIf the velocity of some object is not zero, can its acceleration ever be zero ? 1 - Yes 2 - No correctIf the object is moving at a constant velocity, then the acceleration is zero.Physics 101: Lecture 5, Pg 9Concept QuestionConcept QuestionIs it possible for an object to have a positive velocity at the same time as it has a negative acceleration? 1 - Yes 2 – NocorrectAn object, like a car, can be moving forwardgiving it a positive velocity, but then brake, causing deccelaration which is negative.Physics 101: Lecture 5, Pg 10Kinematics in One DimensionKinematics in One DimensionConstant AccelerationConstant AccelerationSimplifications:In one dimension all vectors in the previous equations can be replaced by their scalar component along one axis.For motion with constant acceleration, average andinstantaneous acceleration are equal.For motion with constant acceleration, the rate with whichvelocity changes is constant, i.e. does not change over time.The average velocity is then simply given asvav = (v0 +v)/2Physics 101: Lecture 5, Pg 11Kinematics in One DimensionKinematics in One DimensionConstant AccelerationConstant AccelerationConsider an object which moves from the initial position x0, at time t0 with velocity v0, with constant acceleration along a straight line. How does displacement and velocity of this object change with time ? a = (v-v0) / (t-t0) => v(t) = v0 + a (t-t0) (1) vav = (x-x0) / (t-t0) = (v+v0)/2 => x(t) = x0 + (t-t0) (v+v0)/2 (2) Use Eq. (1) to replace v in Eq.(2): x(t) = x0 + (t-t0) v0 + a/2 (t-t0) 2 (3)Use Eq. (1) to replace (t-t0) in Eq.(2): v2 = v02 + 2 a (x-x0 ) (4)Physics 101: Lecture 5, Pg 12Summary of Concepts Summary of Concepts kinematics: A description of motionposition: your coordinatesdisplacement: x = change of positionvelocity: rate of change of position•average : x/t•instantaneous: slope of x vs. tacceleration: rate of change of velocity•average: v/t•instantaneous: slope of v vs.