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UB PHY 101 - Lecture 5 - Introduction to Physics

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Lecture 5: Introduction to Physics PHY101Summary of concepts from last lectureConcept QuestionSlide 4Kinematics in One Dimension Constant AccelerationApplication of Eqs. of KinematicsSlide 7Free FallConcept QuestionSlide 10Slide 11Slide 12Kinematics in Two Dimensions Constant AccelerationPhysics 101: Lecture 6, Pg 1Lecture 5Lecture 5: Introduction to Physics: Introduction to PhysicsPHY101PHY101Chapter 2:Equations of Kinematics for Constant Acceleration in 1 Dim. (2.4, 2.5, 2.7)Free Fall (2.6)Chapter 3:Equations of Kinematics for Constant Acceleration in 2 Dim. (3.1, 3.2)Physics 101: Lecture 6, Pg 2Summary of concepts from last Summary of concepts from last lecturelectureposition: your coordinates (just “x” in 1-D)displacement: x = change of position velocity: rate of change of positionaverage : x/tinstantaneous: slope of x vs. t : lim t->0 x/tacceleration: rate of change of velocityaverage: v/tinstantaneous: slope of v vs. t : lim t->0 v/tPhysics 101: Lecture 6, Pg 3Concept Question Concept Question A car is moving along the negative x direction. During part of the trip, the speed increases from 16 to 28 m/s in 1.5 s. Which of the following is correct?1) v>0, a>02) v>0, a<03) v<0, a>04) v<0, a<0•During another part of the trip, the speed decreases from 48 to 26 m/s in 11 s. Which of the following is correct?1) v>0, a>02) v>0, a<03) v<0, a>04) v<0, a<0If speed is increasing, v and a are in same direction.If speed is decreasing, v and a are in opposite direction. correctva+x correctva+xPhysics 101: Lecture 6, Pg 4Which of the following statements is most nearly correct?1 - A car travels around a circular track with constant velocity.2 - A car travels around a circular track with constant speed.3- Both statements are equally correct.Concept QuestionConcept QuestioncorrectOn a circular track, the direction that the car is traveling in is always changing, and since velocity takes into account the direction of travel, the velocity is always changing. Speed, however, is independent of direction and so the speed can stay constant.Physics 101: Lecture 6, Pg 5Kinematics 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 ? aav=a = (v-v0) / (t-t0) => v(t) = v0 + a (t-t0) (1) vav = (x-x0) / (t-t0) = (v+v0)/2 => x = 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 6, Pg 6Application of Eqs. of KinematicsApplication of Eqs. of KinematicsA runner accelerates to a velocity of 5.36 m/s due west in 3.00 s. His average velocity is 0.640 m/s2 due west. What was his velocity when he began accelerating ? [Chapter 2, problem #15]t0= 0 s, v= -5.36 m/s, t=3.00 s, aav=-0.640 m/s2 v0 = ? m/s aav = (v-v0)/(t-t0) => v0= v- aav (t-t0) = -3.44 m/s v0 = 3.44 m/s due westPhysics 101: Lecture 6, Pg 7Application of Eqs. of KinematicsApplication of Eqs. of KinematicsA drag racer starting from rest, speeds up for 402 m with a=+17 m/s2. A parachute then opens, slowing the car down with a=-6.10 m/s2. How fast is the racer after moving 3.50 x 102 m after the parachute opens ? [2-28] 1. Before the parachute opens (car moves +x direction): t0= 0 s, v01 = 0 m/s, x1=+402 m, a1=+17 m/s22. After the parachute opens:t0= 0 s, x2=+3.50 x 102 m, a2=-6.10 m/s2, v=? m/sv2=v022+2 a2 x2 Get v022 from 1.: v02=(2 a1 x1 )1/2=+117 m/s=> v2=(v02+2 a 2 x2)1/2=+96.9 m/sPhysics 101: Lecture 6, Pg 8Free FallFree FallFree fall is the idealized description of the motion of a downward falling body due to gravity:  Air resistance is neglected Acceleration due to gravity is considered to be constantThe acceleration due to gravity is always pointing downward with magnitude g=9.80 m/s2.Physics 101: Lecture 6, Pg 9Concept QuestionConcept QuestionAn object is dropped from rest. If it falls a distance D in time t then how far will it fall in a time 2t ? 1. D/4 2. D/2 3. D 4. 2D 5. 4DCorrect x = 1/2at2Followup question: If the object has speed v at time t then what is the speed at time 2t ? 1. v/4 2. v/2 3. v 4. 2v 5. 4vCorrect v=atPhysics 101: Lecture 6, Pg 10Concept QuestionConcept QuestionA ball is thrown vertically upward. At the very top of its trajectory, which of the following statements is true: 1. velocity is zero and acceleration is zero2. velocity is not zero and acceleration is zero3. velocity is zero and acceleration is not zero4. velocity is not zero and acceleration is not zero correctThe velocity vector changes from moment to moment, buts its acceleration vector does not change. Though the velocity at the top is zero, the acceleration is still constant because the velocity is changing.Physics 101: Lecture 6, Pg 11Free FallFree FallA wrecking ball is hanging from rest from a crane when suddenly the cable breaks. The time it takes the ball to fly half way to the ground is 1.2 s. Find the time for the ball to fall from rest all the way to the ground. [2-45]1. Half way to the ground (-y direction)t0= 0 s, v0 = 0 m/s, t=1.2 s, a=-9.80 m/s2Y1/2=v0 t + ½ a t2 = -7.1 m2. From rest all the way to the ground, y=2 Y1/2t0= 0 s, v0 = 0 m/s, a=-9.80 m/s2, t= ? sY=v0 t + ½ a t2 = ½ a t2 => t= (2 y/a)1/2=1.7 sPhysics 101: Lecture 6, Pg 12Dennis and Carmen are standing on the edge of a cliff. Dennis throws a basketball vertically upward, and at the same time Carmen throws a basketball vertically downward with the same initial speed. You are standing below the cliff observing this strange behavior. Whose ball is moving fastest when it hits the ground? 1. Dennis' ball 2. Carmen's ball 3. Samevv00vv00DennisDennisCarmenCarmenHHvvAAvvBBConcept QuestionConcept Question Correct: v2 = v02 -2gyPhysics 101: Lecture 6, Pg 13Kinematics in Two DimensionsKinematics in Two DimensionsConstant AccelerationConstant AccelerationConsider an object which moves in the (x,y) plane from the initial position r0, at time t0 with velocity v0, with constant acceleration. position: your coordinates (just r=(x,y) in 2-D)displacement: r = r-r0 change of positionvelocity: rate of change of


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