Presentation on theme: "Lecture 5: Introduction to Physics PHY101"— Presentation transcript:
1Lecture 5: Introduction to Physics PHY101 Chapter 2:Equations of Kinematics for ConstantAcceleration in 1 Dim. (2.4, 2.5, 2.7)Free Fall (2.6)Chapter 3:Acceleration in 2 Dim. (3.1, 3.2)1
2Summary of concepts from last lecture position: your coordinates (just “x” in 1-D)displacement: x = change of positionvelocity: rate of change of positionaverage : x/tinstantaneous: slope of x vs. t : lim t->0 x/tacceleration: rate of change of velocityaverage: v/tinstantaneous: slope of v vs. t : lim t->0 v/t
3Concept Question correct v a +x correct v a +x 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 correctva+xDuring 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<0 correctva+xIf speed is increasing, v and a are in same direction.If speed is decreasing, v and a are in opposite direction.
4Concept QuestionWhich 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.correctOn 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.
5Kinematics in One Dimension Constant Acceleration Consider an object which moves from the initial position x0, at time t0with 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) (3)Use Eq. (1) to replace (t-t0) in Eq.(2):v2 = v a (x-x0 ) (4)
6Application 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 m/s2 due west. What was his velocity when he began accelerating ?[Chapter 2, problem #15]t0= 0 s, v= m/s, t=3.00 s, aav= m/s2v0 = ? m/saav = (v-v0)/(t-t0) => v0= v- aav (t-t0) = m/sv0 = 3.44 m/s due west
7Application 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/s
8Free FallFree fall is the idealized description of the motion of a downward falling body due to gravity:Air resistance is neglectedAcceleration due to gravity is considered to be constantThe acceleration due to gravity is always pointingdownward with magnitude g=9.80 m/s2.
9Concept 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/ D/ D D DCorrect x = 1/2at2Followup question: If the object has speed v at time t then what is the speed at time 2t ?1. v/ v/ v v vCorrect v=at
10Concept 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 zero 2. velocity is not zero and acceleration is zero 3. velocity is zero and acceleration is not zero 4. velocity is not zero and acceleration is not zerocorrectThe 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.
11Free FallA 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]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 s
12Concept QuestionDennis 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 Carmen's ball SameCorrect: v2 = v02 -2gyv0DennisCarmenHvAvB
13Kinematics in Two Dimensions Constant Acceleration Consider an object which moves in the (x,y) plane from the initialposition 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 positionvelocity: rate of change of positionaverage : r/tinstantaneous: lim t->0 r/tacceleration: rate of change of velocityaverage: v/tinstantaneous: lim t->0 v/tSame concepts as in one dimension !Equations of kinematics are derived for the x and y componentsseparately. Same equations as in one dimension !