Physics 101: Lecture 03 Kinematics Lecture Notes PDF

Physics 101: Lecture 03 Kinematics Today’s lecture will cover Textbook Sections 3.1-3.3 (and some Ch. 4) Physics 101: Lecture 3, Pg 1  Announcements HW 1 is due Tuesday February 1st at 6 am. No clicker points for Lect. 2? Re-register for this course! Of...

Physics 101: Lecture 03 Kinematics Today’s lecture will cover Textbook Sections 3.1-3.3 (and some Ch. 4) Physics 101: Lecture 3, Pg 1  Announcements HW 1 is due Tuesday February 1st at 6 am. No clicker points for Lect. 2? Re-register for this course! Office hours start Friday. Read the course description & FAQ on the course web site! Physics 101: Lecture 3, Pg 2  Force at Angle Example  A person is pushing a 15 kg block across a floor with k= 0.4 at a constant speed. If she is pushing down at an angle of 25 degrees, what is the magnitude of her force on the block? x- direction: Fx = max Combine: Fpush cos() – Ffriction = 0 Fpush cos() / –mg – FPush sin() = 0 Fpush cos() –  FNormal = 0 Fpush ( cos() /  - sin()) = mg FNormal = Fpush cos() /  Fpush = m g / ( cos()/ – sin()) y- direction: Fy = may Fpush = 80 N FNormal –Fweight – FPush sin() = 0 Normal FNormal –mg – FPush sin() = 0 Pushing y  x  Friction Weight Physics 101: Lecture 3, Pg 3  Homework 2 Example  Calculate the tension in the left string. y x x-direction: F=ma -TL+TR cos()= 0 TR TL  TL = TR cos() y-direction: F=ma W TR sin() – Mg = 0 Combine: TR = Mg / sin() TL = Mg cos()/sin() Physics 101: Lecture 3, Pg 4  Overview  Kinematics: Description of Motion Position and displacement velocity » average » instantaneous Acceleration » average » instantaneous Relative velocity (first pass) Physics 101: Lecture 3, Pg 5  Position vs Time Plots  Gives location at any time.  Displacement is change in position. x (m)  Slope gives instantaneous velocity. 3 Position at t=3, x(3) = 1 t Displacement between t=5 and t=1. x = -1.0 m 4 1.0 m - 2.0 m = -1.0 m -3 Average velocity between t=5 and t=1. v = -0.25 m/s -1 m / 4 s = -0.25 m/s Physics 101: Lecture 3, Pg 6  Velocity vs Time Plots  Gives velocity at any time. v (m/s)  Area gives displacement 3  Slope gives instantaneous acceleration. 6 1.5 t velocity at t=2, v(2) = 3 m/s 4 Displacement between t=0 and t=3: x= 7.5 m -3 t=0 to t=1: ½ (3m/s) (1 s) = 1.5 m t=1 to t=3: (3m/s) (2 s) = 6 m Average velocity between t=0 and t=3? v= 7.5 m / 3s = 2.5 m/s Change in v between t=5 and t=3. v = -2 m/s – 3 m/s = -5 m/s Average acceleration between t=5 and t=3: a = -5 m/s / (2 s) = -2.5 m/s2 Physics 101: Lecture 3, Pg 7  Acceleration vs Time Plots  Gives acceleration at any time.  Area gives change in velocity a (m/s2) 3 Acceleration at t=4, a(4) = -2 m/s2 6 Change in v between t=4 and t=1. v = +4 m/s t t=1-3: v = (3m/s2)(2s) = 6 m/s 24 t=3-4: v = (-2m/s2)(1s) = -2 m/s -3 Physics 101: Lecture 3, Pg 8  Acceleration Preflights Is it possible for an object to have a positive velocity at the same time as it has a negative acceleration? 88% 1 - Yes “the object could be slowing down.” 12% 2 - No If the velocity of some object is not zero, can its acceleration ever be zero ? 87% 1 - Yes “The velocity could be non-zero and constant. A constant velocity has no 13% 2 - No acceleration..” Physics 101: Lecture 3, Pg 9  Velocity ACT 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? A - Yes B - No Drive north 5 miles, put car in reverse and drive south 2 miles. Average velocity is positive. Physics 101: Lecture 3, Pg 10  Dropped Ball y A ball is dropped from a height of x two meters above the ground.  Draw vy vs t v v v 9 A 9 B 9 C 0.5 t 0.5 t 0.5 t -6 -6 -6 v v 9 9 E D 0.5 t 0.5 t -6 -6 Physics 101: Lecture 3, Pg 11  Dropped Ball x A ball is dropped for a height of two meters above the ground. t v  Draw v vs t  Draw x vs t t  Draw a vs t a t Physics 101: Lecture 3, Pg 12  Tossed Ball A ball is tossed from the ground up a height of two meters above the ground. And falls back down y x  Draw v vs t v v v 9 A 9 B 9 C 1 t 1 t 1 t -6 -6 -6 v v 9 9 D E 1 t 1 t -6 -6 Physics 101: Lecture 3, Pg 13  Tossed Ball x A ball is tossed from the ground up a height of two meters above the ground. t And falls back down v  Draw v vs t  Draw x vs t t  Draw a vs t a t Physics 101: Lecture 3, Pg 14  ACT A ball is thrown straight up in the air and returns to its initial position. During the time the ball is in the air, which of the following statements is true? A - Both average acceleration and average velocity are zero. B - Average acceleration is zero but average velocity is not zero. C - Average velocity is zero but average acceleration is not zero. D - Neither average acceleration nor average velocity are zero. Vave = Y/t = (Yf – Yi) / (tf – ti) = 0 aave = V/t = (Vf – Vi) / (tf – ti) Not 0 since Vf and Vi are not the same ! Physics 101: Lecture 3, Pg 15  Relative Velocity (first pass) You are on a train traveling 40 mph North. If you walk 5 mph toward the front of the train, what is your speed relative to the ground? A) 45 mph B) 40 mph C) 35 mph 40 mph N + 5 mph N = 45 mph N 40 5 45 Physics 101: Lecture 3, Pg 16  Relative Velocity You are on a train traveling 40 mph North. If you walk 5 mph toward the rear of the train, what is your speed relative to the ground? A) 45 mph B) 40 mph C) 35 mph 40 mph N - 5 mph N = 35 mph N 40 5 35 Physics 101: Lecture 3, Pg 17  Relative Velocity You are on a train traveling 40 mph North. If you walk 5 mph sideways across the car, what is your speed relative to the ground? A) < 40 mph B) 40 mph C) >40 mph 40 mph N + 5 mph W = 41 mph N 5 | v | 402  52 40 Physics 101: Lecture 3, Pg 18  Relative Velocity  Sometimes your velocity is known relative to a reference frame that is moving relative to the earth. Example 1: A person moving relative to a train, which is moving relative to the ground. Example 2: a plane moving relative to air, which is then moving relative to the ground.  These velocities are related by vector addition: » vac is the velocity of the object relative    to the ground vac  vab vbc » vab is the velocity of the object relative to a moving reference frame » vbc is the velocity of the moving reference frame relative to the ground Physics 101: Lecture 3, Pg 19  Tractor Demo 1 Which direction should I point the tractor to get it across the table fastest? A) 30 degrees left B) Straight across C) 30 degrees right 1 2 3 Physics 101: Lecture 3, Pg 20 Tractor Demo (moving table)  Which direction should I point the tractor to get it across the table fastest? A) 30 degrees left B) Straight across C) 30 degrees right 1 2 3 Physics 101: Lecture 3, Pg 21  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. t  relative velocity: vac = vab + vbc Physics 101: Lecture 3, Pg 22