Physics 231 Topic 1b: One dimensional motion Wade Fisher August 31 2012
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Physics 231 Topic 1b: One dimensional motion Wade Fisher August 31 2012
Physics 231 Topic 1b: One dimensional motion Wade Fisher August 31 2012 MSU Physics 231 Fall 2012 1 Nuances of Lon-CAPA Reporting sig figs and units is nuanced. Lon-CAPA is not always 100% consistent. Example: You calculate your answer to be 590.212 cm2 The problem gives you 3 sig figs (eg, area of a circle is 74.4 cm2) You should report 590 cm2 (3 sig figs) Lon-CAPA may not accept 590 cm2. But it accepts 5.90E2 cm2, 5.90E-2 m2, and 590.2 cm2 (??) My advice: Think carefully about sig figs and try a few different ways to report answers MSU Physics 231 Fall 2012 2 Recall from Last Lecture: Syllabus SI Units Mass: kilograms (kg) Length: meters (m) Time: seconds (s) Units and conversions Dimensional Analysis Scientific Notation (e.g., 4.3x1012 and 2.3E4) Significant Figures MSU Physics 231 Fall 2012 3 Clicker Quiz: MSU Physics 231 Fall 2012 4 Key Concepts: 1D Motion Position and coordinate systems Displacement vs. Distance Velocity Distinguish from speed Average vs instantaneous velocity Acceleration Relationship to velocity Impact on position in 1D motion Constant velocity and constant acceleration equations How to set up & solve problems Understanding graphical representations of distance, velocity, acceleration Covers chapter 2 in Rex & Wolfson MSU Physics 231 Fall 2012 5 Galileo (1564 – 1642) What mathematical forces govern accelerated motion? A wooden and stone sphere are dropped from the tower of pisa. Which one reaches the earth first? Answer: they hit the ground at the same time! MSU Physics 231 Fall 2012 6 Coordinate Systems We use coordinate systems as a graphical representation of the relationships of measurements Distance vs time Espressos sold per shift Cell phone minutes used per day MSU Physics 231 Fall 2012 7 Coordinate Systems MSU Physics 231 Fall 2012 8 Position & Displacement Displacement is a vector and thus has direction (sign) Displacement is not equal to Distance MSU Physics 231 Fall 2012 9 Coordinate Systems MSU Physics 231 Fall 2012 10 Position vs. Time Equally-spaced photographs of a runner reveal his position as a function of time MSU Physics 231 Fall 2012 11 Position vs Time Graph These measurements can be translated to a 2D “Position vs Time” coordinate system. MSU Physics 231 Fall 2012 12 Average Velocity & Speed c Average velocity= Displacement Time interval x = 13.6m x = 20.4m t = 2.0s t = 2.0s x x f xi v t t f ti MSU Physics 231 Fall 2012 13 Vectors vs Scalars X=-30 t=1 X=0 t=0 X=30 x x f xi =-30 v t t f ti speed x t x f xi t f ti X=45 Average velocity: vector =+30 Average speed: scalar MSU Physics 231 Fall 2012 14 Speed vs Velocity A driver commutes from home in the suburbs to work in the city (each way = 60 km). The morning trip takes 60 minutes (1 hour). The drive home takes 90 minutes (1.5 hours). What was his average speed and average velocity over The whole trip? a) b) c) d) e) Average speed: 60 km/h Average speed: 48 km/h Average speed: 60 km/h Average speed: 0 km/h No way to tell. Average velocity: 48 km/h Average velocity: 0 km/h Average velocity: 0 km/h Average velocity: 60 km/h MSU Physics 231 Fall 2012 15 Instantaneous Velocity “But officer, my average speed was only 25 miles per hour...” Sometimes we want to know the speed at one particular point in time. MSU Physics 231 Fall 2012 16 Instantaneous Velocity The velocity at a single point in time is given by the slope of the line tangent to the x-t curve at that time. Conceptualize this condition as a distance measurement in a very tiny time interval MSU Physics 231 Fall 2012 x v lim t 0 t 17 x Acceleration A change in slope in the x-t graph… t v …means a change in velocity: Acceleration MSU Physics 231 Fall 2012 t 18 Acceleration vf Average acceleration: average change in velocity v vi ti tf t MSU Physics 231 Fall 2012 19 Instantaneous Acceleration acceleration at one point in time: v a lim t 0 t v Slope of the tangent to the v-t curve at that point in time t MSU Physics 231 Fall 2012 20 Question v t=0 v = 20 t=1 v=5 t What is the average acceleration between t=0 and t=1 A) 15 B) -15 C) 0 MSU Physics 231 Fall 2012 D) infinity 21 Question GALILEO! Which v-t diagram matches which a-t diagram? MSU Physics 231 Fall 2012 22 Clicker Question Which v-t diagram matches the x-t diagram on the right? MSU Physics 231 Fall 2012 23 Overview up to this point VECTORS • Displacement • Average Velocity • Instantaneous velocity • Average acceleration • Instantaneous acceleration SCALARS • Distance •Average speed •Instantaneous speed MOTION DIAGRAMS MSU Physics 231 Fall 2012 24 Constant Acceleration Velocity at time t equals… velocity v (m/s) v(t ) v0 at v(t) v0 t0 Velocity at t=0… t time (s) Plus the gain in velocity per second multiplied by the time elapsed (every second, the velocity increases with a m/s) MSU Physics 231 Fall 2012 25 Constant Acceleration II x(t ) xo v t xo ( v0 vt ) 2 t xo vot at 1 2 2 Start position plus average velocity multiplied by time Substitute v 0 vt v 2 Substitute MSU Physics 231 Fall 2012 v(t ) v0 at 26 Constant motion x(m) x(t)=x0 t v m/s v(t)=0 t v(t)=v0+at t a m/s2 a(t)=0 t v m/s v(t)=v0 t a m/s2 Constant acceleration 2 x(t)=x +v t x(t)=x +v t+½at 0 0 0 0 x(m) x(m) t v m/s Constant velocity t a m/s2 a(t)=0 t MSU Physics 231 Fall 2012 a(t)=a0 t 27 v (m/s) t=1,v=2 2 0 v (m/s) 0 T (s) t=1,v=2 2 0 0 1)What is the distance covered in 1 second? 2)What is the area indicated by ? MSU Physics 231 Fall 2012 T (s) Q 1. 2. a) b) c) d) 1. 2. 1. 2. 1. 1. 2. 2. 28 Gravitational Free Fall 5 kg A B 1 kg 100 m x(t ) xo vo t at v(t ) v0 at 1 2 2 a is the acceleration felt due to gravitation (it is called ‘g’ and its value is 9.81 m/s2) g does not depend on the mass of falling object MSU Physics 231 Fall 2012 29 Clicker Question A person throws a ball straight up in the air. Which of the following is true about the acceleration and velocity at the highest point the ball reaches? a) b) c) d) Velocity is 0, acceleration is not 0. Velocity is not 0, acceleration is 0. Both velocity and acceleration are 0. Neither velocity nor acceleration are 0. MSU Physics 231 Fall 2012 30 Galileo’s Free Fall Experiment MSU Physics 231 Fall 2012 31 Galileo’s Free Fall Experiment Galileo throws a canon ball from the tower of Pisa. Ignoring friction what is the distance covered between t=1 and t=3 seconds? The initial velocity was 0 m/s. MSU Physics 231 Fall 2012 32 For Next Week Rex & Wolfson Chapter 3: Motion in 2 Dimensions Homework Set 1: Covers Chapters 1 & 2 Due Friday 9/7 @ 11PM MSU Physics 231 Fall 2012 33 Point Particle Treatment The skier can be represented by a simple, point-like object MSU Physics 231 Fall 2012 x 34 velocity acceleration motion + 0 traveling with constant speed in the positive direction + + traveling with increasing speed in the positive direction + - traveling with decreasing speed in the positive direction 0 0 0 + standing still 0 - accelerating towards the negative direction - 0 traveling with constant speed in the negative direction - + traveling with decreasing speed in the negative direction - - traveling with increasing speed in the negative direction accelerating towards the positive direction PHY 231 MSU Physics 231 Fall 2012 35 35 V vs T Diagram Example The figure shows velocity versus time. a) What is the displacement after 4 s? b) What is the average velocity? a) The distance covered is the area under the v-t diagram. x= ½x2x4 + 1x4 + (2+½x1x2) = 11 m b) Average velocity = x/t = 11m/4s = 2.75 m/s MSU Physics 231 Fall 2012 36 Instantaneous Velocity Example x x f xi v t t f ti The velocity at one point in time is the slope of the tangent to the x-t curve at that time x t What is the velocity at point C? v = x/t = (105-85)/(3-1) = 20/2 = 10 m/s x t What is the velocity at point E? v = x/t = (35-75)/(4.5-3.5) = -40/1 = -40 m/s Be careful with the sign: velocity is a vector and can be negative! MSU Physics 231 Fall 2012 37 Ball Toss Problem Dave throws a ball straight up with a velocity of 5.0 m/s. 1) How high does it go relative to the height at which it was released? 2) After how much time does it reach that height? x(t) = x(0) +v(0)t + ½at2 v(t) = v(0) + at x(t) = 0 + 5t + ½(-9.8)t2 v(tmax) = 0 = 5 - 9.8tmax tmax = 5/9.8 = 0.51 s it decelerates at it highest point, the velocity is zero time to reach highest point xmax = x(t=0.51s) = 50.51 - ½9.8(0.51) 2 = 1.3 m MSU Physics 231 Fall 2012 38 A vs T Diagram Example a (m/s2) Given this a-t diagram, calculate: a) Velocity after 4 s b) Distance traveled in 4 s The object was originally at rest. 3 2 1 0 1 2 3 4 t (s) a) v(t) = v(0)+at After 2 seconds: v(2) = 0 + 1x2 = 2 m/s After 4 seconds: v(4) = v(2) + 2x2 = 2+4 =6 m/s b) x(t) = x(0) + v(0)t + ½at2 After 2 seconds: x(2) = 0+0 + ½x1x22 = 2 m After 4 seconds: x(4) = x(2) + v(2)t + ½at2 = 2 + 2x2 + ½x2x22 = 10 m MSU Physics 231 Fall 2012 39 Area of triangle: ½ base x height = ½x4ux5u = 10u2 (u: unit of length in drawing above) 10u2 = 68.4cm2 1 u2 = 6.84cm2 Full square: 9x9u2 = 81u2 corresponds with 81x6.84 = 554 cm2 MSU Physics 231 Fall 2012 40