Lec2 - Physics 2A Lecture 2 Today: Estimation. Variables of...

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Unformatted text preview: Physics 2A Lecture 2 Today: Estimation. Variables of motion. Olga Dudko UCSD Physics Estimation Sometimes there will not be an easily found, exact answer available for a given problem. In this case, you must estimate certain values in order to continue. Example: what if you were asked to find the volume of a cow? A first order estimate: assume that the cow is a sphere => Vcow = 4/3 R3 Estimates are designed to get you close to the answer with ease (say within a factor of ten). How many piano tuners are there in Chicago? Physicist Enrico Fermi was known for his estimation prowess. At a party (in the late 60's), he was approached by someone and asked: "How many piano tuners are there in Chicago?" Solution First we should find out how many pianos there were in Chicago... Enrico Fermi, Nobel Prize in Physics (1938) Solution (cont'd) Estimation Fermi first estimated how many people lived in Chicago, at the time it was 700,000. He then guessed that 1 out of 3 families owns a piano. Assuming a family is about 4 people this means 1 out of 12 people have a piano. We round this to 1 out of 10 for ease. This means that there were about 70,000 pianos in Chicago at that time. Solution (cont'd) Estimation Next, Fermi turned to how many piano tuners could get work for 70,000 pianos. He first determined that it took 1 to 2 hours to tune a piano. Thus, a piano tuner could service about four pianos a day. Assume a typical work schedule, 5 days per week and 50 weeks per year. So, in a year a piano tuner will tune: (4 pianos/day)x(5 days/week)x(50 weeks/year) = 1,000 pianos/year Answer Estimation A piano should be tuned about once a year. (1,000 pianos/year) x (1 year/tuner) = 1,000 pianos/tuner Thus, 70,000 pianos = 70 tuners 1,000 pianos tuner Actual listing in the phone book: 50 piano tuners Motion Chapter 2 will focus on motion in one dimension. Any description of motion involves three concepts: 1) Displacement 2) Velocity 3) Acceleration First, define a coordinate system Nearly the first thing you must do in every Physics problem is define a coordinate system. Choosing the proper coordinate system can make a huge difference. For one dimensional motion it is rather easy: 1) Choose an origin. 2) Choose a positive direction. Displacement and Distance x is defined as the position compared to the origin. Displacement , x, is a difference between positions. x = x2-x1 0 x1 x2 x x Distance , d, is the total length traveled. d can only be a positive quantity. x, can be a positive or negative quantity. Velocity and Speed The velocity of an object is its displacement over a period of time. Average velocity , vavg, is v avg x x 2 - x1 = = t t 2 - t1 The speed of an object is its distance traveled over a period of time. Average speed is speedavg = d d = t t 2 - t1 Average speed is only a positive quantity. Average Velocity Graphically, we find average velocity by examining the rise ( x) over the run ( t) in an x vs. t graph. between points A and B is v avg x 2 - x1 = t 2 - t1 = 50m - 30m 10s - 0s Average velocity = 20m = 2m s 10s Instantaneous Velocity Instantaneous velocity is the velocity at a given instant of time. x dx Instantaneous velocity, v, is v = lim t = dt t 0 i ma nitesi nfi ls Instantaneous speed will just be the magnitude (only positive) of the instantaneous velocity. We usually "drop" instantaneous when talking about velocity or speed. Velocity Graphically, we find instantaneous velocity by examining the slope of an x vs. t graph at a particular point. from average velocity in that you do not care what happens over a time period. You would like to know what happens at a time instant . It is different Example Velocity Runner A is initially 4.0 miles west of a flagpole and is running with a constant velocity of 6.0 mph due east. Runner B is initially 3.0 miles east of the flagpole and is running with a constant velocity of 5.0 mph due west. How far are the runners from the flagpole when they meet? Solution First, define a coordinate system. Let's choose the flagpole as the origin with east being the positive x-direction. W E B A 0 x Solution (cont'd) Velocity A B Each runner needs their own separate variables measured from the flagpole: xA vA = For Runner A => t Similarly, for Runner B => But recall that displacement is 0 x xA = vA t xB = vB t x B = x B, final - x B,initial We can thus rewrite our equations for Runner A and Runner B as: x A, final = x A,initial + v A t x B, final = x B,initial + v B t Solution (cont'd) Velocity When the runners meet their positions will be equal: x B, final = x A, final x B,initial + v B t = x A,initial + v A t We know every variable here except for time, let's solve for time: x A,initial - x B,initial t = -4.0 miles - 3.0 miles , t= , -5.0 mi - 6.0 mi vB - vA hr hr 7.0 t= hr = 0.636 hr 11.0 Answer Velocity We wanted to know how far from the flagpole the runners meet. We can use either position equation to solve: x B, final = x B,initial + v B t , x B, final = 3.0 miles+ (-5.0 mi )(0.636 hr) hr x B, final = -0.18 miles W B E ... the negative sign means that they meet to the west of the flagpole. A 0 x Acceleration The acceleration of an object tells us how much its velocity changes over a period of time. Average acceleration , aavg, is aavg= Instantaneous acceleration is the acceleration of an object at a given instant of time. Acceleration is v v2 - v1 = t t -t 2 1 v dv a = lim t = t 0 dt Acceleration is usually measured in units of m/s2. But you can also find it in other units so be careful. Acceleration Watch out for the sign of acceleration. Positive acceleration does not always mean "speeding up." Nor does negative acceleration always mean "slowing down." When acceleration and velocity are in the same direction, the object "speeds up." When acceleration and velocity are in opposite directions, the object "slows down." For Next Time: Finish up the homework for Chapter 1 Read chapter 2 Start working on the homework for Chapter 2 Discussion Sessions will start today (04/02) and will be at WLH 2001 from 5:00 p.m. to 5:50 p.m. ...
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