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The International System of Units Quantity Unit Name Symbol Length meter m • Many SI Derived Units: Time second s 1 Newton = 1 N = 1 kg · m/s2 Mass kilogram kg • Three Basic Units (SI) 1 Watt = 1 W = 1 N · m = 1 kg · m2/s2 • Prefixes for SI Units: 1 kg = 1 × 103 grams 1 ps = 1 × 10-12 Conversion Factor seconds • Changing Units: 1 min =1 1 min = 60 s → 60 s 60 s =1 1 min ⎛ 60 s ⎞ 2 min = ( 2 min)(1) = ( 2 min)⎜ ⎟ = 120 s ⎝ 1 min ⎠ R. Field 1/9/2014 University of Florida PHY 2053 Factor Prefix Symbol 1012 tera- T 109 giga- G 106 mega- M 103 kilo- k 10-2 centi- c 10-3 milli- m 10-6 micro- μ 10-9 nano- n 10-12 pico- p Page 1 distance: meter 1791: one ten-millionth of a quadrant of the Earth… --1889: platinum-iridium bar --1983: The meter is the length of the path traveled by light in vacuum during a time interval of 1 / (299,792,458) of a second (i.e., by definition, c = 299,792,458 m/s) R. Field 1/9/2014 University of Florida PHY 2053 Page 2 time: second Since long ago (Egyptians and Greeks): 1 day = 24 hours 1 s = (1/24) x (1/60) x (1/60) of the full Earth turn However, the Earth rotation period varies by a few ms… 1967: The second is the duration of 9,192,631,770 periods of the radiation emitted by caesium-133 atom R. Field 1/9/2014 University of Florida PHY 2053 Page 3 mass: kilogram 1799: 10x10x10 cm3 of water 1 u = 1.66053886 × 10 --1899: platinum-iridium cylinder (d=h=39 mm) -27 kg atomic mass unit (u): • 12 u = mass of 12C atom • universal, reproducible, nor requiring an artifact Density: The density of a material, ρ, is the mass per unit volume: ρ = m/V ρ(water) = 1.00 g/cm3 R. Field 1/9/2014 University of Florida PHY 2053 Page 4 1-d Motion: Position & Displacement • The x-axis: We locate objects by specifying their position along an axis (in this case x-axis). The positive direction of an axis is in the direction of increasing numbers. The opposite is the negative direction. • Displacement: x(t) The change from position x1 to position x2 is called the displacement, Δx. Δx = x2 –x1 The displacement has both a magnitude, |Δx|, and a direction (positive or negative). time t • Graphical Technique: A convenient way to describe the motion of an object is to plot the position x as a function of time t (i.e. x(t)). R. Field 1/9/2014 University of Florida PHY 2053 Page 5 1-d Motion: Average Velocity • Average Velocity v = vave The average velocity is defined to be the displacement, Δx, that occurred during a particular interval of time, Δt (i.e. vave = Δx/Δt). Δx x(t 2 ) − x(t1 ) x2 − x1 = = = t 2 − t1 t 2 − t1 Δt • Average Speed The average speed is defined to be the magnitude of total distance covered during a particular interval of time, Δt (i.e. save = (total distance)/Δt). R. Field 1/9/2014 University of Florida PHY 2053 Page 6 Average Velocity: Example Problem Final Position 3:56 pm Initial Position 3:14 pm xf = -26 km x=0 W xi = 3 km x-axis xE = 10 km E A train that is initially at the point xi = 3 km at 3:14 pm travels 7 km to the East to the point xE = 10 km. It then reverses direction and travels 36 km to the West to the final point xf = -26 km arriving at 3:56 pm. What is the train’s average velocity (in km/h) for this trip? v = vave Δx x(t f ) − x(ti ) x f − xi − 26km − (3km) − 29km = = = = = ≈ −41.43km / h Δt t f − ti t f − ti 42 min× (1h / 60 min) 0. 7 h Note that the displacement Δx is equal to the average velocity times Δt. Δx = vave Δt = (−41.43km / h)(0.7h) ≈ −29km R. Field 1/9/2014 University of Florida PHY 2053 Page 7 Average Speed: Example Problem Final Position 3:56 pm Initial Position 3:14 pm xf = -26 km x=0 W xi = 3 km x-axis xE = 10 km E A train that is initially at the point xi = 3 km at 3:14 pm travels 7 km to the East to the point xE = 10 km. It then reverses direction and travels 36 km to the West to the final point xf = -26 km arriving at 3:56 pm. What is the train’s average speed (in km/h) for this trip? save = d tot 7km + 36km 43km = ≈ 61.43km / h = Δt 42 min× (1h / 60 min) 0.7 h Note that the total distance d is equal to the average speed times Δt. d tot = save Δt = (61.43km / h)(0.7 h) ≈ 43km R. Field 1/9/2014 University of Florida PHY 2053 Page 8 Exam 1 Fall 2013: Problem 5 Initial Position t= 0 • A train traveling along the x-axis is initially at the point xi at t = 0. The train then x-axis travels 10 km to the East (i.e. right) as shown in the figure. It then reverses dE = 10 km direction and travels 30 km to the West (i.e. left) to the final point xf. If the train’s dW = 30 km average speed for this trip was 20 km/h what was its average velocity for the trip x f = xi + d E − dW (in km/h)? d E + dW d tot d E + dW Δt = = save = Answer: -10 save Δt Δt % Right: 64% xi W vave E Δx x f − xi d E − dW d E − dW − 20km = = = = save = (20km / h) = −10km / h Δt Δt Δt d E + dW 40km R. Field 1/9/2014 University of Florida PHY 2053 Page 9 1-d Motion: Instantaneous Velocity x x(t+Δt) x(t) R. Field 1/9/2014 University of Florida { shrink Δt t t+Δt t Δt PHY 2053 Page 10 1-d Motion: Instantaneous Velocity x x(t+Δt) x(t) R. Field 1/9/2014 University of Florida { shrink Δt t+Δt t t Δt PHY 2053 Page 11 1-d Motion: Instantaneous Velocity x x(t+Δt) x(t) R. Field 1/9/2014 University of Florida t+Δt { shrink Δt t t Δt PHY 2053 Page 12 1-d Motion: Instantaneous Velocity x x(t) x(t+Δt) R. Field 1/9/2014 University of Florida { shrink Δt t t+Δt Δt t PHY 2053 Page 13 1-d Motion: Instantaneous Velocity x tangent line at t x(t) Δx dx = v(t ) = lim dt Δt →0 Δt R. Field 1/9/2014 University of Florida Instantaneous velocity v(t) is slope of x-t tangent line at t t t The velocity v(t) is the rate of change of x(t) with respect to t at time t. PHY 2053 Page 14 1-d Motion: Acceleration • Acceleration a = aave When a particles velocity changes, the particle is said to undergo acceleration (i.e. accelerate). v Δv v(t 2 ) − v(t1 ) v2 − v1 = = = Δt t 2 − t1 t 2 − t1 v(t) v2 • Average Acceleration Hello Δv “rise” v1 The average acceleration is defined to be the change in velocity, Δv, that occurred during a particular interval of time, Δt (i.e. aave = Δv/Δt). • Instantaneous Acceleration a Δv v The acceleration a(t) is the rate of change of v(t) with respect to t at time t. v(t) Δv dv = a (t ) = lim Δt →0 Δt dt R. Field 1/9/2014 University of Florida Instantaneous acceleration a(t) is slope of v-t tangent line at t PHY 2053 Page 15 1-d Motion: Summary • Instantaneous Velocity The velocity v(t) is the rate of change of x(t) with respect to t at time t. Δx dx v(t ) = lim = dt Δt →0 Δt a (t ) = • Instantaneous Acceleration The acceleration a(t) is the rate of change of v(t) with respect to t at time t. Δv dv = a (t ) = lim Δt →0 Δt dt R. Field 1/9/2014 University of Florida dv d dx d 2 x = = dt dt dt dt 2 v v(t) Instantaneous acceleration a(t) is slope of v-t tangent line at t PHY 2053 Page 16