lec02-203-08-1D-mot-mod

lec02-203-08-1D-mot-mod - N i i=1 x A = x A A = area under...

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1 Dimensional Kinematics 2-1
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2-1a Instantaneous velocity
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2-2 2-2
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2-3
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2-3a General summary for 1D motion
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2-4 before derivation summarize important results & do examples
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2 0 0 t x = x + v t +a 2 0 v = v +at 2 2 o o v - v = 2a (x- x ) 0 (x - x ) = v t 0 0 (v + v ) v = 2 For constant acceleration (a) 0 0 t = 0, x = x , v = v Δ Δ x = v t Δ Δ x v = t Δ Δ v = a t Δ Δ v a = t Δ → Δ = Δ t 0 x dx v = Lim t dt Δ x = area under v(t) curve v(t) = slope of x(t) curve a(t) = slope of v(t) curve 1D Motion in general in general Const. a only
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2-5
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ball dropped (from rest) h y = 0 + y dir. g t = 0 : y=h : v 0 =0 2 t g - h y 2 = gt - v = Q2- velocity when it hits ground? 2 t g - h 0 y 2 = = g 2h t ± = correct vel. - = down what about – time !! 0 y = g 2h t = and Q1- time it hits ground? 2gh g 2h g - gt - v = = = (2) (1) (2) y) 2a v v 2 o 2 = -g (0-h) 0 2gh v 2 = 2gh v ± = correct time another way one way 2gh v + = at • another possible t<0 history !!! • ball could have been thrown upward at this –t • from t=0 on problem the same 2-6-better
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2-7 Δ Δ Δ Δ Δ Δ ≅ Δ = Δ = 1 2 3 N N N N i i i i=1 i=1 i=1 x = x + x + x + . .. x x x v t A N Lim i v Δ Δ i i i x v t = A area
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Unformatted text preview: N i i=1 x A = x A A = area under v(t) curve displacement = area under v(t) curve True in general ! 2-7a 2-8 ball thrown upward y = 0 + y dir. g=9.8 m/s 2 t = 0 : y=0 : v i =v 2 t y = v t - g 2- maximum height ? - time to get to top ? (1) (2) 2 2 o v - v = 2a ( y)-g or another way 2-9 v i at top v=0 !!! v v - gt = at top (1) put t top into (2) 2 top max 0 top t y = v t- g 2 2 max v v g y = v- g 2 g 2 max v y = 2g 2 2 o- v = 2(-g) ( y) 2 o v y = 2g- How long till ground ? 2 t 0 = y = v t - g 2 at ground y=0 !!! [ ] t 0 = v - g t 2 t 0 = v - g 2 t = 0 Started at ground t=0 2v t = g back at ground Ball thrown upward graphical interpretation 2-10 2-11...
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lec02-203-08-1D-mot-mod - N i i=1 x A = x A A = area under...

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