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lec02-203-12-1D-motfixreord1-23-12

lec02-203-12-1D-motfixreord1-23-12 - 1D Motion in general x...

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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 ) f i (v + v ) v = 2 For constant acceleration (a=constant) [Important special case] 0 0 t = 0, x = x , v = v x = v t x v = t v = a t v a = t   t 0 x dx v = Lim t dt x = area under v(t) curve v(t) = slope of x(t) curve a(t) = slope of v(t) curve 1D Motion in general Const. a only 0 (v + v ) v = 2 x = v t equivalent relations to attack problems Will derive later - 1 st experiment and problems ACTUALLY WILL OFTEN USE “y” NOT “x” 2-0

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x i t i x f t f x = 0 x t = t f - t i units [m/s] v + x + x x = 0 3 m t=2 s 5 m t=6 s v = = = +0.5 m/s x f - x i 5 - 3 t f - t i 6 - 2 positive (+) => to the right! + x x = 0 3 m t=2 s 1 m t=6 s 1 Dimensional Kinematics x = x f - x i displacement units [m] -how fast -what direction x = x f - x i =1 3 = -2 m average velocity x = t v x = v t time interval units [s] example 1 t = 6-3= 3s -how far -what direction positive (+) => to the right! example 2 x = x f - x i =5 3 = +2 m i f i f i f negative (-) => to the left! t = 6-2= 4s 4 Δx -2 = = = - 0. Δt v 5 m/ s negative (-) => to the left! 2-1
2-2 Instantaneous velocity Average acceleration m / s s

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2-2 2-3