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# You know from integral calculus that area under the

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Unformatted text preview: of area under the curve (or line) of vx vs t graph between t = 0 and t = t? v (m/s) vx Area = vxi t + vxi 23 6 8 t (s) Δt t 3 2 12 ax t 2 12 ax t 2 vxi t t t Kinema#c Equa#ons using calculus We can derive the same results by integra#ng velocity and accelera#on: Let’s start with dvx ax = For constant accelera#on ax: vxf dt vxi ∫ dvx = ax ∫ dt 0 t ' dx Using: v x = , dt xf vxf − vxi = ax t vxf = vxi + ax t xi ∫ dx = ∫ v 0 t x t dt ' + ax t ' dt ' 1 ax t 2 2 x f − xi = ∫ (v 0 xi ) x f = xi + vxi t + Ques#ons: •  Derive the third kinema#c equa#on using using calculus. •  You know, from integral calculus, that area under the curve of a func#on f(x) between x = a and x = b is given by ∫ f ) dx . Relate that out integra#on on the right ( x above. Again, what does this area mean...
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