Vxf vxi equaon1 gives us t ax now plugging this in

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Unformatted text preview: #me at t = 0, Δt = t, this becomes vx , avg = t x f − xi vxi + vxf = Therefore, we have t 2 Here we can eliminate vxf, from Equa#on(1): x f − xi vxi + vxi + ax t t 2 1 x f = xi + vxi t + ax t 2 2 = Equa#on (2) This is another widely used kinema#c equa#on. Note that this true for the case of constant accelera#on. Mo#on of par#cle with constant accelera#on (contd…) Finally, we obtain another useful kinema#c equa#on by elimina#ng t in instead of elimina#ng vxf, as we did earlier. vxf − vxi Equa#on(1) gives us: t= ax Now, plugging this in t 2 vxf − vxi v 2f − v 2i x x We get 2( x f − xi ) = (vxi + vxf )t = (vxi + vxf ) = = x f − xi vxi + vxf ax ax This can be wriXen as v 2f = v 2 i + 2 ax ( x f − xi ) x x Equa#on (3) This is the third kinema#c equa#on we use quite frequently. Lets put them together vxf = vxi + ax t 1 2 2 v 2f = v 2i + 2 ax ( x f − xi ) x x x f = xi + vxi t + ax t Velocity vs #me graphs Let’s interpret the same graphs but x replaced by vx. slope = lim vx Δv vx Δvx dvx t = Δt → 0 Δt dt dv slope of posi#on ­#me graph = accelera#on ax = x We know dt What is the meaning...
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This note was uploaded on 11/16/2010 for the course PHY 2170 taught by Professor Blank during the Spring '08 term at Wayne State University.

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