Physics 1 Problem Solutions 313

# Physics 1 Problem Solutions 313 - Appendix 1 Introductory...

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Unformatted text preview: Appendix 1 Introductory Physics Equation Sheet 309 If 2 0 = ax + bx + c , then x= −b ± √ b b2 − 4ac =− ± 2a 2a b 2a 2 − c a 5 Vector Formulae a = (ax , ay , az ) = ax x + ay y + az z ˆ ˆ ˆ a + b = (ax + bx , ay + by , az + bz ) a · b = ab cos θ = ax bx + ay by + az bz a × b = (ab sin θ)ˆ n 6 One-Dimensional Kinematics vavg = ∆x ∆t ∆x ∆t v = lim ∆t→0 v = v0 + at aavg = 1 x = x0 + v0 t + at2 2 1 x = x0 + (v0 + v )t 2 ∆v ∆t a = lim ∆t→0 ∆v ∆t 2 v 2 = v0 + 2a(x − x0 ) 1 x = x0 + vt − at2 2 g = 9.8 m/s2 7 Two- and Three-Dimensional Kinematics: General vavg = ∆r ∆t ∆r ∆t→0 ∆t v = lim acentripetal = aavg = v2 (−r) ˆ r ∆v ∆t→0 ∆t ∆v ∆t a = lim acentripetal = v2 r 8 Projectile Motion x = vx,0 t 1 y = y0 + vy,0 t − gt2 2 xmax = vx,0 = v0 cos θ 2 v0 sin(2θ) g ymax = vy,0 = v0 sin θ 2 v0 sin2 θ 2g 9 Very Basic Newtonian Physics Fnet = ma Fopp = −F Ff static = min[Fapplied , Ff static max ] Fcentripetal = m Fg = mg g = 9.8 m/s2 F = −kx Ff static max = µstatic FN v2 (−r) ˆ r Fcentripetal = m Ff kinetic = µkinetic FN v2 r ...
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## This note was uploaded on 11/16/2011 for the course PHY 2053 taught by Professor Buchler during the Fall '06 term at University of Florida.

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