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Unformatted text preview: ds 2 = δξ 2 + δψ 2 + δζ 2 ds 2 = δξ 2 + δψ 2 + δζ 2 + δτ 2 c = 299  79  2458 μ σ1 εξαχτλψ ds 2 = δξ 2 + δψ 2 Minkowski Special Relativity: 1908 Sep 21 ds 2 = + δΞ ( 29 δΞ ( 29 + δΨ ( 29 δΨ ( 29 + δΖ ( 29 δΖ ( 29 + δτ ( 29 δτ ( 29 dX = ι δξ dY = ι δψ dZ = ι δζ ds 2 = + ι δξ ( 29 ι δξ ( 29 + ι δψ ( 29 ι δψ ( 29 + ι δζ ( 29 ι δζ ( 29 + δτ ( 29 δτ ( 29 ds 2 =  δξ 2 δψ 2 δζ 2 + χ 2 δτ 2 ds 2 =  δξ 2 δψ 2 δζ 2 + δτ 2 ds 2 = δξ 2 + δψ 2 + δζ 2 χ 2 δτ 2 Chapter 3, #12 6 CLICKER QUESTIONS For the vectors and a = 3.0μ ( 29 ι + 4.0μ ( 29 ϕ b = 5.0μ ( 29 ι +2.0μ ( 29 ϕ 1) give a + b in unitvector notation 2) give a + b as a magnitude 3) give a + b as a direction relative to i 4) give b a in unitvector notation 5) give b a as a magnitude 6) give b a as a direction relative to i ion in Two and Three Dimensions Motion in Two and Three Dimensions x y z r r 1 r r 2 ∆ r ur u ion in Two and Three Dimensions Displacement Vector r Δ x y z r r 1 r r 2 ∆ r ur u r r 2 = ρ ρ 1 + ∆ρ υρ υ ∆ r ur u = r r 2 r r 1 = x 2 i + y 2 j + z 2 k ( ) x 1 i + y 1 j + z 1 k ( ) = x 2 x 2 ( ) i + y 2 y 2 ( ) j + z 2 z 2 ( ) k ∆ r ur u = D x i + D y j + D z k ion in Two and Three Dimensions Average velocity & Instantaneous velocity r v avg = ∆ ρ ρ ∆ τ = ∆ ξ ι + ∆ ψϕ + ∆ ζ κ ∆ τ = ∆ ξ ∆ τ ι + ∆ ψ ∆ τ ϕ + ∆ ζ ∆ τ κ x y ∆ ρ r ∆ ρ r r v = δ ρ ρ δτ ion in Two and Three Dimensions Average acceleration & Instantaneous acceleration r a = δ ρ ω δτ r a avg = ∆ ρ ω ∆ τ = ρ ω 2 ρ ω 1 ∆ τ r a = δ ρ ω δτ = δ δτ ω ξ ι + ω ψ ϕ + ω ζ κ ( 29 = δω ξ δτ ι + δω ψ δτ ϕ + δω ζ δτ κ a x = δω ξ δτ , α ψ = δω ψ δτ , α ζ = δω ζ δτ , To find the scalar components of a , differentiate the scalar components of v ion in Two and Three Dimensions Chapter 4, #19 The acceleration of a particle moving only on a horizontal x , y plane is given by where a is in m/s/s and time t is in seconds. r = 20.0μ ( 29 ι + 40.0μ ( 29 ϕ At t = 0, the position vector locates the particle, which then has the velocity vector v = 5.0 μ /σ ( 29 ι + 2.0 μ /σ ( 29 ϕ At t = 4.00 s, what are a) its position vector in unit vector notation b) the angle between its direction of travel and the positive direction of the x axis? The motions in x , and y , are independent! a = 3 ξ τι + 4 ξ ψ ϕ a = 3τι + 4τϕ a = 3τι + 4τϕ THIS IS NOT A CONSTANT ACCELERATION PROBLEM! d 2 x d t 2 = 3τ d 2 y d t 2 = 4 τ dy d t = 4 τ 2 2 + 2 = 34 dx d t = 3 τ 2 2 + 5 = 29 x = 3 2 τ 3 3 + 5τ+ 20 = 72 y = 4 2 τ 3 3 + 2τ+ 40 = 90.67 r = 72.0 μ ( 29 ι + 90.67 μ ( 29 ϕ ion in Two and Three Dimensions Chapter 4, #19 b) the angle between its direction of travel and the positive direction of the x axis?...
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This note was uploaded on 10/01/2008 for the course PHYSICS 101 taught by Professor Bennet during the Spring '08 term at Johns Hopkins.
 Spring '08
 bennet
 Physics

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