PhysI_09242009

# PhysI_09242009 - B 4km 2.5km 22km/hr 4m/s 6m/s 50.0m...

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m v r M PB BA PA -20.3cm -20.3cm -20.3cm -20.3cm 60.9cm 40.6cm 20.3cm 0cm x 2 x 1 50.0m 1.00m 1.00m 6m/s 6m/s x θ θ 4m/s 4m/s 4m/s 4m/s A 22km/hr 4km 40km/hr 37° 2.5km B 09-24-2009 171.101, General Physics for Physical Science Majors I Acceleration of Uniform Circular Motion: Reminder a = v 2 /r r The r is supposed to represent a unit of the magnitude and direction of the radius, showing that the direction of acceleration is along the same line as the radius and that the velocity vector is perpendicular to the acceleration of Uniform Circular Motion. Because of this: v a = vacos( ) = vacos(90) = 0 θ r x a = rasin( ) = rasin(0) = 0 θ Relative Motion in One Dimension: Alex, jumping rope in one place, sees Barbara driving away at a constant speed. Paul, who is ahead of Barbara, is moving away from Alex at an accelerating speed. x PA = x PB + x BA v PA = v PB + v BA a PA = a PB Special Relativity: v PA = (v PB + v BA )/(1 + v PB v BA /c) If v PB =c , v PA = (c + v BA )/(1 + v BA /c) = (c 2 + cv BA )/(c + v BA ) = c(c + v BA )/(c + v BA ) = c Relative Motion in Two Dimensions: x PA = (x PB î + x PB ) + (x ĵ BA î + x BA ) ĵ v PA = (v PB î + v PB ) + (v ĵ BA î + v BA )…. ĵ

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## This note was uploaded on 10/01/2009 for the course DOGEE 171.101 taught by Professor Henry during the Fall '09 term at Johns Hopkins.

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PhysI_09242009 - B 4km 2.5km 22km/hr 4m/s 6m/s 50.0m...

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