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CHAPTER 3 - 18

# CHAPTER 3 - 18 - v RB v HB θ v HR v WS v BS θ v BW 58 The...

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Unformatted text preview: v RB v HB θ v HR v WS v BS θ v BW 58. The position is r = (2.0 m) cos (3.0 s –1 ) t i + (2.0 m) sin (3.0 s –1 ) t j . ( a ) The magnitude of r is r = {[(2.0 m) cos (3.0 s –1 ) t ] 2 + [(2.0 m) sin (3.0 s –1 ) t ] 2 } 1/2 = 2.0 m. Thus the particle is always 2.0 m from the origin, so it is traveling in a circle. ( b ) We find the velocity by differentiating: v = d r /d t = – (2.0 m)(3.0 s –1 ) sin (3.0 s –1 ) t i + (2.0 m)(3.0 s –1 ) cos (3.0 s –1 ) t j = – (6.0 m/s) sin (3.0 s –1 ) t i + (6.0 m/s) cos (3.0 s –1 ) t j . We find the acceleration by differentiating: a = d v /d t = – (6.0 m)(3.0 s –1 ) cos (3.0 s –1 ) t i – (6.0 m)(3.0 s –1 ) sin (3.0 s –1 ) t j = – (18.0 m/s 2 ) cos (3.0 s –1 ) t i – (18.0 m/s 2 ) sin (3.0 s –1 ) t j ....
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CHAPTER 3 - 18 - v RB v HB θ v HR v WS v BS θ v BW 58 The...

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