notes001 (45)

# notes001 (45) - v ( t ) = j v ( t ) j = q (3 t ) 2 + (4 t )...

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September 24, 2010 NAME_____________________________ The motion of a particle is given by x = 3 2 t 2 y = 2 t 2 t > 0 . Find the position function: r ( t ) the velocity: v ( t ) ( 4 points if you get to here ) the speed: v ( t ) = j v ( t ) j the acceleration: a ( t ) ( 10 points if you get to here ) the unit tangent vector: T ( t ) the curvature: ( t ) ( 16 points if you get to here ) the tangential component of acceleration: a T ( t ) the normal component of acceleration: a N ( t ) ( 20 points if you get to here ) Solution: r ( t ) = 3 2 t 2 i ± 2 t 2 j . The velocity is v ( t ) = d r dt = 3 t i + 4 t j 1

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and the speed is
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Unformatted text preview: v ( t ) = j v ( t ) j = q (3 t ) 2 + (4 t ) 2 = 5 t . The acceleration is a ( t ) = d v dt = 3 i + 4 j . The unit tangent vector is T ( t ) = 1 v ( t ) v ( t ) = 1 5 t (3 t i + 4 t j ) = 3 5 i + 4 5 j . Also d T dt = and &amp; &amp; &amp; &amp; d T dt &amp; &amp; &amp; &amp; = 0 so the curvature is &amp; ( t ) = 1 v ( t ) &amp; &amp; &amp; &amp; d T dt &amp; &amp; &amp; &amp; = 1 5 t (0) = 0 . The tangential component of acceleration is a T ( t ) = dv dt = d dt (5 t ) = 5 and the normal component of acceleration is a N ( t ) = &amp; ( t ) ( v ( t )) 2 = (0) (5 t ) 2 = 0 . 2...
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## notes001 (45) - v ( t ) = j v ( t ) j = q (3 t ) 2 + (4 t )...

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