CHAPTER 3 - 27

# CHAPTER 3 - 27 - When the jogger is moving toward the stern...

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landing point could be above or below the point directly across the river, that is, positive or negative.  This is also a  good problem for a numerical solution on a spreadsheet, where absolute values can be used. 92. If  v SW  is the velocity of the ship with respect to the water,  v JS  the velocity of the jogger with respect to the ship, and  v JW  the velocity of the jogger with respect to the water, then  v JW  =  v JS  +  v SW  . We choose the direction of the ship as the positive direction.  Because all vectors are parallel, in each case the  motion is one-dimensional. When the jogger is moving toward the bow, we have v JW  =  v JS  +  v SW  = 2.0 m/s + 8.5 m/s =          10.5 m/s in the direction of the ship's motion
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Unformatted text preview: . When the jogger is moving toward the stern, we have v JW = – v JS + v SW = – 2.0 m/s + 8.5 m/s = 6.5 m/s in the direction of the ship's motion . 93. The centripetal acceleration must equal g : g = v 2 / R = (2p R / T ) 2 / R = 4p 2 R / T 2 ; 9.80 m/s 2 = 4p 2 (0.55 × 10 3 m)/ T 2 , which gives T = 47.1 s. For the rotation speed we have revolutions/day = (86,400 s/day)/(47.1 s/rev) = 1.8 × 10 3 rev/day . 94. The centripetal acceleration is a R = v 2 / r , or r = v 2 / a R . Thus r is minimal when a R is maximal: r min = [(700 km/h)/(3.6 ks/h)] 2 /(6.0)(9.80 m/s 2 ) = 6.4 × 10 2 m = 0.64 km ....
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## This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at Berkeley.

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