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Unformatted text preview: (0 , ), so the only value of t in this interval for which s ( t ) = 0 is t = 2, which means that s ( t ) 6 = 0 for t (0 , 2) and t (2 , ). To determine where s ( t ) is positive and negative, we can make a sign chart for the factors in s ( t ) when restricted to these two intervals: interval 1 t 2 t-2 t + 2 s ( t ) (0 , 2) +-+-(2 , ) + + + + The signs + or - in the table can determined by checking the value of the given factor at a single point in the interval. The conclusions to be drawn from the table are as follows: The particle is moving in the positive direction when the velocity satises s ( t ) > 0, or when t (2 , ). It is moving in the negative direction when the velocity satises s ( t ) < 0, or when t (0 , 2). It is at rest when s ( t ) = 0, or at t = 2 sec. 2...
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- Fall '10