Ch02-p073 - (e As implied by our answer to part(c it moves leftward for times immediately after t = 2 s In fact the expression found in part(a

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73. (a) The derivative (with respect to time) of the given expression for x yields the “velocity” of the spot: v ( t ) = 9 – 9 4 t 2 with 3 significant figures understood. It is easy to see that v = 0 when t = 2.00 s. (b) At t = 2 s, x = 9(2) – ¾(2) 3 = 12. Thus, the location of the spot when v = 0 is 12.0 cm from left edge of screen. (c) The derivative of the velocity is a = – 9 2 t which gives an acceleration (leftward) of magnitude 9.00 m/s 2 when the spot is 12 cm from left edge of screen. (d) Since v >0 for times less than t = 2 s, then the spot had been moving rightwards.
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Unformatted text preview: (e) As implied by our answer to part (c), it moves leftward for times immediately after t = 2 s. In fact, the expression found in part (a) guarantees that for all t > 2, v < 0 (that is, until the clock is “reset” by reaching an edge). (f) As the discussion in part (e) shows, the edge that it reaches at some t > 2 s cannot be the right edge; it is the left edge ( x = 0). Solving the expression given in the problem statement (with x = 0) for positive t yields the answer: the spot reaches the left edge at t = 12 s ≈ 3.46 s....
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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