p02_011

Fundamentals of Physics,Vol 1 (Chapters 1 - 20)

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11. We use Eq. 2-4. (a) The velocity of the particle is v = dx dt = d dt ( 4 12 t +3 t 2 ) = 12 + 6 t. Thus, at t = 1 s, the velocity is v =( 12 + (6)(1)) = 6m/s . (b) Since v< 0, it is moving in the negative x direction at t =1s. (c) At t =1s,the speed is | v | =6m/s. (d) For 0 <t< 2s , | v | decreases until it vanishes. For 2 <t< 3s , | v
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Unformatted text preview: | v | is larger than that value for t &gt; 3 s. (e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to positive (note that as t → + ∞ , we have v → + ∞ ). One can check that v = 0 when t = 2 s. (f) No. In fact, from v = − 12 + 6 t , we know that v &gt; 0 for t &gt; 2 s....
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