15. We use Eq. 2-4. to solve the problem. (a) The velocity of the particle is vdxdtddtttt==−+=−+().412 31262Thus, at t= 1 s, the velocity is v= (–12 + (6)(1)) = –6 m/s. (b) Since v<0, it is moving in the negative xdirection at t= 1 s. (c) At t= 1 s, the speedis |v| = 6 m/s. (d) For 0 <t<2 s, |v| decreases until it vanishes. For 2 <t<3 s, |v| increases from zero to the value it had in part (c). Then, |
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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.