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Unformatted text preview: v 2 = – 0.15. Thus, the velocity of block C after the second explosion is v 2 = –(0.15 m/s)i ^ . (b) Between t = 0 and t = 0.80 s, the block moves v 1 Δ t = (0.60 m/s)(0.80 s) = 0.48 m. Between t = 0.80 s and t = 2.80 s, it moves an additional v 2 Δ t = (– 0.15 m/s)(2.00 s) = – 0.30 m. Its net displacement since t = 0 is therefore 0.48 m – 0.30 m = 0.18 m....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Momentum

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