ch04-p018 - 18. We make use of Eq. 4-16. (a) The...

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() 2 ˆˆ ˆ 6.0 4.0 i + 8.0 j 6.0 8.0 i dv d at t t dt dt == = G G in SI units. Specifically, we find the acceleration vector at 3.0 s t = to be 2 6.0 8.0(3.0) i ( 18 m/s )i. −= (b) The equation is G =− 60 80 .. # bg i=0 ; we find t = 0.75 s. (c) Since the y component of the velocity, v y = 8.0 m/s, is never zero, the velocity cannot vanish. (d) Since speed is the magnitude of the velocity, we have || vv = G 2 2 2 6.0 4.0 8.0 10 tt += in SI units (m/s). To solve for t , we first square both sides of the above equation, followed by some rearrangement: 22 6.0 4.0 64 100
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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.

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