(d) The direction of avgvGis1tan [( 0.22 km/h)/(0.67 km/h)]18θ−=−=−°, or 18°southof east. (e) The average speed is distinguished from the magnitude of average velocity in that it depends on the total distance as opposed to the net displacement. Since the camel travels 140 km, we obtain (140 km)/(90 h) = 1.56 km/h 1.6 km/h≈.(f) The net displacement is required to be the 90 km East from Ato B. The displacement from the resting place to Bis denoted 3.rΔGThus, we must have123ˆ+ + = (90 km)irrrΔΔΔGGGwhich produces 3ˆˆ(30 km)i(20 km)jrΔ=+Gin unit-vector notation, or (36 33 )∠°in magnitude-angle notation. Therefore, using Eq. 4-8 we obtain avg36km|| = = 1.2km/h.(120 90) hv−G(g) The direction of vGis the same as Gr3(that is, 33° north of east). 121. On the one hand, we could perform the vector addition of the displacements with a vector-capable calculator in polar mode ((75 37 ) + (65 90 ) = (63 18 )),∠−°∠−°but in keeping with Eq. 3-5 and Eq. 3-6 we will show the details in unit-vector notation.
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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.