F10-PHYS115-UAE Assignment 2 Solutions1/4 27. INTERPRETWe interpret this as a problem involving the addition of three displacements in two dimensions and finding the magnitude and direction of the resultant vector. The key concepts here are displacement and average velocity. DEVELOPUsing Equation 3.1, we see that in two dimensions, a displacement vector 1rcan be written as, in unit vector notation, 111111ˆˆˆˆ(cossin)xyrr irjrij where22111()()xyrrr and 1111tan(/).yxrrOne may write down a similar expression for 2rand 3.rThe displacement vector1ris related to the velocity vector 1vby 11.rvtWe shall take a coordinate system with x axis east, y axis north, and origin at the starting point. EVALUATE(a)The first segment of the trip which last, for 110 min(1/6) h,tcan be written, in unit-vector notation, as 11110ˆˆ(40 mi/h)h(6.67 mi)60rvtjjSimilarly, we have 2ˆ(5.0 mi) ,riand the time spent on this segment is 222/(5.0 mi)/(60 mi/h)0.083 h5 min.tr v Finally, the third segment has length 333(30 mi/h) 6h/603.0 mi.rvtA unit vector in the southwest direction is 1ˆˆˆˆcos225sin 225()2ijij Therefore,31ˆˆˆˆ(3.0 mi)()( 2.12 mi)( 2.12 mi) .2rijij These displacements and their sum are shown in the figure. The total displacement is 12311ˆˆˆˆˆˆ(6.67 mi)(5.0 mi)( 2.12 mi)( 2.12 mi)(2.88 mi)(4.55 mi)totrrrrvtjiijij (b)Since the total travel time is 10 min5 min6min21 min(21/60) h,t the average velocity for the trip is ˆˆ(2.88 mi)(4.55 mi)ˆˆ(8.22 mi/h)(13.0 mi/h)(21/60) htotrijvijtASSESSWe expect bothtotrandvto be in the first quadrant since their components are all positive. Instead of unit-vector notation,totrandvcould be specified by their magnitudes 22(2.88 mi)(4.55 mi)5.38 mitotrand 15.4 mi/h,vrespectively, and common direction, 1tan[(4.55 mi)/(2.88 mi)]57.7N of E.
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