(b) Using the result from part (a), we find vbg= vbwcosθ= 5.5 km/h. Thus, traveling a distance of A= 6.4 km requires a time of (6.4 km)/(5.5 km/h) = 1.15 h or 69 min. (c) If her motion is completely along the yaxis (as the problem implies) then with vwg= 3.2 km/h (the water speed) we have total= + = 1.33 h+ bwwgbwwgDDtvvvv−whereD= 3.2 km. This is equivalent to 80 min. (d) Since +bwwgbwwgbwwgbwwgDDvvvv+=+−−+the answer is the same as in the previous part, i.e., total= 80 mint.(e) The shortest-time path should have 0.=°This can also be shown by noting that the case of general leads to ˆˆcos i (sin + ) jbgbwwgbwbwwgvvvvvvθθ=+=+GGGwhere the xcomponent of Gvbgmust equal l/t. Thus, = cosbwltvwhich can be minimized using dt/d= 0. (f) The above expression leads to t= (6.4 km)/(6.4 km/h) = 1.0 h, or 60 min. 89. We establish coordinates with
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