PTake a Test Joshua gibson Google Chrome httpswwwmathxlcom

PTake a Test - Joshua gibson - Google Chrome https://www.mathxl.com/Student/PlayerTest.aspx?testId=123098733&centerwin-yes; Joshua gibson 2/28/16 5:08 PM Test: Chapters 7 & 8 Test Overview 22 23 24 25 26 27 28 This Question: 1 pt This Test: 28 pts 19 of 28 complete A river has a constant current of 3 kilometers per hour. At what angle to a boat dock should a motorboat, capable of maintaining a constant speed of 20 kilometers per hour, be headed in order to reach a point directly opposite the dock? If the river is -kilometer wide, how long will it take to cross? Boat Direction of boa due to current At what angle should the boat be heading relative to the boat dock which is orthogonal to the current? 8.60 (Do not round until the final answer. Then round to one decimal place as needed.) How long will it take for the boat to cross the river? More hours (Do not round until the final answer. Then round to two decimal places as needed.) Enter your answer in each of the answer boxes Previous Question Next Question Submit Test 5:09 PM 2/28/2016 Search the web and Windows

Solution

One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I\'ll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have:

a) theta = arcsin(3/20) = approx. 8.63°

The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line.

x = sqrt(20^2 - 3^2)
= sqrt(400 - 9)
= sqrt 391

b) The boat\'s crossing time =
(3/4) km/(sqrt 391 km/hr)
= (0.75)/sqrt 391) hr
= approx. 0.038 = 0.04 hr