https://www.chegg.com/homework-help/questions-and-answers/birds-evolved-ability-
ID: 2001082 • Letter: H
Question
https://www.chegg.com/homework-help/questions-and-answers/birds-evolved-ability-control-motions-air-controlling-forces-exerted-bodies-air-contact-fo-q10188667
the first question were answered on that link but to answer the last questions it seems that the answers from the others,which are on the link, are needed to answer the last three questions that are posted here.
Birds have evolved to the ability to control their motions through the air by controlling the forces exerted on their bodies by the air; the contact forces on a bird by the air plus the force of gravity on the bird by the Earth determine the change in motion of the bird at any instant of time. In this problem, you will be asked to use what you know about acceleration to track the motion of a diving hawk. A hawk is flying horizontally Southward at an altitude of 143.0 m and at a speed of 1.70 m/s when he spots a slow-flying finch ahead of his current location and below his current path. At time zero, the hawk starts a dive by controlling the contact forces on him by the air so as to produce an acceleration of 4.70 m/s2 in a direction 73.0 degrees below horizontally Southward for a time of 0.750 seconds.
1) Once again, please assume that the hawk's acceleration during the 0.9-s interval is constant. What is the magnitude of the hawk's displacement during the 0.9-s interval? Give your answer as an ordered pair, with magnitude first, followed by a comma, then followed by the direction. Give the direction in terms of an angle measured below the horizontally Southward direction.(You have now been tracking the hawk's motion for a total of 0.750 + 8.5 + 0.9 = 10.15 seconds.)
2)What is the total Southward displacement of the hawk during those 10.15 seconds? Specify the direction with a plus or minus sign assuming that Southward is positive and Northward is negative.
3)Finally, what is the final altitude of the hawk at the end of those 10.15 seconds?
Explanation / Answer
(1) displacement in 0.9 s interval = (2.73i + 3.37k)*0.9 + 0.5(-4.156cos74.008i - 4.156sin74.008k)*0.9*0.9
= 2.457i + 3.033k - 0.463i - 1.618k
= 1.994i +1.415k
=> magnitude of displacement = 2.445 m
=> direction =tan-1(1.415/1.994)
= 35.36 degrees below the horizontally Southward direction
(2) total Southward displacement of the hawk during those 10.15 seconds = 1.661 + 23.205 + 1.994
= 26.86 m towards southwards
(3) final altitude of the hawk at the end of those 10.15 seconds = 143 - (1.264+28.64+1.415)
= 111.681 m
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