Ta The Expert TAI Human-lik x C https:// 49 Tube pertt Assignment Status Click h
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Ta The Expert TAI Human-lik x C https:// 49 Tube pertt Assignment Status Click here for detailed view Problem Status Completed Completed Completed Completed Completed Completed Completed Completed 10 Class Management Help Chapter 3 HW Begin Date: 10/5/2015 11:00:00 AM Due Date: 10/11/2015 9:00 PM End Date: 12/11/2015 11:59:00 PM (10%) Problem 10: An athlete crosses a 22 m wide river by swimming perpendicular to the water current at a speed of 0.15 m/s relative to the water. He reaches the opposite side at a distance 38 m downstream from his starting point. Randomized Variables 22 38 0.15 m/s 50% Part (a) How fast is the water in the river flowing with respect to the ground in m/s? Grade Summary 0.1118 Potential 7 8 9 HOME Submissions acoso ITA A 4 5 6 Attempts remaining: cotan0 asino (696 per attempt) at ano acotan0 sinh0 cos ho tanho cotanh0 BACKSPACE CLEAR Degrees O Rad deduction. Hints remaining Feedback: 09 deduction per feedback. -A diagram might help you work out this problem -How are the displacement and the velocity related in this problem? The velocity triangle is similar to the displacement triangle. 4 50% Part b) What is the speed of the swimmer with respect to a friend at rest on the ground in m/s? 0/11/20Explanation / Answer
Time to cross t = w/v =22/0.15=146.66 sec
a) Obviously speed of the river current = 38/146.66=0.259 m/sec
b) The swimmer's velocity now has two perpendicular components. You can compute the magnitude of his velocity the same way you compute hypotenuse of a right angled triangle.
Compared to an observer at rest, the swimmer's speed is V = [Vr² + v²]
V =sqrt(0.259^2+0.15^2)=0.299 m/sec
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