Dave rows a boat across a river at 3.9 m/s. The river flows at 5.8 m/s and is 340 m across.?

Question

(a) In what direction, relative to the shore, does Dave's boat go?
°
(b) How long does it take Dave to cross the river?
s
(c) How far downstream is Dave's landing point?
m


I am trying to use the pathag therom, but my anwers are still coming out wrong. The trig functions are also confusing so if someone could explain them to me that would be great

Please help

Answer 1

He rows 3.9 meters a second? Wow, he's cookin'.

A.) 56 degrees, more or less

B.) 87.1 seconds - 340 meters across the river at 3.9 meters per second.

C.) 505.18 Meters - the distance traveled in 87.1 seconds at 5.8 meters per second

Answer 2

(a) A^2 + B^2 = C^2 (3.4m/s)^2 + (2.5m/s)^2 = C^2 C = 4.22 m/s The degree extremely relies upon which direction the river is flowing, notwithstanding the basics of that is Arctan (Vfy/Vix) so enable's assume that 3.4m/s is going North via the river and the present is flowing at 2.5m/s due East.. Arctan(3.4/2.5) = fifty 3.sixty seven degrees North of East. (b) Y = ViyT Y/Viy=T 39m/3.4m/s=T T= 11.47seconds (c) X=VixT X= (2.5m/s)(11.47s) X= 28.68m downstream