Respuesta :
Answer: part a: v_river=1.78 m/s and part b:
According to his friend he is moving in at 4.05m/s in at an angle theta with the river's edge
Explanation:The velocity v_s of the swimmer in the rivers direction ( take it as x) is
[tex]v_x= 1.1 m/s\\[/tex]
width= w =24
hence the time taken to cross the width can be found by
t=[tex]w/v_x[/tex]
t=21.81 secs
this is due to the fact that rivers velocity don't effect the time taken by the swimmer to reach the other side but only moves the swimmer downstream
downstream distance =d=39m
The swimmer stays for the same time t in the river and is move 39m
the river velocity can be then found by using this distance
[tex]v_y=d/t\\v_y=1.7875[/tex]
as the swimmer now has both the x and y components of velocity the net velocity is
[tex]v=\sqrt{v_x^2+v_y^2} \\v=2.1m/s[/tex]
According to his friend he is moving in at 4.05m/s in at an angle theta with the river's edge
theta= [tex]tan^{-1} (v_y/v_x)[/tex]
for the values you mentioned the problem statement i.e.
w=29
d=37
vs=1
Using the same procedure as above the
t=w/vs
t=29
[tex]vr=v_y\\v_y=d/t\\v_y=1.276[/tex]
Also for part b
[tex]v=\sqrt{v_x^2+v_y^2} \\v=1.6m/s[/tex]
