V = 6.83m/s @ (-90)o + 2.89m/s @ 0o.
X=6.83*cos(-90) + 2.89*cos(0)=2.89m/s.
Y=6.83*sin(-90) + 2.89*sin(0)=-6.83m/s.
tanA = Y/X = -6.83/2.89 = -2.36332
A = -67.1 = 67.1o Below the desired direction. Therefore, the swimmer must head in a direction 67.1o above the desired direction.
A swimmer swims in still water at a speed of 6.83 m/s. He intends to swim directly across a river that has a downstream current of 2.89 m/s.
(a) What must the swimmer's direction be?
3 answers
Vsc = Vs + Vc = 6.83i + 2.89
Tan A = 6.83/2.89 = 2.36332
A = 67.1o CCW = 22.9o E. of N.(90-67.1).
Direction = 22.9o W. 0f N.
Tan A = 6.83/2.89 = 2.36332
A = 67.1o CCW = 22.9o E. of N.(90-67.1).
Direction = 22.9o W. 0f N.
1) sin^-1 (2.89/6.8)
2) 2.89^2 + x^2 = 6.8^2
2) 2.89^2 + x^2 = 6.8^2
1 answer