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?
1) sin^-1 (2.89/6.8)
2) 2.89^2 + x^2 = 6.8^2
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.
To find the swimmer's direction, we need to consider the speed and direction of the river current. Let's break down the forces acting on the swimmer:
1. The swimmer's swimming speed in still water is 6.83 m/s. This is the speed at which the swimmer can swim without any external force acting on them.
2. The river has a downstream current with a speed of 2.89 m/s. This means the water is flowing in the same direction the swimmer wants to cross the river.
Since the swimmer wants to swim directly across the river, they need to compensate for the downstream current. To do this, they should point themselves slightly upstream. By doing so, the downstream current will carry them diagonally downstream while they swim diagonally upstream.
To calculate the angle at which the swimmer should point, we can use trigonometry. The tangent of the angle between the swimmer's intended direction and the river's direction is equal to the ratio of the river's current speed to the swimmer's swimming speed:
tan(θ) = (River current speed) / (Swimmer's swimming speed)
θ = arctan((River current speed) / (Swimmer's swimming speed))
Plugging in the numbers:
θ = arctan(2.89 / 6.83)
Using a scientific calculator or a trigonometric table, we can find the arctan value:
θ ≈ 0.396 radians
The swimmer's direction should be approximately 0.396 radians upstream from directly across the river.
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.