Swimming at 2.00 m/s relative to still water, a swimmer heads directly across a river of width 80 m. He arrives 35 m downstream from the point directly across the river from his starting point. (a) What is the speed of the current in the river? (b) At what angle relative to the line perpendicular to the shore should he head to arrive at a point directly opposite his starting point?

Question

Swimming at 2.00 m/s relative to still water, a swimmer heads directly across a river of width 80 m. He arrives 35 m downstream from the point directly across the river from his starting point. (a) What is the speed of the current in the river? (b) At what angle relative to the line perpendicular to the shore should he head to arrive at a point directly opposite his starting point?

Answer 1

Tan A = 35/80 = 0.4375

A = 23.6o

a. Tan23.6 = Vc/2
Vc = 2*Tan23.6 = 0.874 m/s = Velocity of
the current.

b. Direction = 23.6o to the left.

Answer 2

To find the speed of the current in the river and the angle the swimmer needs to head, we can use the concept of vector addition.

Let's break down the motion of the swimmer into two components: the swimming speed relative to still water and the speed of the current.

(a) To find the speed of the current, we need to consider the horizontal displacement of the swimmer. The swimmer travels 35 m downstream, which is the effect of both their swimming speed and the current. Since the current only affects motion horizontally, we can assume that the swimmer's relative speed to still water is entirely vertical.

Using the Pythagorean theorem, we can determine the swimmer's actual displacement while swimming across the river. The horizontal displacement is the distance across the river, which is 80 m. The vertical displacement is zero since the swimmer is not moving vertically with respect to the river.

Using the formula for displacement, we have:

Displacement^2 = Horizontal Displacement^2 + Vertical Displacement^2

Displacement^2 = 80^2 + 0^2
Displacement^2 = 6400

Since the displacement is the sum of the swimmer's velocity relative to still water and the velocity of the current, we can write the equation as:

(2 m/s)^2 + (Current Speed)^2 = 6400

4 + (Current Speed)^2 = 6400
(Current Speed)^2 = 6400 - 4
(Current Speed)^2 = 6396

Taking the square root of both sides, we find:

Current Speed = √6396 ≈ 79.98 m/s ≈ 80 m/s (rounded to two decimal places)

Therefore, the speed of the current in the river is approximately 80 m/s.

(b) To determine the angle relative to the line perpendicular to the shore that the swimmer should head, we can use trigonometry. The swimmer's velocity vector relative to still water and the velocity vector of the current form the resultant velocity vector.

Let θ be the angle the swimmer needs to head. We can find θ using the tangent function:

tan(θ) = (Current Speed) / (Swimming Speed)

tan(θ) = (80 m/s) / (2 m/s)
tan(θ) = 40

Taking the inverse tangent (arctan) of both sides, we find:

θ = arctan(40)

Using a calculator, we find that θ ≈ 87.14° (rounded to two decimal places).

Therefore, the swimmer should head at an angle of approximately 87.14° relative to the line perpendicular to the shore to arrive directly opposite their starting point.