A swimmer heads directly across a river, swimming at 1.9 m/s relative to the water. She arrives at a point 50 m downstream from the point directly across the river, which is 79 m wide.

(a) What is the speed of the river current?
m/s

(b) What is the swimmer's speed relative to the shore?
m/s

(c) In what direction should the swimmer head so as to arrive at the point directly opposite her starting point?
° upstream from straight ahead

A man can swim directly across a stream of width 300 m in 4 minutes, when there is no current and in 5 minutes when there is current. Find velocity of the current.

To determine the answers, let's break down the problem step by step:

Step 1: Establish the given information.
- The swimmer's speed relative to the water is 1.9 m/s.
- The distance the swimmer is carried downstream is 50 m.
- The width of the river is 79 m.

Step 2: Calculate the speed of the river current.
- The swimmer's speed relative to the shore is composed of two components: the swimmer's speed relative to the water and the speed of the river current.
- Since the swimmer arrives 50 m downstream, it means that the time it took to cross the river is equal to the distance divided by the swimmer's speed: time = distance / speed.
- The swimmer's speed relative to the shore is equal to the width of the river divided by the time to cross the river: relative speed = distance / time.
- Plugging in the values, we get relative speed = 79 m / (50 m / 1.9 m/s).
- Simplifying, we find that the relative speed is 1.9 m/s * (79 m / 50 m) ≈ 3.03 m/s.
- Since the relative speed consists of the swimmer's speed relative to the water plus the speed of the river current, we can conclude that the speed of the river current is 3.03 m/s - 1.9 m/s = 1.13 m/s.

Step 3: Calculate the swimmer's speed relative to the shore.
- The swimmer's speed relative to the shore is already calculated in step 2 and is 3.03 m/s.

Step 4: Determine the direction the swimmer should head.
- To reach the point directly opposite her starting point, the swimmer needs to account for the river current and swim slightly upstream.
- Since the river current is carrying the swimmer downstream, she needs to compensate by swimming upstream. The direction she should head is upstream from straight ahead.

So, the answers to the questions are:

(a) The speed of the river current is 1.13 m/s.
(b) The swimmer's speed relative to the shore is 3.03 m/s.
(c) The swimmer should head upstream from straight ahead.

To answer the questions, we will use the concept of vector addition. The resultant velocity of the swimmer can be found by adding the velocity of the river current to the swimmer's velocity relative to the water.

Let's calculate the answers step-by-step:

(a) To find the speed of the river current, we need to determine the component of the swimmer's velocity that is perpendicular to the river's flow. We know that the swimmer traveled 50 m downstream while crossing the river, which is the same as the component of the river's current.

Using the formula: Velocity = Distance / Time

The time taken to cross the river is given by: Time = Distance / Velocity

Time = 79 m / 1.9 m/s, which is approximately 41.5789 seconds.

Therefore, the component of the river's current is 50 m / 41.5789 s ≈ 1.2052 m/s.

So, the speed of the river current is approximately 1.2052 m/s.

(b) To find the swimmer's speed relative to the shore, we need to calculate the resultant velocity by adding the swimmer's velocity relative to the water to the river current.

The resultant velocity (v_resultant) can be determined using the Pythagorean theorem.

v_resultant^2 = (v_swimmer^2) + (v_current^2)

v_resultant^2 = (1.9 m/s)^2 + (1.2052 m/s)^2

v_resultant^2 ≈ 3.61 m^2/s^2 + 1.4512 m^2/s^2

v_resultant^2 ≈ 5.0612 m^2/s^2

Taking the square root of both sides, we get:

v_resultant ≈ √5.0612 m/s

v_resultant ≈ 2.25 m/s (approx)

Therefore, the swimmer's speed relative to the shore is approximately 2.25 m/s.

(c) To determine the direction the swimmer should head to arrive at the point directly opposite her starting point, we need to consider the angle between the resultant velocity and the direction of the river flow.

The angle can be found using the formula: tan(angle) = (v_swimmer / v_current)

tan(angle) = (1.9 m/s / 1.2052 m/s)

angle = tan^(-1)(1.9 / 1.2052)

angle ≈ 58.6° (approx)

Therefore, the swimmer should head approximately 58.6° upstream from straight ahead to arrive at the point directly opposite her starting point.

tan A = 50m/79m = 0.63291

A = 32.3o East of North = 57.7o CCW.

b. Vs = 1.9m/sin57.7 = 2.25 m.[57.7o] = Velocity of swimmer.

Vc = 2.25m/s[57.7o] - 1.9[90o]

X = 2.25*cos57.7 - 1.9*cos90=1.20 m/s.
Y=2.25*sin57.7 - 1.9*sin90 = 0 m/s.

b. Vc = 1.20 + i0 = 1.20 m/s =Velocity of the current.

c. 32.3oWest of North.

Vc = X/cosB = 1.20/cos0.0878=13.67m/s.