Question

An athlete crosses a 24 m wide river by swimming perpendicular to the water current at a speed of 1.1 m/s relative to the water. He reaches the opposite side at a distance 39 m downstream from his starting point.

w= 29m

d=37m

vs= 1 m/s

A) How fast is the water in the river flowing with respect to the ground in m/s?

B)What is the speed of the swimmer with respect to a friend at rest on the ground in m/s?

Answer 1

The river's current flows at approximately 1.83 m/s with respect to the ground, and the swimmer's speed with respect to a friend at rest on the ground is approximately 2.14 m/s.

Step-by-step explanation:

To determine the speed of the river and the speed of the swimmer with respect to a friend on the ground, we need to find two components: the swimmer's velocity perpendicular to the current and the river's current velocity. Given that the swimmer crosses the river and ends up 40 meters downstream, we can set up a right triangle where the width of the river is one side (24 m), and the distance downstream (where the swimmer ends up) is the other side (40 m).

The swimmer's speed perpendicular to the current is 1.1 m/s. To find the time it takes for the swimmer to cross the river, we can use the equation:

Time = Distance / Speed

Time = 24 m / 1.1 m/s = 21.82 seconds (approximately).

Now, we can calculate the current's speed by using the downstream distance and the time:

Current Speed = Distance downstream / Time

Current Speed = 40 m / 21.82 s = 1.83 m/s (approximately).

For part B, calculating the speed of the swimmer with respect to a friend at rest on the ground involves combining the swimmer's speed across the river and the current's speed down the river to find the resultant velocity:

Resultant Speed = √(Swimmer's Speed² + Current Speed²)

Resultant Speed = √(1.1² + 1.83²)

Resultant Speed = √(1.21 + 3.35)

Resultant Speed = √(4.56)

Resultant Speed = 2.14 m/s

Answer 2

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

`v_x= 1.1 m/s\\`

width= w =24

hence the time taken to cross the width can be found by

t=
`w/v_x`

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

`v_y=d/t\\v_y=1.7875`

as the swimmer now has both the x and y components of velocity the net velocity is

`v=√(v_x^2+v_y^2) \\v=2.1m/s`

According to his friend he is moving in at 4.05m/s in at an angle theta with the river's edge

theta=
`tan^(-1) (v_y/v_x)`

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

`vr=v_y\\v_y=d/t\\v_y=1.276`

Also for part b

`v=√(v_x^2+v_y^2) \\v=1.6m/s`