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

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

$v=\sqrt{v_x^2+v_y^2} \\v=2.1m/s$

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

$v=\sqrt{v_x^2+v_y^2} \\v=1.6m/s$