Question

A camera operator is filming a nature explorer in the Rocky Mountains. The explorer needs to swim across a river to his campsite. By watching debris flowing down the river, the operator estimates that the stream is flowing at
0.665
m
/
s
. In still water, the explorer can swim at
0.759
m
/
s
. At what angle, less than 90°, with respect to the shoreline should the operator advise him to swim so that he travels directly across the stream to his campfire?
angle:

°
The water is near freezing in temperature. Typically a human can only swim in such water for about
300
s
(or
5
min
) before hypothermia sets in. Calculate the time the explorer spends in the water if the river is
29.3
m
wide.
time in the water:

s
Based on the results, what should the camera operator's decision be about the explorer's swim?

Approved. He will get cold but he should be able to make it across.

Sorry, but the swim must be cancelled. He will never make it across in time.

Answer 1

Answer:

a. Angle= 28.82°

b. Approved. He will get cold but he should be able to make it across

Explanation:

Velocity Vector

The velocity is a physical quantity that measures how fast or slow at a particular direction some object is moving. It must be expressed as a vector with both a magnitude and direction. If the object is confined to move in one direction, then we can use the speed as the scalar (magnitude only) equivalent of the velocity.

a.

The explorer wants to swim across a river to his campsite, as shown in the image below. The river has a velocity vr and the explorer can swim at ve in still water. If he swam directly to the campsite, he would end up in a point below it because the river would push him down. He must swim with a velocity such that he overcomes the stream but he advances to its objective. Let's call the angle he must swim at respect to the shoreline to achieve his goal. The explorer's velocity can be decomposed in its rectangular components vx and vy. To overcome the river's velocity:

$v_{ey}=v_r$

We can compute the vertical component of the explorer's velocity as

$v_{ey}=|v_e|cos\alpha$

Thus

$v_r=|v_e|cos\alpha$

Solving for $\alpha$

$\displaystyle cos\alpha=\frac{v_r}{|v_e|}$

$\displaystyle cos\alpha=\frac{0.665}{0.759}=0.876$

Then we have the angle is

$\alpha=28.82^o$

b.

The horizontal component of the explorer's velocity is

$v_{ex}=0.759sin28.82^o$

$v_{ex}=0.366\ m/s$

This is the real velocity the explorer is having directly to the campsite

Knowing that

$\displaystyle v=\frac{x}{t}$

Solving for t

$\displaystyle t=\frac{x}{v}$

Calculating the time it takes the explorer to cross the river

$\displaystyle t=\frac{29.3}{0.366}$

$t=80\ sec$

Since this value is less than the limit value of hypothermia (300 sec), the decision is

Approved. He will get cold but he should be able to make it across