Relative Motion (Swimmer Crossing River Question)

[Aristata]

Homework Statement

A swimmer who can swim at a speed of 0.80m/s in still water heads directly across a river 86m wide. The swimmer lands at a position on the far bank 54 m downstream from the starting point. Determine:

(C) The direction of departure that would have taken the swimmer directly across the river.

(s - swimmer
g - ground
w - water

Vsw = 0.8 m/s
d across stream = 86m)

Homework Equations

(n/a - see bellow)

The Attempt at a Solution

From the previous two parts of the question I determined that it took the swimmer 107.5s to cross the river and thus the speed of the current is 0.5 m/s [E]. And that the velocity of the swimmer relative to the shore was 0.94 m/s [58 N of E]. (Which according to my textbook is correct.)

(t=107.5s
Vwg = 0.5 m/s [E]
Vsg=0.94 m/s [58 N of E])

Now, I figured that in order to end up straight across where you start from, you would have to swim [58 N of W] since the current resulted in the swimmer following a path of [58 N of E]. (So this would negate the effect of the current?) However, according to the textbook the answer is [W 51 N] and I have no clue how else to approach this question. Any help/tips please?

 

[denverdoc]

the situation isn't symmetric as one might intuit at first glance. In the first case, all the velocity goes to crossing the river, in the second some is lost; ie

t=86/.8=107.5 as you posted. The velocity of the current downstream then just 54/107.5=0.5 m/s

so now you need to aim upwards to compensate for the drift--in other words total velocity^2=0.64=Vx^2+0.5^2 solve for Vx, then angle should fall out.

 

[m2003]

Relative motion is a concept that is often used in physics to describe the motion of an object with respect to another moving object. In this case, the swimmer is moving with a constant velocity of 0.8 m/s in still water, while the river's current is moving at a velocity of 0.5 m/s to the east. This creates a relative motion between the swimmer and the shore, resulting in the swimmer landing 54 m downstream from the starting point.

To determine the direction of departure that would have taken the swimmer directly across the river, we can use vector addition to find the resultant velocity. The swimmer's velocity relative to the ground is 0.94 m/s [58 N of E], and the current's velocity is 0.5 m/s [E]. Using the Pythagorean theorem, we can find the magnitude of the resultant velocity, which is 1.06 m/s.

Now, to find the direction of the resultant velocity, we can use trigonometry. The angle between the swimmer's velocity relative to the ground and the resultant velocity is the same as the angle between the river's current and the shore. This angle can be found using the inverse tangent function, which gives us approximately 26.6 degrees. Therefore, the direction of departure that would have taken the swimmer directly across the river is 26.6 degrees west of north, or [W 51 N]. This takes into account the effect of the current and allows the swimmer to cross the river in a straight path.

It is important to note that relative motion can be a complex concept, and it is important to fully understand the velocities and directions involved in order to accurately determine the resultant velocity. I would recommend reviewing the concepts and equations used in this problem and practicing with similar problems to solidify your understanding.