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.