# 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?

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.