Relative Velocity of a swimmer

[yowatup]

Homework Statement

In an anniversary celebration of Marilyn Bell's 1954 crossing of Lake Ontario a swimmer set out from the shores of New York and maintained a velocity of 4m/s [N]. As the swimmer approached the Ontario shore, she encountered a cross current of 2m/s [E 25deg S]. Find her velocity with respect to the crowd observing from the beach.

The Attempt at a Solution

Firstly, am I to understand that this is a right-angle triangle? When representing it graphically, it certainly does not look like a right triangle:

http://i543.photobucket.com/albums/gg464/yowatupguystill/vector.jpg

However, when I endeavor to solve this by converting from polar to cartesian co-ordinates, it seems that I have to assume a right-triangle.

Let S be the swimmer, W be the water, and G the ground.

sVw = 4 m/s [N] = (4, 90*)
wVg = 2 m/s [E25*S] = (2, -335*)
sVg = ?
.: sVg = sVw + wVg
= [0, 4] + [1.8, 0.84]
= [1.8, 4.84]
= (5.2, 69.5*)

I am not very confident in my answer. For starters, I am not supposed to really solve this using polar-cartesian conversion, but I was at a standstill when attempting another solution. Any light shed on a solution for this would be much appreciated.

 

[Redbelly98]

Yo. Welcome to PF

[1.8, 0.84]

Double-check the +/- signs in those numbers. Things look fine otherwise.

I am not supposed to really solve this using polar-cartesian conversion ...

That's weird, because that is by far the standard and preferred way to solve problems like this. Alternatively, but more cumbersome, is to use the law of cosines and law of sines from trig.

 

[thomate1]

I would like to point out that the relative velocity of the swimmer with respect to the crowd observing from the beach is not a simple addition of the swimmer's velocity and the cross current's velocity. This is because the swimmer and the crowd are both moving in different directions and at different speeds. Therefore, we need to consider the concept of relative velocity, which is the velocity of an object with respect to another object in motion.

In this case, the relative velocity of the swimmer with respect to the crowd can be found by subtracting the velocity of the crowd from the velocity of the swimmer. We can break down the swimmer's velocity into its horizontal and vertical components, and then add the cross current's velocity to the horizontal component. This will give us the swimmer's velocity with respect to the crowd.

The swimmer's horizontal velocity can be calculated as 4m/s * cos(90°) = 0m/s, and the vertical velocity can be calculated as 4m/s * sin(90°) = 4m/s. Adding the cross current's velocity, which has a magnitude of 2m/s and a direction of 25° south of east, we can find that the horizontal component is 2m/s * cos(25°) = 1.84m/s and the vertical component is 2m/s * sin(25°) = 0.84m/s.

Therefore, the swimmer's velocity with respect to the crowd can be calculated as (0m/s + 1.84m/s, 4m/s + 0.84m/s) = (1.84m/s, 4.84m/s). This means that the swimmer is approaching the crowd at a speed of 1.84m/s and moving upwards at a speed of 4.84m/s.

It is important to note that this calculation assumes that the crowd is stationary and not moving with the current. If the crowd is also moving with the current, then the relative velocity of the swimmer would be different. In order to find the exact relative velocity, we would need to know the velocity of the crowd as well.