Relative motion of a swimmer crossing a river

[chengbin]

A swimmer can swim at 2.0 m/s in still water. She attempts to swim due north across a river flowing eastward at 1.5 m/s.

a) What is her velocity relative to the ground?

b) Suppose she wants to arrive directly across from her starting point. In what direction must she head? What will be her velocity with respect to the ground?

My teacher wants the solution in this format and notation.

V_{s,w} = 2.0 m/s [N]

V_{w,g} = 1.5 m/s [E]

V_{m,g} = V_{s,w} + V_{w,g}

V_{m,g} = V_{s,w} + V_{w,g}

= \sqrt {(2.0 m/s)^2 + (1.5 m/s)^2}

=2.5 m/s

\tan \theta = \frac {1.5 m/s}{2.0 m/s}

\theta = 37^{\circ}

The relative velocity to the ground is 2.5 m/s [N 37 E]

I don't know how to setup the solution for part b. The teacher only told us only use the equation V_{a,c} = V_{a,b} + V_{b,c}. How do I do this question? I know I have to do = \sqrt {(2.0 m/s)^2 - (1.5 m/s)^2} to get the answer, but how would I show that in the above notation?

 

[JaredJames]

I don't think there is any need for for notation for part B.

In order to travel in a straight line and arrive directly across from her starting point she would have to compensate for the eastward flowing river. Which means swimming 2.5 m/s 37 degrees west (relative to north).

If you really need to show notation, you could show that:

<br /> 1.5 = \sqrt {(2.5)^2 - (2.0)^2}<br />

Which is the horizontal component of swimming 2.5 m/s 37 degrees west (relative to north) - it compensates for the eastward flow of 1.5m/s.

And then with Sin\theta = O / H = 1.5 / 2.5 = 37^{\circ}

This is the angle (relative to north) she needs to swim at.

Jared

 

[chengbin]

I don't think there is any need for for notation for part B.

In order to travel in a straight line and arrive directly across from her starting point she would have to compensate for the eastward flowing river. Which means swimming 2.5 m/s 37 degrees west (relative to north).

If you really need to show notation, you could show that:

<br /> 1.5 = \sqrt {(2.5)^2 - (2.0)^2}<br />

Which is the horizontal component of swimming 2.5 m/s 37 degrees west (relative to north) - it compensates for the eastward flow of 1.5m/s.

And then with Sin\theta = O / H = 1.5 / 2.5 = 37^{\circ}

This is the angle (relative to north) she needs to swim at.

Jared

Actually you do. The answer is actually 1.3 m/s at 49 degrees. Our teacher said that's a classic problem, and most people think the answer is 37 degrees, while it is actually 49.

Therefore I need to show notation to show my understanding of how it works.

 

[JaredJames]

But that would only give you 1.0m/s westward, the stream would still pull you at 0.5m/s eastward. I must be missing something here.

If she swam directly into the flow she would have to maintain 1.5m/s to remain in one place. In order to travel northwards she has to turn so some of her swimming action provides the vertical motion. For every degree she turns northwards, she has to increase the speed she swims at in order to maintain the horizontal 1.5m/s component.

An aircraft with an airspeed of 60 knots, flying into a headwind of 60 knots does not have any groundspeed. It is not moving, it is stationary. I don't see how this is different.

Jared

 

[chengbin]

But that would only give you 1.0m/s westward, the stream would still pull you at 0.5m/s eastward. I must be missing something here.

If she swam directly into the flow she would have to maintain 1.5m/s to remain in one place. In order to travel northwards she has to turn so some of her swimming action provides the vertical motion. For every degree she turns northwards, she has to increase the speed she swims at in order to maintain the horizontal 1.5m/s component.

Jared

The hypotenuse is 2 m/s, the horizontal side (water) is 1.5 m/s, and the vertical side is the velocity relative to the ground.

\sqrt {(2.0)^2 - (1.3)^2} = 1.3

To find the angle,

\sin \theta = \frac {1.5}{2}

\theta = 49^{\circ}

 

[JaredJames]

Ah, I misread the question for my first answer (thought it wanted her northward speed the same) and took the 1.3 to be the hypotenuse because you said

1.3 m/s at 49 degrees

In which case, 49 degrees from north is correct.

Jared

 

[chengbin]

I know how to find the answer, but I need to show it, because that's half the marks.

 

[JaredJames]

Well you know the hypotenuse = 2.0m/s and you know the required westward speed = 1.5m/s. Plug those two into your equation and you come out with the vertical velocity. Once you have that, do as you have before and you get the angle wrt north.

The answer will be 2.0m/s at 49 degrees from north.

To get the 49 degrees, it's as you did before. Sin(theta) = O / H = 1.5 / 2.0 = 0.75

Inverse Sin 0.75 = 49 degrees wrt north.

Jared