Magnitude problem (velocity of ship relative to water)

[Erin0702]

A ship is sailing due south at 20 miles per hour. A man walks west (i.e., at right angles to the side of the ship) across the deck at 3 miles per hour. What are the magnitude and direction of his velocity relative to the surface of the water?

magnitude = sqrt(u1^2+u2^2)
I'm not sure where to get the points for u from...do I need to draw a picture to get them?

The part that's throwing me off is about everything being relative to the surface of the water...does this mean this picture is going to have to look different?

Thanks!

 

[skeeter]

the man's velocity relative to the surface of the water is just the vector sum of the ship's velocity and the man's velocity.

magnitude = sqrt(20² + 3²)

let the positive x-axis be due East ...

direction relative to the positive x-axis = arctan(-20/-3) + 180

 

[soroban]

Re: Magnitude problem

Hello, Erin0702!

A ship is sailing due south at 20 miles per hour.
A man walks west (i.e., at right angles to the side of the ship) across the deck at 3 miles per hour.
What are the magnitude and direction of his velocity relative to the surface of the water?

They had to say "relative to the surface of the water".
If they asked: "What are his speed and direction relative to the ship?",
\(\displaystyle \;\;\)the answer is embarrassingly simple: 3 mph west . . . duh!

Imagine looking down from a GPS satellite.
You see the ship moving south at 20 mph and the man moving 3 mph west across the ship.
In your view (relative to the earth), the man is walking diagonally toward the southwest.
And that is the direction and magnitude (speed) they are asking for.

         3    A
    *---------*
             /|
            / |
           /  |
          / θ |
         /    | 20
        /     |
       /      |
      /       |
     /        |
  B * - - - - *
         3

The magnitude is the length of \(\displaystyle AB:\;\;\sqrt{3^2\,+\,20^2}\:=\:\sqrt{409} \:\approx\:20.22\)

\(\displaystyle \;\;\)Therefore, his speed is about \(\displaystyle 20.22\) mph.


Since \(\displaystyle \,\tan\theta\:=\:\frac{3}{20}\), we have: \(\displaystyle \,\theta\:=\:\arctan(0.15) \:\approx\:8.53^o\)

\(\displaystyle \;\;\)Therefore, his direction is: \(\displaystyle S\,8.53^o\,W\)