Vector problem....?

A ship heads in a direction N 25° W. The speed of the ship relative to the water is 50mi/hr. The water is flowing derectly south. It is observed that the true direction of the ship is directly west.


a) Express the velocity of the ship as a vector in componet form.


b) Find the speed of the water.


c) Express the velocity of the water as a vector in componet form.


d) Find the true speed of the ship.

Vector problem....?
It would be nice to include a diagram, but I will try without one, using the component forms of vectors in the following form: %26lt;mcosƏ, msinƏ%26gt;, where m is the magnitude of the vector and Ə is the direction from the positive x-axis. Speed is used to convey magnitude in these applications.





ship's component vector = %26lt;50cos115°, 50 sin 115°%26gt;


current's component vector = %26lt;wcos270°, wsin270°%26gt; = %26lt;0, -w%26gt;, where w is the speed of the water





The actual direction of the ship is the resultant (sum) vector of the above 2 vectors. We know that it's due west, so Ə=180° and its magnitude is the actual speed of the ship (let's call it S) . The sum of the 2 vectors will therefore actually equal the vector %26lt;-S,0%26gt;, where S is the true speed of the ship. As such,





%26lt;50cos115°, 50 sin 115°%26gt; + %26lt;0, -w%26gt; = %26lt;-S,0%26gt;





so S = -50cos115° = 21.2 mph


and W = 50sin 115° = 45.3 mph





Really fast current!!
Reply:you're welcome! Report It

Reply:What's helpful is to draw a picture. There should be two vectors, one for the speed and direction of the boat, one for the speed and direction of the water. The boat vector, we'll call it B, has a magnitude of 50 and a direction 25 degrees to the left of north (we'll assume north is straight up on the paper). The water vector, we'll call W, is going straight down, and has an unknown magnitude. Both vectors are coming out of a single point, where the boat is. To find B's components, draw a right triangle with B as the hypotenuse. Find the lengths of the two sides, which will be B's two components, one for the x direction, Bx, and one for the y direction, By. Make sure you use the correct angle when the sides, 25 degrees was measured from the y-axis, not the x-axis. A common way to write this in component form is %26lt;Bx i^ + By j^%26gt; (put the carots " ^ " on top of the i and j, instead of next to them i can't get the computer to do that)





Since the boat is not moving North or South, the Wy=By The y-components of each velocity vector cancel each other, thus the boat does not move in the y (N-S) direction. Since W has no x-component (it's not moving East or West) the velocity of the water must equal Wy.





The y component of W is already figured out, that's Wy, and the x component we already know is 0, so you can just write %26lt;-Wy j^%26gt; You put the minus because the water is going south.





The true speed of the ship, since the y components of the two velocity vectors cancel, is the sum of the x components. But since W has no x component, the true speed is just the magnitude of Bx.





I hope this helps!! =)

pansy