# Find the Sea Doo speed.A Sea Doo travels due north at velocity of 45 kmh and encounters a wave of a velocity of 10 kmh. Find the resulting velocity of the Sea Doo.

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### 1 Answer

This question does not specify what direction the wave is traveling. In that case, let's just do 2 things. Let's assume one of four cases:

1) The wave is travelling the opposite or same direction.

2) The wave is travelling at an angle `theta` relative to our hero on the Sea Doo. (the general case)

In each case, let's define a set of coordinate axes such that the **positive y direction is north, and the positive x direction is east**.

**Case 1:**

If the wave is travelling in the opposite or same direction, then clearly, it will reduce or increase the effective speed of the Sea Doo.

If you want to think in terms of vectors, whenever we have a situation like this, we are adding the vertical and horizontal components of vectors. To combine the Sea Doo and wave's velocities, we get the following equation:

`vecv = vecv_(s)+vecv_w`

where `vecv` is our resultant velocity, `vecv_s` is the velocity of the Sea Doo, and `vecv_(w)` is the velocity of the wave. Keep in mind, each velocity vector will have two components: the horizontal and vertical. So, our Sea Doo vector will be:

`v_s = (0, 45)`

because it is not travelling east or west, and it is travelling north at 45 mph. The wave vector in Case 1 will be:

`vecv_w=(0, +-10)`

because it also is not travelling east or west, and it is travelling north or south at 10 mph (the direction south is negative, remember!).

Now, we combine them with a vector sum to get our resultant vector:

`vecv = vecv_s+vecv_w = (0,45) + (0,+-10) = (0+0, 45+-10) = (0,35) or (0,55)`

So, in the case that the wave opposes our hero's motion, he ends up with an overall velocity of 35 mph North. In the case that the wave goes with our hero's direction, he goes north at 55 mph.

**Case 2: The General Case!**

In the case that we are unsure what the angle of approach of the wave is, let's just say it approaches at an angle `phi` **relative to the direction due east**. We then know that the vector components will be related to trigonometric functions of `phi` multiplied by the speed of the wave! For more information on this, see the link below. Our wave velocity becomes:

`vecv_w = (10cosphi,10sinphi)`

To get our hero's resultant velocity vector, we just add the components as we did before:

`vecv = vecv_s+vecv_w = (0,45)+(10cosphi,10sinphi)=(10cosphi,45+10sinphi)`

We should have our answer. However, we more often want the speed (magnitude) and direction of travel.

To get the speed, we take the magnitude (pretty much by applying the pythagorean theorem to the vector components), seen below:

`s =|vecv|=sqrt(100cos^2phi+45^2 + 900sinphi + 100sin^2phi)`

Now, to simplify this, we'll need to use our trig identities and find that the first and last term combine to 100, and that 45^2 = 2025:

`s = sqrt(2125+900sinphi) = 5sqrt(85+36sinphi)`

This is about as much as we can reduce the square root, unfortunately. Let's move on to finding direction.

Normally, the angle at which we're travelling can be found simply by taking the arctan of the vertical over the horizontal component:

`theta=tan^-1((45+10sinphi)/(10cosphi))`

This will work find for us, except in the case that the arctan becomes negative. If this is the case, to get the correct direction we need to add 180 degrees. The reason for this is because of the defined range of arctan (-90 to 90 degrees instead of 0 to 180) and the face that the wave cannot reverse the direction of our hero on the Sea Doo because it simply isn't powerful enough!

Hope that helps!

**Sources:**