A ship travels 3.20 hours with a velocity of 21.1 mph at theta= 123º, including the effects of the

current. The pilot then turns to a heading of 190.0° with a speed of 18.2 mph, including the effects of

the current. If the pilot stays on this heading for 1.10 hours, what is the ship's final displacement from

its original starting point?

The vector addition and kinematics of uniform motion allows to find the result for where the starting point is:

Given parameters

To find

Velocity is a vector magnitude, therefore it has direction and module, to make the addition of these magnitudes vector algebra must be used.

There are two methods for adding vectors:

In the attached we can see a diagram of the displacement of this ship. We use a Cartesian reference system and using trigonometric to decompose the velocities.

First speed

            cos 123 = [tex]\frac{v_1_x}{v_1}[/tex]  

            sin 123 = [tex]\frac{v_1_y}{v_1}[/tex]  

            v₁ₓ = v₁ cos 123

            [tex]v_{1y}[/tex] = v₁ sin 123

            v₁ₓ = 21.1 cos 123 = -11.49 mph

            [tex]v_{1y}[/tex] = 21.1 sin123 = 17.70 mph

Second speed

            v₂ₓ = v₂ cos 190

            [tex]v_{ 2y}[/tex] = v₂ sin190

            v₂ₓ = 18.2 cos 190 = -17.92 mph

            [tex]v_{2y}[/tex] = 18.2 sin190 = -3.16 mph

Now we can use kinematics of uniform motion.

            v = [tex]\frac{d}{t}[/tex]  

             d = v t

Where v is the velocity, d the distance and t the time

Let's find every distance traveled.

For first speed

x-axis

         x₁ = v₁ₓ t₁

         x₁ = -11.49  3.20

         x₁ = - 36,768 miles

y-axis  

         y₁ = [tex]v_{1y} \ t_1[/tex]

        y₁ = 17.70 3.20

        y₁ = 56.64 miles

Travel for 2nd speed

       x₂ = v₂ₓ t₂

       x₂ = -17.92 1.10

       x₂ = -19,712 miles

       y₂ = -3.16 1.10

       y₂ = - 3,476 miles

Let's find the distance traveled on each axis

x-axis

    [tex]x_{total}[/tex] = x₁ + x₂

    [tex]x_{total}[/tex] = -36,768 - 19,712

     [tex]x_{total}[/tex] = -56.48 miles

y-axis

     [tex]y_{total}[/tex] = y₁ + y₂

     [tex]y_{total}[/tex] = 56.64 - 3.476

     [tex]y_{total}[/tex] = 53,164 miles

Let's construct the total displacement vector, for the module let's use the Pythagorean theorem

       [tex]d_{total} = \sqrt{x^2 +y^2} \\d_{total} = \sqrt{56.48^2 +53.164^2}[/tex]

       [tex]d_{total}[/tex] = 77,565 miles

Let's used trigonometry for the direction

      [tex]tan \theta ' = \frac{y_{total} }{ x_{total} } \\\theta ' = tan^{-1} \frac{y_{total} }{ x_{total} }[/tex]  

      θ’= 43.3º

This angle is in the second quadrant to indicate the angle with respect to the positive side of the x-axis in a counterclockwise direction.

   θ = 180 - θ'

    θ = 180 - 43.3

   θ = 136.7º

In conclusion using the addition vectors and kinematics of uniform motion we can find the result for where the starting point is:

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