Question

Add the following two vectors if \vec{C} = \vec{B} - \vec{A}
Keep a few digits.
|\vec{A}| = 8 m/s
\theta_{A} = 45^{\circ}

|\vec{B}| = 4 m/s
\theta_{B} = 20^{\circ}

\vec{C} = Answer m/s
at ___ degrees r^{\circ} counterclockwise from the +x-axis.

Answer 1

The addition of vectors allows us to find that the resulting vector is

         C = 4.69 m/s with an angle of θ = 246.2º

Giving Parameters

• The vector A    A = 9 m / s θ = 45º

• Vector B           B = 4 m / s θ = 20º

To find

• The vector subtracts

Vectors are quantities that have magnitude and direction, so their addition can be done by graphical and analytical methods.

Analytical method is one of the best ways to perform vector addition, it has several parts:

• Decompose vectors into a coordinate system

• Perform the algebraic sum of the components

• Find the resultant vector.

Let's decompose the two vectors into a coordinate system with the horizontal x axis and the vertical y axis, in the attachment we can see a diagram of the vectors

Vector A

            cos 45 = $\frac{A_x}{A}$

            sin 45 = $\frac{A_y} {A}$

            Aₓ = A cos 45

            A_y = A sin 45

            Aₓ = 8 cos 45

            A_y = 8 sin 45

            Aₓ = 5,657 m / s

            A_ = 5,657 m / s

Vector B

            cos 20 = $\frac{B_x}{B}$

            sin 20 = $\frac{B_y}{B}$

            Bₓ = B cos 20

            B_y = B sin 20

            Bₓ = 4 cos 20 = 3,759 m / s

            B_y = 4 sin 20 = 1,368 m / s

They ask for the difference of the two vectors for which we must perform the algebraic subtraction

x-axis

            x = Bx -Aₓ

            x = 3,759 -5,657

            x = -1.894 m / s

y-axis

           y = B_y - A_y

           y = 1,368 - 5,657

           y = -4.289 m / s

We construct the resulting vector

For the module we use the Pythagorean theorem

          c = $\sqrt{x^2 + y^2}$

           c =   $\sqrt{1.894^2 +4.289^2}$      

           c = 4.689 N

For the direction we use trigonometry

         tan θ’= $\frac{y}{x}$

         θ'= tan⁻¹ $\frac{y}{x}$

         θ'= tan⁻¹ $\frac{4.289}{1.894}$

         θ’= 66.2º

Note that the two magnitudes are negative therefore they are in the third quadrant, to measure from the first quadrant in a counterclockwise direction from the positive side of the x axis

          θ = 180 + θ'

          θ = 180 + 66.2

          θ = 246.2º

In conclusion using vector addition we can find the resulting vector is

         c = 4.69 m / s with an angle of θ = 246.2º

Learn more about vector addition here:

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