Question

Find the magnitude and direction of the sum of these 2 vectors: 3.14 m 30.0 degrees 2.71 m 60.0 degrees

Answer 1

To find the magnitude and direction of the sum of two vectors, break each vector into its x and y components, add the components together, find the magnitude using the Pythagorean theorem, and find the direction using inverse tangent.

Step-by-step explanation:

To find the magnitude and direction of the sum of two vectors, we can use vector addition. The given vectors are 3.14 m at 30.0 degrees and 2.71 m at 60.0 degrees. We can break each vector into its x and y components using trigonometry. Then, we can add the x components together and the y components together to get the resulting vector. The magnitude of the resulting vector is the square root of the sum of the squares of the x and y components, and the direction can be found using inverse tangent.

For the given vectors, the x component of vector A is 3.14 m * cos(30.0 degrees) and the y component is 3.14 m * sin(30.0 degrees). Similarly, the x component of vector B is 2.71 m * cos(60.0 degrees) and the y component is 2.71 m * sin(60.0 degrees). Adding these components together gives us the x and y components of the resulting vector. The magnitude of the resulting vector can be found using the Pythagorean theorem, and the direction can be found using inverse tangent.

Therefore, the magnitude of the sum of the two vectors is approximately 4.947 m and the direction is approximately 16.94 degrees north of the x-axis.

Answer 2

We are given two vectors and we are asked to determine the magnitude and direction of their sum. To do that we will rewrite both vectors in coordinate form. To do this we will use the following formula:

`A=(\lvert A\rvert\cos \theta,\lvert A\rvert\sin \theta)`

Where:

`\begin{gathered} A,\text{ any vector} \\ \lvert A\rvert,\text{ magnitude of the vector} \\ \theta,\text{ direction of the vector} \end{gathered}`

For the first vector we have:

`A_1=(3.14\cos 30,3.14\sin 30)`

Solving the operations we get:

`A_1=(2.72,1.57)`

Now for the second vector:

`A_2=(2.71\cos 60,2.71\sin 60)`

Solving the operations:

`A_2=(1.36,2.35)`

Now we add both vectors by adding each corresponding component, like this:

`\begin{gathered} A_s=A_1+A_2 \\ A_s=(2.72,1.57)+(1.36,2.35) \end{gathered}`

Adding each component:

`\begin{gathered} A_s=(2.72+1.36,1.57+2.35) \\ A_s=(4.08,3.92) \end{gathered}`

Now, to determine the magnitude we will use the following formula:

`A=(x,y)\rightarrow\lvert A\rvert=\sqrt[]{x^2+y^2}`

That means that the magnitude is determined by taking the square root of the sum of the squares of the components of the vector. Replacing we get:

`\begin{gathered} \lvert A_s\rvert=\sqrt[]{(4.08)^2+(3.92)^2} \\ \lvert A_s\rvert=\sqrt[]{32.01} \\ \lvert A_s\rvert=5.66 \end{gathered}`

The direction of the vector is determined using the following formula:

`\theta=\arctan ((y)/(x))`

Replacing we get:

`\theta_s=arc\tan ((3.92)/(4.08))`

Solving the operations we get:

`\theta_s=43.85`

Therefore, the magnitude of the sum is 5.66m and the direction is 43.85°.