Mastering Vector Addition and Subtraction

Vector addition is the process of combining two or more vectors to obtain their resultant vector. It is performed by lining up the vectors and adding their corresponding components.

The result of vector 1 plus vector 2 is the vector obtained by lining up vector 1 and vector 2 and adding their corresponding components.

The result of vector 1 minus vector 2 is the vector obtained by lining up vector 1 and the flipped version of vector 2 (to subtract it away) and adding their corresponding components.

<p>To calculate the total force experienced by an object when subjected to multiple forces, you add up the vectors representing those forces by lining them up and adding their corresponding components.</p> Signup and view all the answers

<p>The formula for finding the magnitude of a resultant vector is $c = \sqrt{a^2 + b^2}$, where $a$ and $b$ are the components of the vector in the x and y direction respectively.</p> Signup and view all the answers

<p>The formula for finding the angle of a resultant vector is $\theta = \arctan\left(\frac{b}{a}\right)$, where $a$ and $b$ are the components of the vector in the x and y direction respectively.</p> Signup and view all the answers

<p>The pythagorean theorem can be used to find the magnitude of a resultant vector by squaring the components of the vector in the x and y direction, adding them together, and taking the square root of the sum: $c = \sqrt{a^2 + b^2}$.</p> Signup and view all the answers

<p>The total displacement of the slow velocity eraser can be found by adding the vectors a, b, and c. To find the resultant vector, we can use the law of cosines. Let's call the total displacement vector R. The magnitude of R can be calculated using the equation: $|R| = \sqrt{(|a| + |b|\cos(\theta))^2 + (|b|\sin(\theta) + |c|)^2}$, where $|a|$ is the magnitude of vector a, $|b|$ is the magnitude of vector b, $|c|$ is the magnitude of vector c, and $\theta$ is the angle between vectors b and c. Plugging in the values, we get $|R| = \sqrt{(50 + 40\cos(35))^2 + (40\sin(35) + 30)^2}$. Evaluating this expression gives us the total displacement magnitude. Since the displacement is in vector form, we also need to find the direction of the resultant vector. We can use the equation $\phi = \tan^{-1}\left(\frac{|b|\sin(\theta) + |c|}{|a| + |b|\cos(\theta)}\right)$ to find the angle $\phi$. Plugging in the values, we get $\phi = \tan^{-1}\left(\frac{40\sin(35) + 30}{50 + 40\cos(35)}\right)$. Therefore, the total displacement of the slow velocity eraser is the magnitude of the resultant vector R and its direction is given by the angle $\phi$.</p> Signup and view all the answers

<p>To find the magnitude of vector b in the y direction, we can use the equation $b_y = b \sin(\theta)$, where $b$ is the magnitude of vector b and $\theta$ is the angle between vector b and the positive y-axis. Plugging in the values, we get $b_y = 40 \sin(35)$, which evaluates to approximately 22.943 centimeters.</p> Signup and view all the answers

<p>To find the magnitude of vector b in the x direction, we can use the equation $b_x = b \cos(\theta)$, where $b$ is the magnitude of vector b and $\theta$ is the angle between vector b and the positive x-axis. Plugging in the values, we get $b_x = 40 \cos(35)$, which evaluates to approximately 32.766 centimeters.</p> Signup and view all the answers

<p>To find the angle $\phi$ for the resultant vector, we can use the equation $\phi = \tan^{-1}\left(\frac{c + b_y}{a + b_x}\right)$, where $a$ is the magnitude of vector a, $b_x$ is the magnitude of vector b in the x direction, $b_y$ is the magnitude of vector b in the y direction, and $c$ is the magnitude of vector c. Plugging in the values, we get $\phi = \tan^{-1}\left(\frac{30 + 22.943}{50 + 32.766}\right)$. Evaluating this expression gives us the angle $\phi$, which is approximately 32 degrees.</p> Signup and view all the answers