Question

Vector \(\overrightarrow{\mathbf{a}}\) has components $a_{x}=-3.0 \mathrm{m} / \mathrm{s}^{2}\( and \)a_{y}=\( \)+4.0 \mathrm{m} / \mathrm{s}^{2} .$ (a) What is the magnitude of \(\overrightarrow{\mathbf{a}} ?\) (b) What is the direction of \(\overrightarrow{\mathbf{a}}\) ? Give an angle with respect to one of the coordinate axes.

Short answer

Answer: The magnitude of $\overrightarrow{\mathbf{a}}$ is $5.0\,\mathrm{m}/\mathrm{s}^2$, and the direction of $\overrightarrow{\mathbf{a}}$ is $126.87^{\circ}$ counterclockwise from the positive x-axis.

Step-by-step solution

Find the magnitude of the vector

To find the magnitude of the vector \(\overrightarrow{\mathbf{a}}\), we use the Pythagorean theorem for 2D vectors: \(|\overrightarrow{\mathbf{a}}| = \sqrt{a_{x}^2 + a_{y}^2}\). We have \(a_x = -3.0\,\mathrm{m}/\mathrm{s}^2\) and \(a_y = 4.0\,\mathrm{m}/\mathrm{s}^2\), so we can plug those values in to calculate the magnitude: \(|\overrightarrow{\mathbf{a}}| = \sqrt{(-3.0\,\mathrm{m}/\mathrm{s}^2)^2 + (4.0\,\mathrm{m}/\mathrm{s}^2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0\,\mathrm{m}/\mathrm{s}^2\). The magnitude of the vector is \(5.0\,\mathrm{m}/\mathrm{s}^2\).

Find the angle of the vector with respect to a coordinate axis

To find the angle of the vector with respect to either the x-axis or the y-axis, we can use the inverse tangent function. In this case, we will use the tangent of the angle between the vector and the x-axis, defined as \(\tan{\theta} = \frac{a_y}{a_x}\). We can then find the angle using the inverse tangent function, \(\theta = \arctan{\frac{a_y}{a_x}}\). With the given values for \(a_x\) and \(a_y\), we can calculate the angle: \(\theta = \arctan{\frac{4.0\,\mathrm{m}/\mathrm{s}^2}{-3.0\,\mathrm{m}/\mathrm{s}^2}} \approx -53.13^{\circ}\). The negative angle means that the vector is located in the second quadrant. To express this angle with respect to the positive x-axis, we can add \(180^{\circ}\): \(180^{\circ} - 53.13^{\circ} = 126.87^{\circ}\). The direction of the vector \(\overrightarrow{\mathbf{a}}\) is \(126.87^{\circ}\) counterclockwise from the positive x-axis. So, the magnitude of \(\overrightarrow{\mathbf{a}}\) is \(5.0\,\mathrm{m}/\mathrm{s}^2\), and the direction of \(\overrightarrow{\mathbf{a}}\) is \(126.87^{\circ}\) counterclockwise from the positive x-axis.