A position vector has components \(x=34.6 \mathrm{~m}\) and \(y=-53.5 \mathrm{~m} .\) Find the vector's length and angle with the \(x\) -axis.

To find the length (magnitude) of the position vector, we can use the Pythagorean theorem, which relates the magnitudes of the components \((x, y)\) to the magnitude of the vector \(R\) by the formula: \(R^2 = x^2 + y^2\). In our case, \(x = 34.6\) m, and \(y = -53.5\) m, so we need to plug these values into the equation and find the square root of the result to get the magnitude of the vector. \(R = \sqrt{x^2 + y^2} = \sqrt{34.6^2 + (-53.5)^2}\) \(R \approx \sqrt{(34.6)^2 + (-53.5)^2}\) \(R \approx \sqrt{1197.96 + 2862.25}\) \(R \approx \sqrt{4060.21}\) \(R \approx 63.7\) m So, the magnitude of the position vector is approximately \(63.7\) m.

Now, we need to find the angle \(\theta\) between the position vector and the x-axis. To find this angle, we can use the arctangent function: \(\theta = \arctan(\frac{y}{x})\). It is important to note that we need to consider in which quadrant the vector is located. In our case, \(x = 34.6\) and \(y = -53.5\) (which is in the 4th quadrant). So, we should add \(360^\circ\) or \(2\pi\) radians (depending on units) to the arctangent to get the correct angle. \(\theta = \arctan(\frac{-53.5}{34.6})\) \(\theta \approx -56.96^\circ\) Since the calculated angle is negative, we can add \(360^\circ\) to get the angle in the range of \(0^\circ \leq \theta < 360^\circ\): \(\theta \approx 360^\circ - 56.96^\circ\) \(\theta \approx 303.04^\circ\) So, the angle with the x-axis is approximately \(303.04^\circ\).

The length of the position vector is approximately \(63.7\) m, and its angle with the x-axis is approximately \(303.04^\circ\).