/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 The velocity vector of a sprinti... [FREE SOLUTION] | 91Ó°ÊÓ

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The velocity vector of a sprinting cheetah has \(x\) - and \(y\) -components \(v_{x}=+16.4 \mathrm{m} / \mathrm{s}\) and $v_{y}=-26.3 \mathrm{m} / \mathrm{s}$ (a) What is the magnitude of the velocity vector? (b) What angle does the velocity vector make with the \(+x\) - and \(-y\) -axes?

Short Answer

Expert verified
Based on the information provided: (a) The magnitude of the velocity vector is approximately 30.99 m/s. (b) The angle the velocity vector makes with the +x-axis is approximately -58.12 degrees, and with the -y-axis is approximately 58.12 degrees.

Step by step solution

01

Find the magnitude of the velocity vector

The magnitude of the velocity vector can be found by applying the Pythagorean theorem to its components, which are \(v_x = +16.4 \text{m/s}\) and \(v_y = -26.3 \text{m/s}\). The magnitude is represented by \(v\), so we have \(v = \sqrt{v_x^2 + v_y^2}\). Calculate the value by substituting the given values. \(v = \sqrt{(16.4)^2 + (-26.3)^2}\) \(v = \sqrt{268.96 + 691.69}\) \(v = \sqrt{960.65}\) \(v \approx 30.99 \text{ m/s}\) Hence, the magnitude of the velocity vector is approximately 30.99 m/s.
02

Find the angle with the +x-axis

Now, we will find the angle between the velocity vector and the +x-axis, denoted as \(\theta\). To do this, we will use the inverse tangent function (arctan) and the components of the velocity vector: \(\theta = \arctan\left(\frac{v_y}{v_x}\right)\), Plugging in the values of \(v_x\) and \(v_y\), we have: \(\theta = \arctan\left(\frac{-26.3}{16.4}\right)\), Now, use a calculator to find the arctan value. \(\theta \approx -58.12^{\circ}\) Since the angle is negative, the angle between the +x-axis and the velocity vector is in the clockwise direction.
03

Find the angle with the -y-axis

To find the angle the velocity vector makes with the -y-axis, we first determine in which quadrant the velocity vector lies. Given that \(v_x = +16.4\) and \(v_y = -26.3\), the vector is in the fourth quadrant. Since the reference angle is 58.12 degrees, the angle between the -y-axis and the velocity vector will be: \(\angle \text{with -y-axis} = 360^{\circ} - \theta\) \(\angle \text{with -y-axis} = 360^{\circ} - (-58.12^{\circ})\) \(\angle \text{with -y-axis} = 418.12^{\circ}\) For convention, we can subtract 360 to represent the angle value within \(0^{\circ}\) to \(360^{\circ}\): \(\angle \text{with -y-axis} = 418.12^{\circ} - 360^{\circ}\) \(\angle \text{with -y-axis} = 58.12^{\circ}\) The angle the velocity vector makes with the -y-axis is approximately 58.12 degrees. In conclusion, (a) The magnitude of the velocity vector is approximately 30.99 m/s. (b) The angle the velocity vector makes with the +x-axis is approximately -58.12 degrees, and with the -y-axis is approximately 58.12 degrees.

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