Question

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

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

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.

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.

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.