Problem 60 - Algebra and Trigonometry

Question

True Velocity of a Jet A jet is flying through a wind that is blowing with a speed of 55 \(\mathrm{mi} / \mathrm{h}\) in the direction \(\mathrm{N} 30^{\circ} \mathrm{E}\) (see the figure). The jet has a speed of 765 \(\mathrm{mi} / \mathrm{h}\) relative to the air, and the pilot heads the jet in the direction \(\mathrm{N} 45^{\circ} \mathrm{E}\) (a) Express the velocity of the wind as a vector in component form. (b) Express the velocity of the jet relative to the air as a vector in component form. (c) Find the true velocity of the jet as a vector. (d) Find the true speed and direction of the jet.

Step-by-Step Solution

Answer

The true velocity of the jet is \(\langle 588.57, 568.44 \rangle\) and its true speed is approximately 816.27 mi/h in the direction N 44.08° E.

Step 1: Calculate Wind Velocity Vector

The wind is blowing at 55 mi/h in the direction N 30° E. We need to express the wind's velocity as a vector. The components are calculated as follows:\[v_w = \langle 55 \cos(30^{\circ}), 55 \sin(30^{\circ}) \rangle\]\[v_w = \langle 55 \times \frac{\sqrt{3}}{2}, 55 \times \frac{1}{2} \rangle\]\[v_w \approx \langle 47.63, 27.5 \rangle\]

Step 2: Calculate Jet's Relative Velocity Vector

The jet is moving at 765 mi/h in the direction N 45° E. We express the jet's velocity in vector form:\[v_j = \langle 765 \cos(45^{\circ}), 765 \sin(45^{\circ}) \rangle\]\[v_j = \langle 765 \times \frac{\sqrt{2}}{2}, 765 \times \frac{\sqrt{2}}{2} \rangle\]\[v_j \approx \langle 540.94, 540.94 \rangle\]

Step 3: Determine True Velocity of Jet

The true velocity of the jet is the vector sum of the jet's velocity relative to the air and the wind's velocity:\[v_{\text{true}} = v_j + v_w = \langle 540.94 + 47.63, 540.94 + 27.5 \rangle\]\[v_{\text{true}} = \langle 588.57, 568.44 \rangle\]

Step 4: Compute True Speed and Direction

The true speed is the magnitude of the true velocity vector:\[||v_{\text{true}}|| = \sqrt{(588.57)^2 + (568.44)^2}\]\[||v_{\text{true}}|| \approx \sqrt{346455.77 + 322138.49} \approx 816.27 \text{ mi/h}\]To find the direction, calculate the angle East of North using the tangent function:\[\theta = \tan^{-1}\left(\frac{568.44}{588.57}\right)\]\[\theta \approx \tan^{-1}(0.9658) \approx 44.08^{\circ}\]Thus, the direction is approximately N 44.08° E.

Key Concepts

Coordinate Geometry

Coordinate geometry is about identifying the location of points using coordinate systems. In vector calculations, we often utilize a two-dimensional plane consisting of x (horizontal) and y (vertical) axes. This helps express vectors, which have both magnitude and direction, in a more comprehensible form.

• Vectors: A vector is represented as an arrow pointing from one point to another, capturing both direction and magnitude.
• Component form: Vectors can be broken down into their respective x and y components, making them easier to analyze mathematically.

For example, to represent the velocity of the wind in the original exercise, we identify its direction and use trigonometry to express it in component form. By converting vector directions like N 30° E into angles, we can effectively plot and solve problems in coordinate geometry.

Vector Addition

Vector addition is the process of combining two or more vectors to find a resultant vector. This is crucial in determining true velocities, as it incorporates both magnitudes and directions from different sources, like a jet and the wind in our problem.

• Geometric Method: Graphically add vectors by placing them tip-to-tail to form a resultant vector.
• Analytic Method: Use component forms, adding corresponding x and y components to get the final vector's components.

Through vector addition, we found the true velocity of the jet. We combined the velocity components of the jet and the wind, resulting in the final vector that describes the jet's actual path.

Magnitude of a Vector

The magnitude of a vector is its length, representing the size of the quantity. To calculate magnitude, use the Pythagorean theorem in coordinate geometry, where the vector's components form a right triangle.

• Formula: Given a vector \(v = \langle x, y \rangle\), the magnitude \(||v||\) is calculated as \(\sqrt{x^2 + y^2}\).

In our jet problem, the magnitude reflects the jet's true speed over the ground. By applying the formula to the components of the true velocity vector \(v_{true} = \langle 588.57, 568.44 \rangle\), we determined that the speed is approximately 816.27 mi/h.

Direction Cosine

Direction cosine is about expressing the vector's direction using cosine angles from the coordinate axes. These cosines aid in understanding how a vector spreads across the x and y axes.

• Angle calculation: Use trigonometric functions like tangent to find angles between a vector and one of the coordinate axes.
• Application: Helpful for determining the actual direction of velocity in navigation problems.

In the case of the jet, we used the tangent function to calculate the resulting angle for the true direction of the jet's path. This angle, N 44.08° E, helps the pilot understand exactly which path the jet is taking relative to the Earth's surface.