91Ó°ÊÓ

Understanding vector addition is crucial in determining the resultant velocity of the aircraft relative to the ground. When we talk about vectors in physics, we are referring to quantities that have both a magnitude and a direction. In this problem, the aircraft's velocity and the wind's velocity are represented as vectors.

• The aircraft’s vector, traveling east, is given a positive value on the horizontal (x-axis), represented by the unit vector \( \hat{i} \).
• The wind’s vector, blowing south, is represented on the vertical (y-axis) with the unit vector \( \hat{j} \), typically negative due to its direction opposite to the north.

To find the speed of the aircraft relative to the ground, we perform vector addition, combining the vectors of the aircraft and the wind. This is done by adding their corresponding components: the eastward component from the aircraft and the southward component from the wind. The result is a new vector that reflects how fast and in what direction the aircraft is moving across the ground. This process uses the simple vector addition formula:\[\mathbf{V_r} = \mathbf{V_a} + \mathbf{V_w} = 70.0 \, \hat{i} + 20.0 \, \hat{j}\].

The Pythagorean theorem is key in calculating the magnitude of the resultant vector obtained from vector addition. This theorem applies to right-angled triangles, positing that the square of the hypotenuse is equal to the sum of the squares of the other two sides. In our context, we apply this to calculate the resultant speed of the aircraft relative to the ground.

When the vectors \( \mathbf{V_a} \) (70.0 km/h east) and \( \mathbf{V_w} \) (20.0 km/h south) are added, they form a right triangle.

• The eastward velocity acts as one side.
• The southward velocity forms the other side.
• The resultant vector \( \mathbf{V_r} \) represents the hypotenuse.

Thus, to find the speed of the aircraft relative to the ground, we compute the magnitude of the resultant by employing the Pythagorean theorem formula:\[\|\mathbf{V_r}\| = \sqrt{(70.0 \, \text{km/h})^2 + (20.0 \, \text{km/h})^2}\]Upon solving the above, \[\|\mathbf{V_r}\| = \sqrt{4900 + 400} = \sqrt{5300} \approx 72.8 \, \text{km/h}\] This shows us how much and how swiftly the aircraft travels relative to the ground.

Navigating an aircraft involves understanding various factors affecting its course and speed, including wind velocity. Pilots must account for the wind when planning a flight path to ensure the aircraft reaches its intended destination efficiently.

In this problem, the pilot must consider the southward wind pushing against the aircraft. The goal is to calculate the correct heading and airspeed so that the actual path taken by the aircraft over the ground matches the planned route.

• We blend navigational experience with physics and mathematics.
• Accurate navigation ensures minimal fuel consumption and on-time arrivals.
• Pilots use instruments and calculations, like vector addition, to reach this goal.

By understanding how to manage these vectors and wind conditions, pilots can effectively control the direction and speed of the aircraft for safe and precise arrival at the destination.