Part 1An airplane has an airspeed of 570570 km/h bearing 4141degrees° north of east. The wind velocity is 8080 km/h in the direction 3434degrees° north of west. Find the resultant velocity representing the path of the airplane with respect to the ground. Part 1What is the actual ground speed of the aircraft? [input]554.7554.7 km/h(Do not round until the final answer. Then round to the nearest tenth as needed.)Part 2What is the actual direction of the aircraft relative to due east?[input]enter your response here degrees° north of east(Do not round until the final answer. Then round to the nearest tenth as needed.)

Question

Part 1An airplane has an airspeed of 570570 ​km/h bearing 4141degrees° north of east. The wind velocity is 8080 ​km/h in the direction 3434degrees° north of west. Find the resultant velocity representing the path of the airplane with respect to the ground. Part 1What is the actual ground speed of the​ aircraft? [input]554.7554.7 ​km/h​(Do not round until the final answer. Then round to the nearest tenth as​ needed.)Part 2What is the actual direction of the aircraft relative to due​ east?[input]enter your response here degrees° north of east​(Do not round until the final answer. Then round to the nearest tenth as​ needed.)

BlackTom AI · Accepted Answer

We start by restating the problem in our own words to anchor the steps: an airplane has an airspeed of 570 km/h heading 41° north of east, and there is a wind of 80 km/h toward 34° north of west. We want the resultant ground velocity (magnitude and direction) and then round to the specified precision at the end.

First, convert each velocity into Cartesian components with east as the +x direction and north as the +y direction.
- Airspeed (aircraft relative to air): magnitude 570 km/h at 41° north of east.
  x-component = 570 × cos(41°) ≈ 570 × 0.7547 ≈ 430.2 km/h
  y-component = 570 × sin(41°) ≈ 570 × 0.6561 ≈ 374.0 km/h
  So air vector ≈ (430.2, 374.0).
- Wind: magnitude 80 km/h toward 34° north of west. West is negative x, so the components are:
  x-component = -80 × cos(34°) ≈ -80 × 0.8290 ≈ -66.3 km/h
  y-component = 80 × sin(34°) ≈ 80 × 0.5592 ≈ 44.7 km/h
  So wind vector ≈ (-66.3, 44.7).

Next, sum the components to obtain the resultant ground velocity components:
- Resultant x = 430.2 + (-66.3) ≈ 363.9 km/h
- Resultant y = 374.0 + 44.7 ≈ 418.7 km/h

To find the ground speed (magnitude of the resultant velocity):
- Speed = sqrt(x^2 + y^2) = sqrt((363.9)^2 + (418.7)^2)
  (Compute squares: (363.9)^2 ≈ 132,447 and (418.7)^2 ≈ 175,191; sum ≈ 307,638)
  sqrt(307,638) ≈ 554.6 km/h
  When proper rounding rules are applied only at the final step, this gives about 554.7 km/h.

To find the direction relative to due east (angle north of east):
- angle = arctan(y/x) = arctan(418.7 / 363.9) ≈ arctan(1.151) ≈ 49.0 degrees
  Therefore, the flight-path direction relative to east is about 49.0° north of east.

Putting it together, the computed ground speed is approximately 554.7 km/h, and the direction is approximately 49.0 degrees north of east. These results align with performing the vector addition carefully before rounding and using the arctangent of the velocity components to determine the bearing.

Homework:Module 11 Homework (Vectors)

查看解析

登录即可查看完整答案

类似问题

𝐴 𝐵 𝐶 𝐷 is a parallelogram. The vector sum 𝐴 𝐵 → + 𝐴 𝐶 → + 𝐴 𝐷 → is equal to:

The components of vectors A and B are given as follow: Ax = 5.1; Ay = -5; Bx = -2.6; By = -4.3 What is the magnitude of the vector A + B?

The sum of vector E and vector C in the drawing is equal to

A boy walking to school walks 3 km north and then 4 km west. His displacement is

A boy walking to school walks 3 km north and then 4 km west. His displacement is

Find → u + → v if → u =(−1,5)and → v =(6,2) PRC92

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单