Jet Aircraft With A Mass Of 4325 Kg And Engine Exerting

A Jet Aircraft With A Mass Of4325kg Has An Engine That Exerts A Force

A jet aircraft with a mass of 4,325 kg has an engine that exerts a force (thrust) equal to 60,800 N. (a) What is the jet's acceleration? (Give the magnitude.) (b) What is the jet's speed after it accelerates for 7 seconds? (Assume it starts from rest.) (c) How far does the jet travel during these 7 seconds?

Paper For Above instruction

The dynamics of motion concerning a jet aircraft provided with a known mass and force allows us to explore fundamental principles of classical mechanics, particularly Newton's second law of motion. By analyzing the given parameters, we can determine the acceleration, subsequent velocity, and distance traveled, thereby illustrating core concepts of kinematics and dynamics in a practical context.

Introduction

Newton's second law states that the net force acting upon an object is equal to the mass of the object multiplied by its acceleration (F = ma). This principle is central to understanding the behavior of the jet aircraft when subjected to a constant thrust. The specific values provided—mass of 4,325 kg and thrust of 60,800 N—serve as the basis for subsequent calculations of acceleration, velocity after a given time, and displacement during that period.

Calculating the Jet's Acceleration

Given the force (F) and the mass (m), the acceleration (a) can be directly calculated using Newton's second law rearranged as a = F/m.

Plugging in the exact values:

a = 60,800 N / 4,325 kg ≈ 14.055 m/sec²

Thus, the magnitude of the jet's acceleration is approximately 14.055 meters per second squared.

Determining the Jet's Velocity After 7 Seconds

The initial velocity (u) is zero, as the jet starts from rest. Using the kinematic equation for velocity under constant acceleration:

v = u + at

v = 0 + (14.055 m/sec²)(7 sec) ≈ 98.385 m/sec

Therefore, after 7 seconds of acceleration, the jet's speed reaches approximately 98.385 meters per second.

Calculating the Distance Traveled During 7 Seconds

The displacement (s) during this period can be found using the equation:

s = ut + (1/2)at²

Since the initial velocity u is zero:

s = (1/2)(14.055 m/sec²)(7 sec)² = (0.5)(14.055)(49) ≈ 344.65 meters

Thus, the jet travels approximately 344.65 meters during the 7 seconds of acceleration.

Conclusion

Utilizing fundamental principles of physics, we conclude that the jet's acceleration is approximately 14.055 m/sec². After 7 seconds, its speed reaches about 98.385 m/sec, and during this interval, it covers a distance of roughly 344.65 meters. These calculations illustrate the practical application of Newtonian mechanics to real-world scenarios involving aircraft motion, emphasizing the importance of understanding force, mass, and resulting kinematic behavior in aeronautical engineering.

References

  • Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers with Modern Physics (10th ed.). Cengage Learning.
  • Halliday, D., Resnick, R., & Walker, J. (2014). Fundamentals of Physics (10th ed.). Wiley.
  • Tipler, P., & Mosca, G. (2008). Physics for Scientists and Engineers. W. H. Freeman and Company.
  • Young, H. D., & Freedman, R. A. (2019). University Physics with Modern Physics (15th ed.). Pearson.
  • Giancoli, D. C. (2013). Physics: Principles with Applications (7th ed.). Pearson.
  • NASA. (2020). Principles of Flight. NASA Glenn Research Center. https://www.grc.nasa.gov/
  • Grolier Online. (2020). Newton's Laws of Motion. Grolier Multimedia Encyclopedia. https://edu.grolier.com
  • NASA. (2019). How Do Jet Engines Work? NASA Technology. https://www.nasa.gov/
  • Department of Aeronautics and Astronautics, MIT. (2021). Aerodynamics and Propulsion. MIT OpenCourseWare. https://ocw.mit.edu/
  • Scholarly article: Anderson, J. D. (2016). Fundamentals of Aerodynamics. McGraw-Hill Education.

Related Assignments