Jeepney Acceleration Problem Velocity And Position Calculation

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This article delves into the fascinating world of physics, specifically focusing on the concepts of acceleration, velocity, and position. We will explore a practical problem involving a jeepney, a common mode of transportation in the Philippines, and its motion under non-constant acceleration. Understanding these principles is crucial for anyone interested in mechanics, engineering, or simply the science behind everyday movements. This comprehensive guide aims to provide a clear and detailed explanation of the problem, its solution, and the underlying physics concepts, ensuring that readers of all backgrounds can grasp the fundamental principles involved.

Problem Statement

We are presented with a scenario where a jeepney accelerates at a rate that changes with time. The acceleration is given by the equation a = (1.5 m/s³) * t, which means the acceleration increases linearly with time. We are given two specific questions to address:

  • A. If the velocity of the jeepney is 4.0 m/s at time t = 1.0 s, what is its velocity at t = 3.0 s?
  • B. If the position of the jeepney at t = 1.0 s is 5.0 m, what is its position at t = 3.0 s?

These questions require us to apply our knowledge of kinematics, the branch of physics that describes the motion of objects without considering the forces that cause the motion. We will need to use integration to determine the velocity and position of the jeepney as functions of time, given its time-dependent acceleration.

A. Determining Velocity at t = 3.0 s

To find the velocity of the jeepney at t = 3.0 s, we need to integrate the acceleration function with respect to time. Recall that acceleration is the rate of change of velocity, so integrating acceleration with respect to time gives us the velocity. The key formula here is:

v(t) = ∫ a(t) dt

Where:

  • v(t) is the velocity as a function of time.
  • a(t) is the acceleration as a function of time.

Step-by-Step Solution

  1. Integrate the acceleration function: Given a(t) = (1.5 m/s³) * t, we integrate with respect to t:

    ∫ (1.5 m/s³) * t dt = (1.5 m/s³) * (t² / 2) + C = 0.75 * t² + C

    Where C is the constant of integration. This constant represents the initial velocity or the velocity at t = 0.

  2. Determine the constant of integration (C): We are given that the velocity at t = 1.0 s is 4.0 m/s. We can use this information to find C:

    1. 0 m/s = 0.75 * (1.0 s)² + C
    2. 0 m/s = 0.75 m/s + C C = 4.0 m/s - 0.75 m/s = 3.25 m/s

    So, the velocity function is:

    v(t) = 0.75 * t² + 3.25

  3. Calculate the velocity at t = 3.0 s: Now we can plug in t = 3.0 s into the velocity function:

    v(3.0 s) = 0.75 * (3.0 s)² + 3.25 m/s v(3.0 s) = 0.75 * 9.0 s² + 3.25 m/s v(3.0 s) = 6.75 m/s + 3.25 m/s v(3.0 s) = 10.0 m/s

Conclusion for Part A

The velocity of the jeepney at t = 3.0 s is 10.0 m/s. This result shows how the velocity increases over time due to the increasing acceleration. The key takeaway here is the application of integration to derive velocity from acceleration and the use of initial conditions to determine the constant of integration.

B. Determining Position at t = 3.0 s

To find the position of the jeepney at t = 3.0 s, we need to integrate the velocity function with respect to time. Recall that velocity is the rate of change of position, so integrating velocity with respect to time gives us the position. The fundamental formula for this part is:

x(t) = ∫ v(t) dt

Where:

  • x(t) is the position as a function of time.
  • v(t) is the velocity as a function of time.

Step-by-Step Solution

  1. Integrate the velocity function: We found that v(t) = 0.75 * t² + 3.25. Now we integrate with respect to t:

    ∫ (0.75 * t² + 3.25) dt = 0.75 * (t³ / 3) + 3.25 * t + D = 0.25 * t³ + 3.25 * t + D

    Where D is the constant of integration. This constant represents the initial position or the position at t = 0.

  2. Determine the constant of integration (D): We are given that the position at t = 1.0 s is 5.0 m. We can use this information to find D:

    1. 0 m = 0.25 * (1.0 s)³ + 3.25 * (1.0 s) + D
    2. 0 m = 0.25 m + 3.25 m + D D = 5.0 m - 0.25 m - 3.25 m = 1.5 m

    So, the position function is:

    x(t) = 0.25 * t³ + 3.25 * t + 1.5

  3. Calculate the position at t = 3.0 s: Now we can plug in t = 3.0 s into the position function:

    x(3.0 s) = 0.25 * (3.0 s)³ + 3.25 * (3.0 s) + 1.5 m x(3.0 s) = 0.25 * 27.0 s³ + 9.75 m + 1.5 m x(3.0 s) = 6.75 m + 9.75 m + 1.5 m x(3.0 s) = 18.0 m

Conclusion for Part B

The position of the jeepney at t = 3.0 s is 18.0 m. This result indicates the total displacement of the jeepney from its initial position, considering its changing velocity and acceleration. The key concept demonstrated here is the integration of the velocity function to obtain the position function and the importance of initial conditions in determining the complete solution.

Synthesis and Key Concepts

This problem highlights several fundamental concepts in physics:

  • Acceleration: The rate of change of velocity with respect to time. In this case, the acceleration is not constant but varies linearly with time.
  • Velocity: The rate of change of position with respect to time. We found the velocity function by integrating the acceleration function.
  • Position: The location of an object in space. We found the position function by integrating the velocity function.
  • Integration: A mathematical tool used to find the area under a curve. In physics, integration is used to find velocity from acceleration and position from velocity.
  • Initial Conditions: The values of velocity and position at a specific time, which are crucial for determining the constants of integration.

The relationship between acceleration, velocity, and position can be summarized as follows:

  • Acceleration is the derivative of velocity with respect to time.
  • Velocity is the derivative of position with respect to time.
  • Velocity is the integral of acceleration with respect to time.
  • Position is the integral of velocity with respect to time.

This problem demonstrates the power of calculus in solving physics problems. By using integration, we were able to find the velocity and position functions, which describe the motion of the jeepney at any time.

Practical Applications and Real-World Significance

The concepts explored in this problem have numerous practical applications in the real world. Understanding acceleration, velocity, and position is crucial in various fields, including:

  • Engineering: Designing vehicles, machines, and structures that move safely and efficiently.
  • Aerospace: Calculating the trajectories of aircraft and spacecraft.
  • Sports: Analyzing the motion of athletes and sports equipment.
  • Robotics: Programming robots to move and interact with their environment.
  • Traffic Management: Optimizing traffic flow and preventing accidents.

The jeepney problem, although simplified, provides a realistic example of how these concepts apply to everyday life. The motion of any vehicle, whether it's a car, a train, or a jeepney, can be described using the principles of kinematics. By understanding these principles, we can design safer and more efficient transportation systems.

Conclusion

In conclusion, we have successfully determined the velocity and position of a jeepney accelerating at a time-dependent rate. By applying the principles of kinematics and using integration, we found that the velocity of the jeepney at t = 3.0 s is 10.0 m/s, and its position at t = 3.0 s is 18.0 m. This problem illustrates the fundamental relationship between acceleration, velocity, and position and the importance of initial conditions in solving motion problems.

The key takeaway from this article is the application of calculus, specifically integration, in solving physics problems related to motion. Understanding these concepts is essential for anyone pursuing a career in science, engineering, or any field that involves the analysis of motion. Moreover, the jeepney example provides a tangible connection between theoretical physics and real-world applications, demonstrating the relevance of these concepts in our daily lives. We hope this comprehensive guide has provided you with a clear and insightful understanding of the problem and the underlying physics principles.