Constant Force Motion Distance Traveled In Time Intervals
Understanding the Motion of a Particle Under Constant Force
When we delve into the fascinating world of physics, the motion of objects under the influence of forces is a fundamental concept. Imagine a particle, initially at rest, subjected to a constant force. This scenario, seemingly simple, unveils a wealth of information about the relationship between force, motion, and time. To truly grasp the dynamics at play, let's embark on a comprehensive exploration, focusing on the specific scenario where a particle moves under a constant force for 20 seconds. Our investigation will center around calculating the distances covered by the particle during two distinct time intervals: the first 10 seconds (denoted as s₁) and the subsequent 10 seconds (denoted as s₂). By meticulously analyzing these distances, we aim to establish a definitive relationship between s₁ and s₂, thereby providing valuable insights into the particle's motion.
At the heart of this analysis lies the understanding of constant force. A constant force, as the name suggests, implies that the force acting on the particle remains unchanged in both magnitude and direction throughout its motion. This constancy simplifies our calculations significantly, allowing us to apply the well-established principles of kinematics, which provide a framework for describing motion without delving into the causes of that motion. In this context, kinematics provides us with the essential equations that link displacement, velocity, acceleration, and time.
One of the most crucial equations in our arsenal is the equation of motion that relates the distance traveled to the initial velocity, time, and acceleration. Given that the particle starts from rest, its initial velocity is zero, which further simplifies our calculations. The distance traveled, therefore, becomes directly proportional to the square of the time elapsed. This proportionality underscores a critical aspect of constant force motion: the distance covered increases quadratically with time. This means that the particle covers progressively larger distances in equal time intervals as its velocity increases due to the constant acceleration imparted by the force. With this foundational understanding, we can now delve into the specifics of our problem, calculating s₁ and s₂ and establishing their relationship.
To further enrich our understanding, let's explore the concept of acceleration in the context of a constant force. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In our case, since the force is constant and the mass of the particle remains unchanged, the acceleration is also constant. This constant acceleration is the key to understanding why the distance traveled increases quadratically with time. As the particle accelerates uniformly, its velocity increases linearly with time, and consequently, the distance it covers in each subsequent time interval becomes progressively larger.
To determine the relationship between the distances s₁ and s₂, we need to employ the equations of motion derived from the principles of kinematics. As established earlier, the particle starts from rest and experiences a constant force, leading to constant acceleration. Let's denote this constant acceleration as 'a'. Using the equation of motion that relates distance, initial velocity, time, and acceleration, we can calculate the distance covered in the first 10 seconds (s₁) and the subsequent 10 seconds (s₂).
For the first 10 seconds, the initial velocity (u) is 0, the time (t) is 10 seconds, and the acceleration is 'a'. Applying the equation s = ut + (1/2)at², we get:
s₁ = (0)(10) + (1/2)a(10)² = 50a
Thus, the distance s₁ covered in the first 10 seconds is 50a, where 'a' represents the constant acceleration experienced by the particle. This equation highlights the direct proportionality between the distance covered and the acceleration. A higher acceleration will result in a larger distance covered in the same time interval. Now, let's shift our focus to the subsequent 10 seconds and calculate the distance s₂.
For the next 10 seconds, we need to consider the particle's motion from the 10-second mark to the 20-second mark. To calculate s₂, we can consider the total distance covered in 20 seconds and subtract the distance covered in the first 10 seconds (s₁). Alternatively, we can analyze the motion specifically during this 10-second interval, keeping in mind that the particle now has an initial velocity at the 10-second mark. To use this approach, we must first determine the velocity of the particle at the end of the first 10 seconds.
The velocity (v) at the end of the first 10 seconds can be calculated using the equation v = u + at, where u is the initial velocity (0), a is the acceleration, and t is the time (10 seconds). Therefore,
v = 0 + a(10) = 10a
This velocity, 10a, becomes the initial velocity for the subsequent 10-second interval. Now, we can apply the same equation of motion, s = ut + (1/2)at², to calculate s₂. In this case, u = 10a, t = 10 seconds, and the acceleration remains 'a'. Plugging these values into the equation, we get:
s₂ = (10a)(10) + (1/2)a(10)² = 100a + 50a = 150a
Therefore, the distance s₂ covered in the next 10 seconds is 150a. This value is significantly larger than s₁, which underscores the effect of constant acceleration: the particle covers a greater distance in the same time interval as its velocity increases. Now that we have calculated both s₁ and s₂, we can proceed to establish the relationship between them.
Having meticulously calculated the distances s₁ and s₂ covered by the particle in the first 10 seconds and the subsequent 10 seconds, respectively, we are now poised to establish the crucial relationship between these two distances. This relationship will provide valuable insights into the nature of motion under constant force and how distance traveled varies with time under constant acceleration. As derived in the previous section, we have:
s₁ = 50a s₂ = 150a
Where 'a' represents the constant acceleration experienced by the particle. To uncover the relationship between s₁ and s₂, we can simply divide s₂ by s₁:
s₂ / s₁ = (150a) / (50a) = 3
This result reveals a clear and concise relationship: s₂ = 3s₁. This equation is the key to understanding the dynamics of motion under constant force. It signifies that the distance covered in the subsequent 10 seconds (s₂) is precisely three times the distance covered in the first 10 seconds (s₁). This observation is a direct consequence of the constant acceleration experienced by the particle. As the particle accelerates uniformly, its velocity increases linearly with time. Consequently, the distance it covers in each successive time interval increases at an accelerating rate, resulting in the s₂ being three times s₁.
This relationship, s₂ = 3s₁, is not merely a mathematical curiosity; it embodies a fundamental principle of physics. It underscores the non-linear nature of distance traveled under constant acceleration. Unlike motion with constant velocity, where the distance covered is directly proportional to time, motion with constant acceleration exhibits a quadratic relationship between distance and time. This means that the distance increases much faster as time progresses.
To further solidify our understanding, let's consider a numerical example. Suppose the constant acceleration 'a' is 2 m/s². Using our equations, we can calculate s₁ and s₂:
s₁ = 50a = 50(2) = 100 meters s₂ = 150a = 150(2) = 300 meters
As expected, s₂ (300 meters) is indeed three times s₁ (100 meters), confirming our established relationship. This example vividly illustrates how the particle covers significantly more distance in the second 10-second interval due to its increasing velocity under constant acceleration. This relationship has far-reaching implications in various fields of physics and engineering. It is crucial in analyzing the motion of projectiles, the acceleration of vehicles, and many other real-world scenarios involving constant forces.
In conclusion, our exploration of a particle's motion under the influence of a constant force has unveiled a fundamental relationship between the distances covered in consecutive time intervals. By meticulously applying the principles of kinematics and the equations of motion, we have demonstrated that the distance covered in the subsequent 10 seconds (s₂) is precisely three times the distance covered in the first 10 seconds (s₁), expressed as s₂ = 3s₁. This relationship stems directly from the constant acceleration experienced by the particle due to the constant force acting upon it.
This 3:1 ratio is not merely a numerical coincidence; it encapsulates the essence of motion under constant acceleration. It highlights the quadratic relationship between distance and time, where the distance increases at an accelerating rate as the particle's velocity increases linearly with time. This understanding is crucial in comprehending various physical phenomena, from the trajectory of projectiles to the acceleration of vehicles.
The analysis we have undertaken serves as a cornerstone for more advanced concepts in physics and engineering. It provides a solid foundation for understanding non-uniform motion, where acceleration is not constant, and for analyzing more complex scenarios involving multiple forces and constraints. The principles and techniques employed in this exploration are widely applicable in various fields, making it an essential topic for students and professionals alike.
Furthermore, the investigation of this seemingly simple scenario underscores the power of mathematical modeling in physics. By translating a physical problem into mathematical equations, we can gain profound insights and make accurate predictions about the behavior of the system. The equation s₂ = 3s₁ is a testament to this power, providing a concise and elegant description of a complex physical phenomenon.
In essence, the study of a particle under constant force is a gateway to understanding the intricate world of motion and forces. It lays the groundwork for more advanced topics and provides a framework for analyzing a wide range of real-world scenarios. The relationship s₂ = 3s₁ serves as a constant reminder of the elegance and power of physics in explaining the world around us. The problem that we explored is a classic example in physics that illustrates the application of constant acceleration equations. The question is "A particle under the action of a constant force moves from rest up to 20 seconds. If the distance covered in the first 10 seconds is s₁ and that covered in the next 10 seconds is s₂, then:(A) s₁ = s₂(B) s₂ = 3s₁(C) s₂ = 2s₁(D) s₂ = 4s₁".
The correct answer is (B) s₂ = 3s₁. This problem requires a solid understanding of kinematics and the ability to apply the equations of motion appropriately. By understanding the concepts and applying them diligently, we can successfully unravel the mysteries of motion under constant force.
