Calculating Velocity Of Motion S=3+9t-1.5t² At T=3s A Physics Guide

Introduction

Understanding motion is a fundamental concept in physics, and one of the key aspects of describing motion is velocity. Velocity tells us how fast an object is moving and in what direction. In this article, we will delve into the process of calculating the velocity of an object given its displacement function. Specifically, we will focus on the equation s = 3 + 9t - 1.5t², which describes the displacement (s) of an object as a function of time (t). Our goal is to determine the velocity of the object at a specific time, t = 3 seconds. This exploration will involve understanding the relationship between displacement, velocity, and time, and applying calculus concepts to derive the velocity function from the displacement function. We'll break down the equation, explain the steps involved in differentiation, and then apply the result to find the velocity at the given time. This process is not just a mathematical exercise; it has real-world applications in understanding the motion of vehicles, projectiles, and other objects. By the end of this article, you will have a clear understanding of how to calculate velocity from a displacement function and appreciate the practical significance of this concept.

Understanding Displacement and Velocity

Before diving into the calculations, it's crucial to have a solid understanding of the concepts of displacement and velocity. Displacement, in physics, refers to the change in position of an object. It is a vector quantity, meaning it has both magnitude (how far the object has moved) and direction. In our equation, s = 3 + 9t - 1.5t², 's' represents the displacement of the object from a reference point at a given time 't'. The units of displacement are typically meters (m), and time is measured in seconds (s). This equation tells us how the object's position changes over time.

Velocity, on the other hand, is the rate of change of displacement with respect to time. It also is a vector quantity, and its magnitude represents the speed of the object, while its direction indicates the direction of motion. Velocity is typically measured in meters per second (m/s). In simpler terms, velocity tells us how quickly an object is changing its position. The key relationship to remember is that velocity is the derivative of displacement with respect to time. This means that if we have a function that describes the displacement of an object as a function of time, we can find the velocity function by differentiating the displacement function. This is a fundamental concept in calculus and is widely used in physics to analyze motion.

In the context of our problem, the equation s = 3 + 9t - 1.5t² provides the displacement of the object at any given time 't'. To find the velocity at t = 3s, we need to determine the velocity function, which is the derivative of this displacement function. This will give us an equation that tells us the velocity of the object at any time 't', and we can then plug in t = 3s to find the specific velocity at that moment. Understanding this relationship between displacement and velocity is essential for solving problems involving motion and is a cornerstone of classical mechanics.

Differentiating the Displacement Function

To calculate the velocity, we need to find the derivative of the displacement function, s = 3 + 9t - 1.5t², with respect to time (t). This process involves applying the rules of differentiation from calculus. Differentiation is a mathematical operation that finds the rate at which a function is changing. In this context, it will give us the rate at which the displacement is changing, which is the velocity.

The displacement function, s = 3 + 9t - 1.5t², is a polynomial function, and we can differentiate it term by term using the power rule of differentiation. The power rule states that if we have a term of the form at^n, where 'a' is a constant and 'n' is an exponent, then the derivative of this term with respect to 't' is n*at^(n-1). Let's apply this rule to each term in our displacement function:

1. The first term is 3, which is a constant. The derivative of a constant is always zero. So, d(3)/dt = 0.
2. The second term is 9t, which can be written as 9t^1. Applying the power rule, the derivative is 1 * 9t^(1-1) = 9t^0 = 9. So, d(9t)/dt = 9.
3. The third term is -1.5t², where the exponent is 2. Applying the power rule, the derivative is 2 * (-1.5)t^(2-1) = -3t. So, d(-1.5t²)/dt = -3t.

Now, we combine the derivatives of each term to find the derivative of the entire displacement function:

v = ds/dt = d(3)/dt + d(9t)/dt + d(-1.5t²)/dt = 0 + 9 - 3t

Therefore, the velocity function, v(t), is given by:

v(t) = 9 - 3t

This equation tells us the velocity of the object at any time 't'. The units of velocity are meters per second (m/s), as we discussed earlier. Now that we have the velocity function, we can proceed to calculate the velocity at the specific time of t = 3s.

Calculating Velocity at t = 3s

Now that we have derived the velocity function, v(t) = 9 - 3t, we can easily calculate the velocity of the object at any given time. Our specific goal is to find the velocity at t = 3 seconds. This involves a simple substitution of t = 3 into the velocity function.

Substituting t = 3 into the velocity function, we get:

v(3) = 9 - 3(3)

Now, we perform the arithmetic:

v(3) = 9 - 9

v(3) = 0

Therefore, the velocity of the object at t = 3 seconds is 0 m/s. This result has a significant physical interpretation. It means that at the instant t = 3 seconds, the object is momentarily at rest. The object is neither moving in the positive direction nor the negative direction; its instantaneous velocity is zero. This could be a point where the object changes direction, like the peak of its trajectory if it were thrown upwards, or a brief pause in its motion before continuing in the same direction.

The fact that the velocity is zero at t = 3s doesn't necessarily mean the object is not accelerating. Acceleration is the rate of change of velocity, and even if the velocity is momentarily zero, the object could still be accelerating due to an external force. To determine the acceleration, we would need to differentiate the velocity function with respect to time. However, for this specific problem, we have successfully calculated the velocity at the given time using the velocity function we derived earlier. This demonstrates the power of using calculus to analyze motion and extract valuable information about the behavior of moving objects.

Interpretation of the Result

The result we obtained, a velocity of 0 m/s at t = 3 seconds, has a significant physical interpretation within the context of the object's motion. To fully understand what this means, let's revisit the original displacement function: s = 3 + 9t - 1.5t². This equation describes a parabolic path, where 's' represents the position of the object at time 't'. The negative coefficient of the t² term (-1.5) indicates that the parabola opens downwards, suggesting that the object's motion is influenced by a force that opposes its initial direction, such as gravity in the case of a projectile.

At t = 0, the object's initial position is s = 3 meters. As time progresses, the 9t term initially dominates, causing the object to move in a positive direction (away from the reference point). However, the -1.5t² term becomes more influential as time increases, eventually counteracting the positive velocity. The point where the velocity becomes zero is the turning point of this motion. In other words, at t = 3 seconds, the object momentarily stops before changing its direction. This is analogous to throwing a ball straight up in the air; it slows down, stops at its highest point, and then falls back down.

The velocity of 0 m/s at t = 3s does not mean the object has stopped permanently. It is an instantaneous velocity, representing the speed at that exact moment. Just an instant before t = 3s, the object was moving in one direction, and an instant after, it will be moving in the opposite direction. This turning point is crucial for understanding the overall motion of the object. If we were to analyze the object's acceleration (the rate of change of velocity), we would find that it is constant and negative, indicating a constant force acting against the object's motion.

In summary, the result of 0 m/s at t = 3s tells us that this is the point where the object's direction changes. It's a critical point in the motion, and understanding its significance provides valuable insights into the object's trajectory and the forces acting upon it. This interpretation highlights the power of mathematical analysis in understanding physical phenomena and extracting meaningful information from equations of motion.

Conclusion

In this article, we have successfully calculated the velocity of an object described by the displacement function s = 3 + 9t - 1.5t² at time t = 3 seconds. We began by understanding the fundamental concepts of displacement and velocity, emphasizing their relationship as the rate of change of position with respect to time. We then applied the principles of differentiation to find the velocity function, v(t) = 9 - 3t, by taking the derivative of the displacement function. This process involved using the power rule of differentiation, a crucial tool in calculus for analyzing polynomial functions.

Next, we substituted t = 3s into the velocity function to obtain the velocity at that specific time, which resulted in v(3) = 0 m/s. This result was not merely a numerical answer; it had a significant physical interpretation. We discussed how a velocity of 0 m/s at t = 3s indicates that the object is momentarily at rest, representing a turning point in its motion. This is analogous to the peak of a projectile's trajectory, where it momentarily stops before changing direction due to the force of gravity.

The exercise of calculating velocity from a displacement function demonstrates the power of calculus in understanding and analyzing motion. It allows us to extract valuable information about an object's behavior, such as its instantaneous velocity at a specific time, and to interpret this information in the context of the object's overall motion. This concept has wide-ranging applications in physics and engineering, from designing vehicles and predicting projectile trajectories to analyzing the motion of celestial bodies.

By breaking down the problem step-by-step, from understanding the basic concepts to applying calculus and interpreting the result, we have shown how to approach and solve problems involving motion in a systematic and meaningful way. This approach not only provides a solution but also enhances our understanding of the underlying physical principles and their mathematical representation. The ability to calculate and interpret velocity from displacement is a fundamental skill in physics and a testament to the power of mathematical tools in describing the natural world.