Motion Analysis Of An Object Along A Straight Path With V(t) = 3t^2 - 6t + 4

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Hey guys! Let's dive into a fascinating physics problem today. We're going to explore the motion of an object moving along a straight line. This object's movement is described by a mathematical function, and our mission is to unravel its secrets. We'll break down the problem step by step, so you can follow along easily.

The Velocity Function: Our Guide to Motion

At the heart of our problem lies the velocity function, a powerful tool that tells us how fast the object is moving and in what direction at any given time. In this case, our velocity function is:

\LARGE\displaystyle\text{\begin{gathered} v(t) = 3t^2 - 6t + 4 \end{gathered}}

This equation might look a bit intimidating at first, but don't worry! Let's break it down. The v(t) part simply means "velocity as a function of time." The t represents time, measured in seconds in our problem. The rest of the equation, 3t^2 - 6t + 4, is a mathematical expression that tells us how to calculate the velocity at any time t. The velocity itself is measured in meters per second (m/s).

Understanding the Velocity Function's Components

Now, let's examine the components of this velocity function. You'll notice that it's a quadratic equation, meaning it has a term with t^2. This tells us that the object's velocity isn't constant; it's changing over time. The 3t^2 term indicates that the velocity changes at an increasing rate. The -6t term suggests a deceleration or a change in direction, and the +4 term represents an initial velocity component.

Time Interval: Setting the Stage

Our object's journey isn't infinite; it occurs within a specific time frame. We're interested in the time interval t ∈ [1, 4] seconds. This means we'll be analyzing the object's motion from 1 second to 4 seconds. This interval is crucial because it sets the boundaries for our calculations and interpretations. We're not concerned with what happened before 1 second or after 4 seconds; our focus is solely on this window of time. Understanding the time interval helps us to contextualize the object's motion and make meaningful conclusions about its behavior during this specific period.

Problem Statement: What We Need to Find

Okay, we've got our velocity function and our time interval. Now, what's the actual question we need to answer? This part is missing from the original prompt, so we need to figure out some possibilities. Here are a few common things we might want to determine about the object's motion:

  • Displacement: How far does the object travel during the time interval?
  • Average Velocity: What's the average speed of the object over the time interval?
  • Distance Traveled: What's the total length of the path the object covers, regardless of direction?
  • Acceleration: How is the object's velocity changing over time?

To make this example complete, let's assume the problem asks us to find the displacement of the object during the time interval t ∈ [1, 4] seconds. Displacement, in physics, refers to the change in position of an object. It's a vector quantity, meaning it has both magnitude (how far the object moved) and direction.

Finding Displacement: Connecting Velocity and Position

So, how do we find the displacement of the object given its velocity function? Here's where calculus comes to the rescue! The fundamental relationship between velocity and position is this:

  • Velocity is the rate of change of position.

In mathematical terms, this means that velocity is the derivative of the position function. Conversely:

  • Position is the integral of the velocity function.

This is the key to solving our problem. To find the displacement, we need to integrate the velocity function over the given time interval. The definite integral of the velocity function from t = 1 to t = 4 will give us the displacement.

The Power of Integration: Reversing Differentiation

Integration is a mathematical operation that essentially "undoes" differentiation. Think of it as finding the area under a curve. In our case, we're finding the area under the velocity curve between t = 1 and t = 4. This area represents the net change in position, which is precisely what displacement is. The integration process allows us to move from a function that describes the rate of change (velocity) to a function that describes the accumulated change (displacement).

Solving for Displacement: A Step-by-Step Approach

Alright, let's get our hands dirty with some math! We need to calculate the definite integral of our velocity function, v(t) = 3t^2 - 6t + 4, from t = 1 to t = 4. Here's how we do it:

  1. Find the indefinite integral: First, we need to find the indefinite integral of v(t). This means finding a function whose derivative is v(t). We use the power rule of integration, which states that the integral of t^n is (t^(n+1))/(n+1). Applying this rule to each term in our velocity function:

    • The integral of 3t^2 is t^3.
    • The integral of -6t is -3t^2.
    • The integral of 4 is 4t.

    So, the indefinite integral of v(t) is s(t) = t^3 - 3t^2 + 4t + C, where C is the constant of integration. We'll see why we don't need to worry about C in the next step.

  2. Evaluate the definite integral: Now, we need to evaluate the definite integral. This means plugging in the limits of integration (1 and 4) into our indefinite integral and subtracting the results:

    • s(4) = (4)^3 - 3(4)^2 + 4(4) = 64 - 48 + 16 = 32
    • s(1) = (1)^3 - 3(1)^2 + 4(1) = 1 - 3 + 4 = 2

    The displacement is s(4) - s(1) = 32 - 2 = 30 meters.

The Constant of Integration: Why It Doesn't Matter

You might be wondering, "What happened to the constant of integration, C?" Well, when we evaluate the definite integral, we subtract the indefinite integral at the lower limit from the indefinite integral at the upper limit. The C terms cancel out in this subtraction, so we don't need to worry about them when calculating definite integrals.

The Answer: Unveiling the Object's Journey

We've done it! We've successfully calculated the displacement of the object. The displacement of the object during the time interval t ∈ [1, 4] seconds is 30 meters. This means that the object's final position is 30 meters further along the straight line than its initial position.

Interpreting the Result: What Does 30 Meters Mean?

It's important to understand what this 30-meter displacement tells us. It tells us the net change in position. The object might have moved forward and backward during this time interval, but the overall result is a forward movement of 30 meters. If we wanted to know the total distance traveled, we'd need to consider any changes in direction and calculate the distance traveled in each segment of the motion.

Key Takeaways: Mastering Motion Problems

Let's recap the key concepts we've used to solve this problem. Understanding these concepts will help you tackle similar motion problems in the future:

  • Velocity Function: The velocity function describes how an object's velocity changes over time.
  • Time Interval: The time interval defines the period of motion we're interested in.
  • Displacement: Displacement is the net change in position of an object.
  • Integration: Integration is the mathematical process of finding the area under a curve, and it's the key to connecting velocity and position.
  • Definite Integral: The definite integral of the velocity function over a time interval gives us the displacement during that interval.

By understanding these concepts and practicing problem-solving techniques, you'll become a pro at analyzing motion! Remember, physics is all about understanding the world around us, and these tools help us make sense of how things move.

Further Exploration: Expanding Our Knowledge

This problem was a great starting point, but there's so much more we can explore! Here are some ideas for further investigation:

  • Calculate the Average Velocity: We found the displacement, but what was the average velocity of the object during the time interval? To find the average velocity, divide the displacement by the time interval.
  • Determine the Distance Traveled: As mentioned earlier, the distance traveled is different from displacement if the object changes direction. How would you find the distance traveled in this case? This involves finding when the velocity is zero (potential turning points) and integrating the absolute value of the velocity function.
  • Find the Acceleration: Acceleration is the rate of change of velocity. How would you find the acceleration function given the velocity function? Remember that acceleration is the derivative of velocity.
  • Graph the Motion: Visualizing the motion can be incredibly helpful. Try graphing the position, velocity, and acceleration functions. How do these graphs relate to each other?

By exploring these additional questions, you'll deepen your understanding of motion and develop your problem-solving skills even further. So, keep exploring, keep questioning, and keep learning!