Velocity Calculation: Position Function S = 3t³ + 4t + 9

by Alex Johnson 57 views

Understanding motion is a fundamental concept in physics and mathematics, and it often begins with understanding how position, velocity, and acceleration are related. In this article, we'll dive deep into calculating velocity from a given position function, using the example of s = f(t) = 3t³ + 4t + 9. We will explore the core concepts, the step-by-step process of finding the velocity function, and how to determine the velocity at a specific time. Whether you're a student tackling calculus problems or simply curious about the mathematics of motion, this guide will provide you with a clear and thorough understanding.

Understanding Position, Velocity, and Acceleration

Before we jump into the calculations, let's clarify the relationship between position, velocity, and acceleration. These three concepts are interconnected and form the basis of kinematics, the branch of physics that describes the motion of objects.

  • Position: The position of an object describes its location in space at a particular time. It is often represented by a function s(t), where t represents time. In our case, the position function is given by s = f(t) = 3t³ + 4t + 9.
  • Velocity: Velocity describes how the position of an object changes with time. It is the rate of change of position and is calculated as the derivative of the position function with respect to time. Velocity includes both the speed and direction of motion.
  • Acceleration: Acceleration describes how the velocity of an object changes with time. It is the rate of change of velocity and is calculated as the derivative of the velocity function with respect to time. Acceleration indicates how quickly an object's speed or direction is changing.

In mathematical terms, if we have a position function s(t), then:

  • Velocity, v(t) = s'(t) (the first derivative of s(t))
  • Acceleration, a(t) = v'(t) = s''(t) (the second derivative of s(t))

The Significance of Derivatives in Motion Analysis

The concept of derivatives is crucial in understanding motion. The derivative of a function at a point gives the instantaneous rate of change of the function at that point. In the context of motion:

  • The derivative of the position function gives the instantaneous velocity.
  • The derivative of the velocity function gives the instantaneous acceleration.

This relationship allows us to analyze the motion of an object at any given moment in time, providing valuable insights into its behavior.

(a) Finding the Velocity Function at Time t

Now, let's tackle the first part of our problem: finding the velocity function at time t. Given the position function s = f(t) = 3t³ + 4t + 9, we need to find its derivative with respect to t. This will give us the velocity function v(t).

Applying the Power Rule of Differentiation

To find the derivative of f(t) = 3t³ + 4t + 9, we'll use the power rule of differentiation. The power rule states that if f(x) = axⁿ, where a is a constant and n is any real number, then the derivative f'(x) = naxⁿ⁻¹.

Let's apply this rule to each term in our position function:

  1. Derivative of 3t³:
    • Here, a = 3 and n = 3.
    • Applying the power rule, we get 3 * 3t^(3-1) = 9t².
  2. Derivative of 4t:
    • Here, a = 4 and n = 1 (since t = t¹).
    • Applying the power rule, we get 4 * 1t^(1-1) = 4t⁰ = 4.
  3. Derivative of 9:
    • The derivative of a constant is always 0.

Constructing the Velocity Function

Now, we combine the derivatives of each term to find the velocity function:

v(t) = s'(t) = 9t² + 4 + 0 = 9t² + 4

So, the velocity function at time t is v(t) = 9t² + 4. This function tells us the velocity of the particle at any given time t.

Interpreting the Velocity Function

The velocity function v(t) = 9t² + 4 provides valuable information about the particle's motion. Notice that the velocity is always positive because 9t² is always non-negative (since squaring any real number results in a non-negative value), and we are adding 4 to it. This means the particle is always moving in the positive direction.

Additionally, the velocity increases as time t increases. This indicates that the particle is accelerating, meaning its speed is increasing over time. The 9t² term dominates the velocity function as t grows, showing that the rate of increase in velocity becomes more significant at later times.

(b) Finding the Velocity at Time t = 3 Seconds

Now that we have the velocity function, v(t) = 9t² + 4, we can easily find the velocity at a specific time. In this case, we want to find the velocity at t = 3 seconds. To do this, we simply substitute t = 3 into the velocity function.

Substituting t = 3 into the Velocity Function

v(3) = 9(3)² + 4

Let's calculate this step by step:

  1. Calculate 3²:
    • 3² = 3 * 3 = 9
  2. Multiply by 9:
    • 9 * 9 = 81
  3. Add 4:
    • 81 + 4 = 85

Therefore, the velocity at t = 3 seconds is v(3) = 85.

Interpreting the Result

The velocity at t = 3 seconds is 85 units of distance per unit of time. The units depend on the units used for position and time in the original position function. For example, if the position s is in meters and the time t is in seconds, then the velocity would be 85 meters per second (m/s).

This result tells us that at 3 seconds, the particle is moving at a speed of 85 m/s in the positive direction. This reinforces our earlier observation that the particle's velocity increases over time.

The Importance of Units in Physics

It's crucial to include units when dealing with physical quantities. Units provide context and meaning to the numerical values. In this case, knowing that the velocity is 85 m/s gives us a much clearer understanding of the particle's motion than simply stating the number 85. Always pay attention to units in physics problems and make sure to include them in your answers.

Connecting Velocity to the Graph of the Position Function

A valuable way to understand velocity is to visualize it in the context of the position function's graph. The velocity at any given time t is represented by the slope of the tangent line to the position function's graph at that time.

Visualizing Velocity as Slope

Imagine the graph of s = f(t) = 3t³ + 4t + 9. At any point on this curve, we can draw a tangent line, which is a straight line that touches the curve at that point without crossing it. The slope of this tangent line represents the instantaneous velocity at that time.

  • Positive Slope: A positive slope indicates that the position is increasing with time, meaning the particle is moving in the positive direction. The steeper the slope, the higher the velocity.
  • Negative Slope: A negative slope indicates that the position is decreasing with time, meaning the particle is moving in the negative direction. The steeper the slope (in the negative direction), the higher the velocity in the negative direction.
  • Zero Slope: A zero slope indicates that the position is momentarily not changing, meaning the particle is at rest or at a turning point in its motion.

Relating the Velocity Function to the Graph

Our velocity function, v(t) = 9t² + 4, tells us how the slope of the tangent line changes over time. Since v(t) is always positive, the tangent line to the graph of s(t) will always have a positive slope. Furthermore, as t increases, v(t) increases, meaning the slope of the tangent line becomes steeper, indicating that the particle is moving faster.

Using Graphs to Estimate Velocity

By examining the graph of the position function, we can qualitatively estimate the velocity at different times. For instance, if we draw a tangent line at t = 3 seconds, we would expect its slope to be quite steep, corresponding to our calculated velocity of 85 m/s. This visual connection between the position function's graph and the velocity provides a deeper understanding of the motion.

Conclusion

In this article, we've explored how to calculate velocity from a given position function. We started with the position function s = f(t) = 3t³ + 4t + 9 and successfully found the velocity function v(t) = 9t² + 4 by taking the derivative of the position function. We then calculated the velocity at t = 3 seconds, finding it to be 85 units of distance per unit of time. We also discussed the significance of derivatives in motion analysis, the importance of units, and the graphical interpretation of velocity as the slope of the tangent line to the position function's graph.

Understanding the relationship between position, velocity, and acceleration is crucial in many areas of physics and mathematics. By mastering these concepts and the techniques for calculating them, you'll be well-equipped to tackle more complex problems in motion analysis and related fields. Remember to practice these skills and explore different position functions to further solidify your understanding.

For further reading and to deepen your knowledge, consider exploring resources like Khan Academy's physics section, which offers excellent lessons and practice exercises on kinematics and calculus.