Integrating the Velocity Function

Oct 13, 2025 by recoveryshake.com 116 views

Introduction

In physics, understanding the motion of particles is crucial for analyzing various phenomena, from the trajectory of projectiles to the behavior of complex systems. Given a particle's velocity function, we can determine its position by integrating the velocity function with respect to time. In this article, we will explore how to find the position of a particle traveling along the x-axis, with its velocity described by the function ${$ v(t) = t^{1.1} + 4 \sin(2t - 5) $}$. We will also discuss the significance of the initial position and how it affects the particle's motion.

The Velocity Function

The velocity function ${$ v(t) = t^1.1} + 4 \sin(2t - 5) $}$ describes the rate of change of the particle's position with respect to time. This function consists of two components a polynomial term ${$ t^{1.1 $}$ and a sinusoidal term ${$ 4 \sin(2t - 5) $}$. The polynomial term represents a continuous, non-linear increase in velocity, while the sinusoidal term introduces oscillations in the velocity function.

Integrating the Velocity Function

To find the position of the particle, we need to integrate the velocity function with respect to time. This can be done using the following formula:

${$ x(t) = \int v(t) dt $}$

Substituting the given velocity function, we get:

${$ x(t) = \int (t^{1.1} + 4 \sin(2t - 5)) dt $}$

Evaluating the Integral

To evaluate the integral, we can break it down into two separate integrals:

${$ x(t) = \int t^{1.1} dt + \int 4 \sin(2t - 5) dt $}$

The first integral can be evaluated using the power rule of integration:

${$ \int t^{1.1} dt = \frac{t^{1.1+1}}{1.1+1} + C = \frac{t^{2.1}}{2.1} + C $}$

The second integral can be evaluated using the substitution method:

${$ u = 2t - 5 $}$

${$ du = 2 dt $}$

${$ \int 4 \sin(2t - 5) dt = 2 \int \sin(u) du = -2 \cos(u) + C = -2 \cos(2t - 5) + C $}$

Combining the Results

Combining the results of the two integrals, we get:

${$ x(t) = \frac{t^{2.1}}{2.1} - 2 \cos(2t - 5) + C $}$

Initial Position

We are given that the position of the particle is ${$ x = -4 $}$ when ${$ t = 0 $}$. This information allows us to determine the value of the constant ${$ C $}$.

Substituting ${$ t = 0 $}$ and ${$ x = -4 $}$ into the position function, we get:

${$ -4 = \frac{0^{2.1}}{2.1} - 2 \cos(2(0) - 5) + C $}$

Simplifying the expression, we get:

${$ -4 = -2 \cos(-5) + C $}$

Determining the Constant C

To determine the value of the constant ${$ C $}$, we need to find the value of ${$ \cos(-5) $}$. Since the cosine function is an even function, we can write:

${$ \cos(-5) = \cos(5) $}$

Using a calculator or a trigonometric table, we can find the value of ${$ \cos(5) $}$:

${$ \cos(5) \approx -0.9595 $}$

Substituting this value into the expression for ${$ C $}$, we get:

${$ C = -4 + 2 \cos(5) \approx -4 + 2(-0.9595) \approx -4 - 1.919 \approx -5.919 $}$

Final Position Function

Substituting the value of ${$ C $}$ into the position function, we get:

${$ x(t) = \frac{t^{2.1}}{2.1} - 2 \cos(2t - 5) - 5.919 $}$

This is the final position function of the particle.

Conclusion

In this article, we have determined the position of a particle traveling along the x-axis, with its velocity described by the function ${$ v(t) = t^{1.1} + 4 \sin(2t - 5) $}$. We have also discussed the significance of the initial position and how it affects the particle's motion. The final position function is given by:

${$ x(t) = \frac{t^{2.1}}{2.1} - 2 \cos(2t - 5) - 5.919 $}$

This function describes the position of the particle at any time ${$ t $}$.

References

[1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics (10th ed.). John Wiley & Sons.
[2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (10th ed.). Cengage Learning.

Note: The references provided are for general information and are not specific to the problem discussed in this article.

Introduction

In our previous article, we explored the motion of a particle traveling along the x-axis, with its velocity described by the function ${$ v(t) = t^{1.1} + 4 \sin(2t - 5) $}$. We determined the position of the particle using the velocity function and found the final position function to be:

${$ x(t) = \frac{t^{2.1}}{2.1} - 2 \cos(2t - 5) - 5.919 $}$

In this article, we will address some common questions related to the motion of the particle and provide additional insights into the problem.

Q&A

Q: What is the initial velocity of the particle?

A: To find the initial velocity, we need to evaluate the velocity function at ${$ t = 0 $}$.

${$ v(0) = 0^{1.1} + 4 \sin(2(0) - 5) = 0 - 4 \sin(-5) $}$

Using a calculator or a trigonometric table, we can find the value of ${$ \sin(-5) $}$:

${$ \sin(-5) \approx -0.9595 $}$

Substituting this value into the expression for ${$ v(0) $}$, we get:

${$ v(0) \approx 0 - 4(-0.9595) \approx 3.838 $}$

So, the initial velocity of the particle is approximately ${$ 3.838 $}$.

Q: What is the acceleration of the particle?

A: To find the acceleration, we need to take the derivative of the velocity function with respect to time.

${$ a(t) = \frac{dv}{dt} = \frac{d}{dt} (t^{1.1} + 4 \sin(2t - 5)) $}$

Using the power rule and the chain rule, we can evaluate the derivative:

${$ a(t) = 1.1t^{0.1} + 8 \cos(2t - 5) $}$

Q: What is the maximum velocity of the particle?

A: To find the maximum velocity, we need to find the critical points of the velocity function.

${$ v(t) = t^{1.1} + 4 \sin(2t - 5) $}$

Taking the derivative of the velocity function, we get:

${$ \frac{dv}{dt} = 1.1t^{0.1} + 8 \cos(2t - 5) $}$

Setting the derivative equal to zero, we get:

${$ 1.1t^{0.1} + 8 \cos(2t - 5) = 0 $}$

Solving for ${$ t $}$, we get:

${$ t \approx 1.35 $}$

Substituting this value into the velocity function, we get:

${$ v(1.35) \approx 1.35^{1.1} + 4 \sin(2(1.35) - 5) $}$

Using a calculator or a trigonometric table, we can find the value of ${$ v(1.35) $}$:

${$ v(1.35) \approx 2.43 $}$

So, the maximum velocity of the particle is approximately ${$ 2.43 $}$.

Q: What is the minimum velocity of the particle?

A: To find the minimum velocity, we need to find the critical points of the velocity function.

${$ v(t) = t^{1.1} + 4 \sin(2t - 5) $}$

Taking the derivative of the velocity function, we get:

${$ \frac{dv}{dt} = 1.1t^{0.1} + 8 \cos(2t - 5) $}$

Setting the derivative equal to zero, we get:

${$ 1.1t^{0.1} + 8 \cos(2t - 5) = 0 $}$

Solving for ${$ t $}$, we get:

${$ t \approx 1.35 $}$

Substituting this value into the velocity function, we get:

${$ v(1.35) \approx 1.35^{1.1} + 4 \sin(2(1.35) - 5) $}$

Using a calculator or a trigonometric table, we can find the value of ${$ v(1.35) $}$:

${$ v(1.35) \approx 2.43 $}$

However, we also need to consider the minimum value of the sinusoidal term.

${$ 4 \sin(2t - 5) $}$

The minimum value of the sinusoidal term occurs when ${$ 2t - 5 = -\frac{\pi}{2} $}$.

Solving for ${$ t $}$, we get:

${$ t \approx 1.35 $}$

Substituting this value into the velocity function, we get:

${$ v(1.35) \approx 1.35^{1.1} + 4 \sin(2(1.35) - 5) $}$

Using a calculator or a trigonometric table, we can find the value of ${$ v(1.35) $}$:

${$ v(1.35) \approx -2.43 $}$

So, the minimum velocity of the particle is approximately ${$ -2.43 $}$.

Conclusion

In this article, we have addressed some common questions related to the motion of the particle and provided additional insights into the problem. We have found the initial velocity, acceleration, maximum velocity, and minimum velocity of the particle. These values provide a better understanding of the particle's motion and can be used to analyze the problem further.

References

[1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics (10th ed.). John Wiley & Sons.
[2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (10th ed.). Cengage Learning.

Note: The references provided are for general information and are not specific to the problem discussed in this article.