Let $s(t)=6t^3+45t^2+108t$ Be The Equation Of Motion For A Particle. 1. Find A Function For The Velocity. - $v(t)=18t^2+90t+108$2. Where Does The Velocity Equal Zero? [Hint: Factor Out The GCF.] - $t=$ $\square$

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Introduction

In physics, the equation of motion is a fundamental concept that describes the position of an object as a function of time. Given the equation of motion, we can derive the velocity and acceleration of the object, which are essential in understanding its motion. In this article, we will explore the concept of motion equations and derive the velocity function from a given equation of motion. We will also discuss how to find the critical points where the velocity equals zero.

Equation of Motion

The equation of motion for a particle is given by:

s(t)=6t3+45t2+108ts(t) = 6t^3 + 45t^2 + 108t

where s(t)s(t) is the position of the particle at time tt.

Finding the Velocity Function

To find the velocity function, we need to take the derivative of the equation of motion with respect to time. The velocity function is denoted by v(t)v(t) and is given by:

v(t)=dsdt=ddt(6t3+45t2+108t)v(t) = \frac{ds}{dt} = \frac{d}{dt} (6t^3 + 45t^2 + 108t)

Using the power rule of differentiation, we get:

v(t)=18t2+90t+108v(t) = 18t^2 + 90t + 108

Where Does the Velocity Equal Zero?

To find the critical points where the velocity equals zero, we need to set the velocity function equal to zero and solve for tt. We can factor out the greatest common factor (GCF) to simplify the equation:

v(t)=18t2+90t+108=18(t2+5t+6)=0v(t) = 18t^2 + 90t + 108 = 18(t^2 + 5t + 6) = 0

Factoring the quadratic expression inside the parentheses, we get:

v(t)=18(t+2)(t+3)=0v(t) = 18(t + 2)(t + 3) = 0

Setting each factor equal to zero, we get:

t+2=0ort+3=0t + 2 = 0 \quad \text{or} \quad t + 3 = 0

Solving for tt, we get:

t=βˆ’2ort=βˆ’3t = -2 \quad \text{or} \quad t = -3

Therefore, the velocity equals zero at t=βˆ’2t = -2 and t=βˆ’3t = -3.

Discussion

In this article, we have explored the concept of motion equations and derived the velocity function from a given equation of motion. We have also discussed how to find the critical points where the velocity equals zero. The critical points are essential in understanding the motion of the particle, as they indicate the points where the particle changes direction.

Conclusion

In conclusion, the equation of motion is a fundamental concept in physics that describes the position of an object as a function of time. By deriving the velocity function from the equation of motion, we can gain a deeper understanding of the motion of the object. The critical points where the velocity equals zero are essential in understanding the motion of the particle, as they indicate the points where the particle changes direction.

Mathematical Derivations

Derivation of the Velocity Function

To derive the velocity function, we need to take the derivative of the equation of motion with respect to time. The velocity function is denoted by v(t)v(t) and is given by:

v(t)=dsdt=ddt(6t3+45t2+108t)v(t) = \frac{ds}{dt} = \frac{d}{dt} (6t^3 + 45t^2 + 108t)

Using the power rule of differentiation, we get:

v(t)=18t2+90t+108v(t) = 18t^2 + 90t + 108

Derivation of the Critical Points

To find the critical points where the velocity equals zero, we need to set the velocity function equal to zero and solve for tt. We can factor out the greatest common factor (GCF) to simplify the equation:

v(t)=18t2+90t+108=18(t2+5t+6)=0v(t) = 18t^2 + 90t + 108 = 18(t^2 + 5t + 6) = 0

Factoring the quadratic expression inside the parentheses, we get:

v(t)=18(t+2)(t+3)=0v(t) = 18(t + 2)(t + 3) = 0

Setting each factor equal to zero, we get:

t+2=0ort+3=0t + 2 = 0 \quad \text{or} \quad t + 3 = 0

Solving for tt, we get:

t=βˆ’2ort=βˆ’3t = -2 \quad \text{or} \quad t = -3

Therefore, the velocity equals zero at t=βˆ’2t = -2 and t=βˆ’3t = -3.

References

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

Keywords

  • Motion equations
  • Velocity function
  • Critical points
  • Greatest common factor (GCF)
  • Quadratic expression
  • Power rule of differentiation
    Motion Equations: A Q&A Guide =====================================

Introduction

In our previous article, we explored the concept of motion equations and derived the velocity function from a given equation of motion. We also discussed how to find the critical points where the velocity equals zero. In this article, we will answer some frequently asked questions about motion equations and provide additional insights into this fascinating topic.

Q&A

Q: What is the equation of motion?

A: The equation of motion is a mathematical equation that describes the position of an object as a function of time. It is typically denoted by s(t)s(t) and is used to describe the motion of an object in one dimension.

Q: How do I derive the velocity function from the equation of motion?

A: To derive the velocity function, you need to take the derivative of the equation of motion with respect to time. The velocity function is denoted by v(t)v(t) and is given by:

v(t)=dsdt=ddt(6t3+45t2+108t)v(t) = \frac{ds}{dt} = \frac{d}{dt} (6t^3 + 45t^2 + 108t)

Using the power rule of differentiation, you get:

v(t)=18t2+90t+108v(t) = 18t^2 + 90t + 108

Q: What are critical points?

A: Critical points are points where the velocity equals zero. They are essential in understanding the motion of an object, as they indicate the points where the object changes direction.

Q: How do I find critical points?

A: To find critical points, you need to set the velocity function equal to zero and solve for tt. You can factor out the greatest common factor (GCF) to simplify the equation:

v(t)=18t2+90t+108=18(t2+5t+6)=0v(t) = 18t^2 + 90t + 108 = 18(t^2 + 5t + 6) = 0

Factoring the quadratic expression inside the parentheses, you get:

v(t)=18(t+2)(t+3)=0v(t) = 18(t + 2)(t + 3) = 0

Setting each factor equal to zero, you get:

t+2=0ort+3=0t + 2 = 0 \quad \text{or} \quad t + 3 = 0

Solving for tt, you get:

t=βˆ’2ort=βˆ’3t = -2 \quad \text{or} \quad t = -3

Therefore, the velocity equals zero at t=βˆ’2t = -2 and t=βˆ’3t = -3.

Q: What is the significance of critical points?

A: Critical points are essential in understanding the motion of an object, as they indicate the points where the object changes direction. They are also used to determine the acceleration of an object, which is the rate of change of velocity.

Q: Can you provide an example of a motion equation?

A: Yes, here is an example of a motion equation:

s(t)=2t2+5t+1s(t) = 2t^2 + 5t + 1

To derive the velocity function, you need to take the derivative of the equation of motion with respect to time:

v(t)=dsdt=ddt(2t2+5t+1)v(t) = \frac{ds}{dt} = \frac{d}{dt} (2t^2 + 5t + 1)

Using the power rule of differentiation, you get:

v(t)=4t+5v(t) = 4t + 5

To find critical points, you need to set the velocity function equal to zero and solve for tt:

v(t)=4t+5=0v(t) = 4t + 5 = 0

Solving for tt, you get:

t=βˆ’54t = -\frac{5}{4}

Therefore, the velocity equals zero at t=βˆ’54t = -\frac{5}{4}.

Conclusion

In conclusion, motion equations are a fundamental concept in physics that describe the position of an object as a function of time. By deriving the velocity function from the equation of motion, we can gain a deeper understanding of the motion of an object. Critical points are essential in understanding the motion of an object, as they indicate the points where the object changes direction.

Mathematical Derivations

Derivation of the Velocity Function

To derive the velocity function, you need to take the derivative of the equation of motion with respect to time. The velocity function is denoted by v(t)v(t) and is given by:

v(t)=dsdt=ddt(6t3+45t2+108t)v(t) = \frac{ds}{dt} = \frac{d}{dt} (6t^3 + 45t^2 + 108t)

Using the power rule of differentiation, you get:

v(t)=18t2+90t+108v(t) = 18t^2 + 90t + 108

Derivation of the Critical Points

To find critical points, you need to set the velocity function equal to zero and solve for tt. You can factor out the greatest common factor (GCF) to simplify the equation:

v(t)=18t2+90t+108=18(t2+5t+6)=0v(t) = 18t^2 + 90t + 108 = 18(t^2 + 5t + 6) = 0

Factoring the quadratic expression inside the parentheses, you get:

v(t)=18(t+2)(t+3)=0v(t) = 18(t + 2)(t + 3) = 0

Setting each factor equal to zero, you get:

t+2=0ort+3=0t + 2 = 0 \quad \text{or} \quad t + 3 = 0

Solving for tt, you get:

t=βˆ’2ort=βˆ’3t = -2 \quad \text{or} \quad t = -3

Therefore, the velocity equals zero at t=βˆ’2t = -2 and t=βˆ’3t = -3.

References

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

Keywords

  • Motion equations
  • Velocity function
  • Critical points
  • Greatest common factor (GCF)
  • Quadratic expression
  • Power rule of differentiation