The Position From Its Starting Point Of A Small Plane Preparing For Takeoff Is Given By $x(t) = 1.43t^2$ Meters, Where $t$ Is Measured In Seconds. What Is The Acceleration Of The Plane? $a(t) = \, \text{m/s}^2$

Introduction

When a small plane is preparing for takeoff, its position is given by a mathematical equation that describes its movement over time. In this case, the position of the plane is given by the equation $x(t) = 1.43t^2$ meters, where $t$ is measured in seconds. To understand the movement of the plane, we need to find its acceleration, which is the rate of change of its velocity. In this article, we will discuss how to find the acceleration of the plane using the given position equation.

Understanding the Position Equation

The position equation $x(t) = 1.43t^2$ describes the position of the plane at any given time $t$. The equation is a quadratic function, which means that the position of the plane increases quadratically with time. This means that the plane is accelerating, and its velocity is increasing at a constant rate.

Finding the Velocity

To find the acceleration of the plane, we need to find its velocity first. The velocity of the plane is the rate of change of its position with respect to time. Mathematically, the velocity is given by the derivative of the position equation with respect to time:

$$v(t) = \frac{dx}{dt} = \frac{d}{dt} (1.43t^2)$$

Using the power rule of differentiation, we get:

$$v(t) = 2.86t$$

This means that the velocity of the plane is increasing linearly with time.

Finding the Acceleration

The acceleration of the plane is the rate of change of its velocity with respect to time. Mathematically, the acceleration is given by the derivative of the velocity equation with respect to time:

$$a(t) = \frac{dv}{dt} = \frac{d}{dt} (2.86t)$$

Using the power rule of differentiation, we get:

$$a(t) = 2.86$$

This means that the acceleration of the plane is constant and equal to 2.86 m/s^2.

Discussion

The acceleration of the plane is an important factor in determining its movement. A constant acceleration means that the plane is accelerating at a constant rate, which is a characteristic of a uniformly accelerated motion. This type of motion is commonly observed in objects that are under the influence of a constant force, such as a plane that is accelerating down a runway.

Conclusion

In conclusion, we have found the acceleration of a small plane preparing for takeoff using the given position equation. The acceleration is constant and equal to 2.86 m/s^2, which means that the plane is accelerating at a constant rate. This type of motion is commonly observed in objects that are under the influence of a constant force.

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.

Mathematical Derivations

Derivation of Velocity

The velocity of the plane is given by the derivative of the position equation with respect to time:

$$v(t) = \frac{dx}{dt} = \frac{d}{dt} (1.43t^2)$$

Using the power rule of differentiation, we get:

$$v(t) = 2.86t$$

Derivation of Acceleration

The acceleration of the plane is given by the derivative of the velocity equation with respect to time:

$$a(t) = \frac{dv}{dt} = \frac{d}{dt} (2.86t)$$

Using the power rule of differentiation, we get:

$$a(t) = 2.86$$

Derivation of Position Equation

The position equation is given by:

$$x(t) = 1.43t^2$$

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Introduction

In our previous article, we discussed the position and acceleration of a small plane preparing for takeoff. We found that the acceleration of the plane is constant and equal to 2.86 m/s^2. In this article, we will answer some frequently asked questions related to the position and acceleration of the plane.

Q: What is the position of the plane at time t = 0?

A: The position of the plane at time t = 0 is given by the equation x(t) = 1.43t^2. Substituting t = 0 into the equation, we get x(0) = 1.43(0)^2 = 0. This means that the plane is at rest at time t = 0.

Q: What is the velocity of the plane at time t = 0?

A: The velocity of the plane is given by the derivative of the position equation with respect to time. We found that the velocity is given by v(t) = 2.86t. Substituting t = 0 into the equation, we get v(0) = 2.86(0) = 0. This means that the plane is at rest at time t = 0.

Q: What is the acceleration of the plane at time t = 0?

A: The acceleration of the plane is given by the derivative of the velocity equation with respect to time. We found that the acceleration is given by a(t) = 2.86. This means that the acceleration of the plane is constant and equal to 2.86 m/s^2.

Q: How does the position of the plane change with time?

A: The position of the plane changes quadratically with time. This means that the position of the plane increases rapidly as time increases.

Q: How does the velocity of the plane change with time?

A: The velocity of the plane changes linearly with time. This means that the velocity of the plane increases rapidly as time increases.

Q: What is the relationship between the position, velocity, and acceleration of the plane?

A: The position, velocity, and acceleration of the plane are related by the following equations:

• The position of the plane is given by x(t) = 1.43t^2.
• The velocity of the plane is given by v(t) = 2.86t.
• The acceleration of the plane is given by a(t) = 2.86.

These equations show that the position, velocity, and acceleration of the plane are related by a quadratic, linear, and constant relationship, respectively.

Q: What is the significance of the acceleration of the plane?

A: The acceleration of the plane is an important factor in determining its movement. A constant acceleration means that the plane is accelerating at a constant rate, which is a characteristic of a uniformly accelerated motion. This type of motion is commonly observed in objects that are under the influence of a constant force, such as a plane that is accelerating down a runway.

Conclusion

In conclusion, we have answered some frequently asked questions related to the position and acceleration of a small plane preparing for takeoff. We hope that this article has provided a better understanding of the position and acceleration of the plane and its significance in determining its movement.

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.

Mathematical Derivations

Derivation of Velocity

The velocity of the plane is given by the derivative of the position equation with respect to time:

$$v(t) = \frac{dx}{dt} = \frac{d}{dt} (1.43t^2)$$

Using the power rule of differentiation, we get:

$$v(t) = 2.86t$$

Derivation of Acceleration

The acceleration of the plane is given by the derivative of the velocity equation with respect to time:

$$a(t) = \frac{dv}{dt} = \frac{d}{dt} (2.86t)$$

Using the power rule of differentiation, we get:

$$a(t) = 2.86$$

Derivation of Position Equation

The position equation is given by:

$$x(t) = 1.43t^2$$

This equation describes the position of the plane at any given time $t$. The equation is a quadratic function, which means that the position of the plane increases quadratically with time.