Eliminating the Position Variable

Oct 13, 2025 by pynkocean.com 147 views

Introduction

In physics, the motion of an object can be described using various mathematical equations. One such equation is the velocity equation, which describes the speed of an object as a function of its position. In this article, we will explore the motion of a particle moving along a straight line, with its speed defined as $v=(-4s^2) \text{ m/s}$ , where $s$ is in meters. We will determine the velocity and acceleration as functions of time, using the given initial condition $s=2 \text{ m}$ when $t=0$ .

The Velocity Equation

The velocity equation is given by $v=(-4s^2) \text{ m/s}$ . This equation describes the speed of the particle as a function of its position, $s$ . To determine the velocity as a function of time, we need to eliminate the position variable, $s$ , from the equation.

Eliminating the Position Variable

To eliminate the position variable, $s$ , we can use the fact that the velocity is the derivative of the position with respect to time. Mathematically, this can be expressed as:

v=\frac{ds}{dt}

Substituting the given velocity equation into this expression, we get:

-4s^2=\frac{ds}{dt}

Separating Variables

To solve this differential equation, we can separate the variables, $s$ and $t$ , by dividing both sides of the equation by $-4s^2$ :

\frac{ds}{s^2}=-4dt

Integrating Both Sides

Now, we can integrate both sides of the equation to get:

\int \frac{ds}{s^2}=-4\int dt

Evaluating the integrals, we get:

-\frac{1}{s}=-4t+C

where $C$ is the constant of integration.

Applying the Initial Condition

We are given the initial condition $s=2 \text{ m}$ when $t=0$ . Substituting this condition into the equation, we get:

-\frac{1}{2}=-4(0)+C

Simplifying, we get:

C=-\frac{1}{2}

Substituting the Constant of Integration

Substituting the constant of integration, $C$ , back into the equation, we get:

-\frac{1}{s}=-4t-\frac{1}{2}

Solving for s

To solve for $s$ , we can multiply both sides of the equation by $-1$ :

\frac{1}{s}=4t+\frac{1}{2}

Taking the Reciprocal

Taking the reciprocal of both sides of the equation, we get:

s=\frac{1}{4t+\frac{1}{2}}

Simplifying the Expression

Simplifying the expression, we get:

s=\frac{2}{8t+1}

The Velocity as a Function of Time

Now that we have the position as a function of time, we can determine the velocity as a function of time by taking the derivative of the position with respect to time:

v=\frac{ds}{dt}

Substituting the expression for $s$ , we get:

v=\frac{d}{dt}\left(\frac{2}{8t+1}\right)

Evaluating the Derivative

Evaluating the derivative, we get:

v=-\frac{16}{(8t+1)^2}

The Acceleration as a Function of Time

To determine the acceleration as a function of time, we can take the derivative of the velocity with respect to time:

a=\frac{dv}{dt}

Substituting the expression for $v$ , we get:

a=\frac{d}{dt}\left(-\frac{16}{(8t+1)^2}\right)

Evaluating the Derivative

Evaluating the derivative, we get:

a=\frac{256}{(8t+1)^3}

Conclusion

In this article, we have determined the velocity and acceleration as functions of time for a particle moving along a straight line, with its speed defined as $v=(-4s^2) \text{ m/s}$ , where $s$ is in meters. We have used the given initial condition $s=2 \text{ m}$ when $t=0$ to solve for the position, velocity, and acceleration as functions of time. The results show that the velocity and acceleration are both functions of time, with the velocity given by $v=-\frac{16}{(8t+1)^2}$ and the acceleration given by $a=\frac{256}{(8t+1)^3}$ .

Introduction

In our previous article, we explored the motion of a particle moving along a straight line, with its speed defined as $v=(-4s^2) \text{ m/s}$ , where $s$ is in meters. We determined the velocity and acceleration as functions of time, using the given initial condition $s=2 \text{ m}$ when $t=0$ . In this article, we will answer some frequently asked questions related to this problem.

Q: What is the initial velocity of the particle?

A: To find the initial velocity, we need to find the velocity at $t=0$ . We can do this by substituting $t=0$ into the expression for velocity:

v=-\frac{16}{(8t+1)^2}

Substituting $t=0$ , we get:

v=-\frac{16}{(8(0)+1)^2}=-\frac{16}{1^2}=-16 \text{ m/s}

So, the initial velocity of the particle is $-16 \text{ m/s}$ .

Q: What is the acceleration of the particle at $t=0$ ?

A: To find the acceleration at $t=0$ , we need to find the derivative of the velocity with respect to time and substitute $t=0$ into the expression:

a=\frac{256}{(8t+1)^3}

Substituting $t=0$ , we get:

a=\frac{256}{(8(0)+1)^3}=\frac{256}{1^3}=256 \text{ m/s}^2

So, the acceleration of the particle at $t=0$ is $256 \text{ m/s}^2$ .

Q: How does the velocity of the particle change over time?

A: The velocity of the particle is given by:

v=-\frac{16}{(8t+1)^2}

As time increases, the denominator of the expression increases, which means that the velocity decreases. This is because the particle is slowing down as it moves along the straight line.

Q: How does the acceleration of the particle change over time?

A: The acceleration of the particle is given by:

a=\frac{256}{(8t+1)^3}

As time increases, the denominator of the expression increases, which means that the acceleration decreases. This is because the particle is slowing down as it moves along the straight line.

Q: What is the position of the particle at $t=0$ ?

A: We are given the initial condition $s=2 \text{ m}$ when $t=0$ . This means that the position of the particle at $t=0$ is $2 \text{ m}$ .

Q: How does the position of the particle change over time?

A: The position of the particle is given by:

s=\frac{2}{8t+1}

As time increases, the denominator of the expression increases, which means that the position decreases. This is because the particle is moving along the straight line and its position is changing over time.

Q: What is the velocity of the particle at $t=1$ ?

A: To find the velocity at $t=1$ , we need to substitute $t=1$ into the expression for velocity:

v=-\frac{16}{(8t+1)^2}

Substituting $t=1$ , we get:

v=-\frac{16}{(8(1)+1)^2}=-\frac{16}{9^2}=-\frac{16}{81} \text{ m/s}

So, the velocity of the particle at $t=1$ is $-\frac{16}{81} \text{ m/s}$ .

Q: What is the acceleration of the particle at $t=1$ ?

A: To find the acceleration at $t=1$ , we need to substitute $t=1$ into the expression for acceleration:

a=\frac{256}{(8t+1)^3}

Substituting $t=1$ , we get:

a=\frac{256}{(8(1)+1)^3}=\frac{256}{9^3}=\frac{256}{729} \text{ m/s}^2

So, the acceleration of the particle at $t=1$ is $\frac{256}{729} \text{ m/s}^2$ .

Conclusion

In this article, we have answered some frequently asked questions related to the motion of a particle moving along a straight line, with its speed defined as $v=(-4s^2) \text{ m/s}$ , where $s$ is in meters. We have used the given initial condition $s=2 \text{ m}$ when $t=0$ to solve for the position, velocity, and acceleration as functions of time. The results show that the velocity and acceleration are both functions of time, with the velocity given by $v=-\frac{16}{(8t+1)^2}$ and the acceleration given by $a=\frac{256}{(8t+1)^3}$ .