A Particle Travels Along The \[$x\$\]-axis Such That Its Velocity Is Given By $v(t)=t^{0.9}-3 \cos \left(t^2+1\right$\]. The Position Of The Particle Is \[$x=4\$\] When \[$t=2\$\]. Determine The Position, Velocity, And

**A Particle Travels Along the ${$x\$}$-axis: A Comprehensive Analysis**

In this article, we will delve into the world of mathematical modeling and explore the motion of a particle traveling along the {x$}$-axis. The velocity of the particle is given by the function $v(t)=t^{0.9}-3 \cos \left(t^2+1\right)$, and we are tasked with determining the position, velocity, and acceleration of the particle at any given time $t$. We will also use the given initial condition $x=4$ when $t=2$ to solve for the position function.

To find the position function $x(t)$, we need to integrate the velocity function $v(t)$ with respect to time $t$. This is because the velocity function represents the rate of change of the position function with respect to time.

Q: How do we find the position function $x(t)$?

A: We integrate the velocity function $v(t)$ with respect to time $t$.

Q: What is the formula for the position function $x(t)$?

A: The position function $x(t)$ is given by the formula:

$$x(t) = \int v(t) dt = \int (t^{0.9}-3 \cos \left(t^2+1\right)) dt$$

Q: How do we evaluate the integral?

A: We can evaluate the integral using the power rule of integration and the substitution method.

Q: What is the power rule of integration?

A: The power rule of integration states that:

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

Q: How do we apply the power rule of integration to the position function?

A: We apply the power rule of integration to the term $t^{0.9}$:

$$\int t^{0.9} dt = \frac{t^{1.9}}{1.9} + C$$

Q: What is the substitution method?

A: The substitution method is a technique used to evaluate integrals by substituting a new variable into the integral.

Q: How do we apply the substitution method to the position function?

A: We substitute $u = t^2 + 1$ into the integral:

$$\int -3 \cos \left(t^2+1\right) dt = -3 \int \cos u du$$

Q: What is the antiderivative of $\cos u$?

A: The antiderivative of $\cos u$ is $\sin u$.

Q: How do we evaluate the integral?

A: We evaluate the integral using the antiderivative:

$$-3 \int \cos u du = -3 \sin u + C$$

Q: What is the final position function $x(t)$?

A: The final position function $x(t)$ is:

$$x(t) = \frac{t^{1.9}}{1.9} - 3 \sin \left(t^2+1\right) + C$$

Q: What is the value of the constant $C$?

A: We use the initial condition $x=4$ when $t=2$ to solve for the constant $C$:

$$4 = \frac{2^{1.9}}{1.9} - 3 \sin \left(2^2+1\right) + C$$

Q: How do we solve for the constant $C$?

A: We solve for the constant $C$ by isolating it on one side of the equation:

$$C = 4 - \frac{2^{1.9}}{1.9} + 3 \sin \left(2^2+1\right)$$

Q: What is the final position function $x(t)$?

A: The final position function $x(t)$ is:

$$x(t) = \frac{t^{1.9}}{1.9} - 3 \sin \left(t^2+1\right) + 4 - \frac{2^{1.9}}{1.9} + 3 \sin \left(2^2+1\right)$$

To find the velocity function $v(t)$, we need to differentiate the position function $x(t)$ with respect to time $t$. This is because the velocity function represents the rate of change of the position function with respect to time.

Q: How do we find the velocity function $v(t)$?

A: We differentiate the position function $x(t)$ with respect to time $t$.

Q: What is the formula for the velocity function $v(t)$?

A: The velocity function $v(t)$ is given by the formula:

$$v(t) = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{t^{1.9}}{1.9} - 3 \sin \left(t^2+1\right) + 4 - \frac{2^{1.9}}{1.9} + 3 \sin \left(2^2+1\right) \right)$$

Q: How do we evaluate the derivative?

A: We evaluate the derivative using the power rule of differentiation and the chain rule.

Q: What is the power rule of differentiation?

A: The power rule of differentiation states that:

$$\frac{d}{dx} x^n = nx^{n-1}$$

Q: How do we apply the power rule of differentiation to the velocity function?

A: We apply the power rule of differentiation to the term $t^{1.9}$:

$$\frac{d}{dt} \left( \frac{t^{1.9}}{1.9} \right) = \frac{1.9t^{0.9}}{1.9}$$

Q: What is the chain rule?

A: The chain rule is a technique used to differentiate composite functions.

Q: How do we apply the chain rule to the velocity function?

A: We apply the chain rule to the term $-3 \sin \left(t^2+1\right)$:

$$\frac{d}{dt} \left( -3 \sin \left(t^2+1\right) \right) = -3 \cos \left(t^2+1\right) \frac{d}{dt} \left(t^2+1\right)$$

Q: What is the derivative of $t^2+1$?

A: The derivative of $t^2+1$ is $2t$.

Q: How do we evaluate the derivative?

A: We evaluate the derivative using the chain rule:

$$-3 \cos \left(t^2+1\right) \frac{d}{dt} \left(t^2+1\right) = -3 \cos \left(t^2+1\right) 2t$$

Q: What is the final velocity function $v(t)$?

A: The final velocity function $v(t)$ is:

$$v(t) = t^{0.9} - 6t \cos \left(t^2+1\right)$$

To find the acceleration function $a(t)$, we need to differentiate the velocity function $v(t)$ with respect to time $t$. This is because the acceleration function represents the rate of change of the velocity function with respect to time.

Q: How do we find the acceleration function $a(t)$?

A: We differentiate the velocity function $v(t)$ with respect to time $t$.

Q: What is the formula for the acceleration function $a(t)$?

A: The acceleration function $a(t)$ is given by the formula:

$$a(t) = \frac{dv}{dt} = \frac{d}{dt} \left( t^{0.9} - 6t \cos \left(t^2+1\right) \right)$$

Q: How do we evaluate the derivative?

A: We evaluate the derivative using the power rule of differentiation and the product rule.

Q: What is the power rule of differentiation?

A: The power rule of differentiation states that:

$$\frac{d}{dx} x^n = nx^{n-1}$$

Q: How do we apply the power rule of differentiation to the acceleration function?

A: We apply the power rule of differentiation to the term $t^{0.9}$:

$$\frac{d}{dt} \left( t^{0.9} \right) = 0.9t^{0.1}$$

Q: What is the product rule?

A: The product rule is a technique used to differentiate products of functions.

Q: How do we apply the product rule to the acceleration function?

A: We apply the product rule to the term $-6t \cos \left(t^2+1\right)$:

$$\frac{d}{dt} \left( -6t \cos \left(t^2+1\right) \right) = -6 \cos \left(t^2+1\right) \frac{d}{dt}$$