Question

Find the position x of a particle at time t if its acceleration is$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}\mathit{A}\mathbf{}\mathit{t}\mathbf{}\mathit{\omega }\mathbf{}\mathit{s}\mathit{i}\mathit{n}$.

Short answer

Answer

The position x of a particle at time t is $x\omega =-A\omega t^{-2}\sin +ct+d$when acceleration is $\frac{d^{2}x}{d{t}^2}=At\omega \sin$.

Step-by-step solution

Given information

In this question, the acceleration is given as $\frac{d^{2}x}{d{t}^2}=At\omega \sin$

Meaning of differential equation

In mathematics, an equation with only one independent variable and one or more of its derivatives with respect to the variable is referred to as an ordinary differential equation or ODE. In other words, the ODE is a relation with one independent variable x, a real dependent variable y, and several derivatives $\mathit{y}\mathbf{\text{'}}\mathbf{,}\mathit{y}\mathbf{"}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{y}}^{\mathbf{n}}$in relation to x.

Find the position of a particle

The given acceleration is $\frac{d^{2}x}{d{t}^2}=At\omega \sin$.

Integrate the acceleration two times.

$$\begin{gathered} \frac{dx}{dt}=\int At\omega \sin \left(dt\right) \\ =-A{\omega }^{-1}\cos \omega t+c \end{gathered}$$

The second integration will be

$$\begin{gathered} dx=\left(-A\omega t^{-1}\cos +dt\right) \\ \int dx=\int -At\omega d{t}^{-1}\cos dt+\int \\ x\omega =-A\omega t^{-2}\sin +ct+d \end{gathered}$$

Therefore, the position of a particle at time t is $x\omega =-A\omega t^{-2}\sin +ct+d$.