Find x as a function of t for motion starting from rest at the origin under the influence of a constant Minkowski force in the x direction. Leave your answer in implicit form (t as a function of x).[Answer:

$\frac{\mathbf{2}\mathbf{kt}}{\mathbf{mc}}\mathbf{=}\left[\begin{array}{c}\\ \end{array}\left(Z\right)\sqrt{z^2+1\mathrm{In}}\left[\begin{array}{c}\\ \end{array}z+\sqrt{z^2+1}\right]\right]\mathbf{}\mathbf{,}\mathbf{}\mathbf{where}\mathbf{}\sqrt{\frac{\mathbf{2}\mathbf{kx}}{{\mathbf{mc}}^{\mathbf{2}}}}$

Write the expression for the momentum

$\mathbf{p}\mathbf{=}\frac{\mathbf{mu}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{u}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}}$

Here, m is the mass, u is the velocity, and c is the speed of light.

As the force is always equal to the rate of change of momentum, write the required equation.

$$\begin{gathered} \frac{\mathrm{dp}}{\mathrm{dt}}=\mathrm{k} \\ \frac{\mathrm{dp}}{\mathrm{dt}}\frac{\mathrm{dt}}{\mathrm{d\tau }}=\mathrm{k}......\left(1\right) \\ \mathrm{Here},\frac{\mathrm{dt}}{\mathrm{d\tau }}=\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}} \\ \mathrm{substitute}\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\mathrm{for}\frac{\mathrm{dt}}{\mathrm{d\tau }}\mathrm{and}\frac{\mathrm{mu}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\mathrm{for}\mathrm{p}\mathrm{in}\mathrm{equation}\left(1\right). \\ \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\right)\left(\frac{\mathrm{mu}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\right) \\ \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{u}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\right)=\frac{\mathrm{k}\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}{\mathrm{m}} \\ \mathrm{multiply}\mathrm{by}\frac{1}{\mathrm{u}}\mathrm{for}\frac{\mathrm{dt}}{\mathrm{dx}}\mathrm{on}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ \frac{\mathrm{dt}}{\mathrm{dx}}\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{u}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\right)=\frac{\mathrm{k}\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}{\mathrm{m}} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{u}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\right)=\frac{\mathrm{k}\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}{\mathrm{m}}.......\left(2\right) \\ \mathrm{Let}\mathrm{\omega }=\frac{\mathrm{u}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}} \\ \mathrm{substitute}\frac{\mathrm{u}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\mathrm{for}\mathrm{\omega }\mathrm{in}\mathrm{equation}\left(2\right). \\ \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{\omega }\right)=\frac{\mathrm{k}\left({\displaystyle \frac{\mathrm{u}}{\mathrm{\omega }}}\right)}{\mathrm{m}} \\ \mathrm{\omega }\frac{\mathrm{d\omega }}{\mathrm{dx}}=\frac{\mathrm{k}}{\mathrm{m}} \\ \frac{1}{2}\frac{{\mathrm{d\omega }}^2}{\mathrm{dx}}=\frac{\mathrm{k}}{\mathrm{m}} \\ \frac{{\mathrm{d\omega }}^2}{\mathrm{dx}}=\frac{2\mathrm{k}}{\mathrm{m}} \\ \mathrm{on}\mathrm{futher}\mathrm{solving}, \\ {\mathrm{d\omega }}^2=\frac{2\mathrm{k}}{\mathrm{m}}\mathrm{dx} \\ \int {\mathrm{d\omega }}^2=\int \frac{2\mathrm{k}}{\mathrm{m}}\mathrm{dx} \\ {\mathrm{\omega }}^2=\frac{2\mathrm{k}}{\mathrm{m}}\mathrm{x}+\mathrm{constant}.........\left(3\right) \end{gathered}$$

At the time t = 0, as the constant value is zero, the equation (3).

$w^{\text{2}}=\frac{\text{2}K}{m}x$

On further solving the above equation,

$\begin{array}{rcl}\frac{u^2}{1-\frac{u^2}{c^2}}& =& \frac{2K}{m}x\\ u^2& =& \frac{2K}{m}x-\frac{2Kx{u}^2}{m{c}^2}\\ u^2\left(1+\frac{2Kx}{m{c}^2}\right)& =& \frac{2Kx}{m}\\ u^2& =& \frac{\left(\frac{2Kx}{m}\right)}{\left(1+\frac{2Kx}{m{c}^2}\right)}\end{array}$

Again on further solving,

$$\begin{gathered} u^2=\frac{c^2}{1+\frac{m{c}^2}{2Kx}} \\ u=\frac{c}{\sqrt{1+\frac{m{c}^2}{2Kx}}} \end{gathered}$$

It is known that:

$\frac{dx}{dt}=\frac{c}{\sqrt{1+\frac{m{c}^2}{2Kx}}}$

Integrate the above equation.

$\begin{array}{rcl}\int \frac{dx}{dt}& =& \int \frac{c}{\sqrt{1+\frac{m{c}^2}{2Kx}}}\\ ct& =& \int \sqrt{1+\frac{m{c}^2}{2Kx}}dx\end{array}$

Let $\frac{mc}{\text{2}K}=b^{\text{2}}$

Hence, the above equation becomes,

$ct=\int \sqrt{1+\frac{b^2}{x}}dx$ …… (4)

Let’s assume,

$ct=\int \sqrt{1+\frac{b^2}{x}}dx$

Substitute $2ydy$ for dx and y²for x in equation (4).

$$\begin{gathered} ct=\int \sqrt{1+\frac{b^2}{y^2}}\left(2ydy\right) \\ ct=\int \left(\frac{\sqrt{y^2+b^2}}{y}\right)2ydy \\ ct=\int 2\sqrt{y^2+b^2}dy \\ ct=\left[y\sqrt{y^2+b^2}+b^2\ln \left[y+\sqrt{y^2+b^2}\right]\right]+\text{constant} \end{gathered}$$ …… (5)

$$\begin{gathered} ct=b^2\left[\left(z\right)\sqrt{z^2+1}\ln \left[z+\sqrt{z^2+1}\right]\right]a \\ \frac{ct}{\left(\frac{mc}{2K}\right)}=\left[\left(z\right)\sqrt{z^2+1}\ln \left[z+\sqrt{z^2+1}\right]\right] \\ \frac{2Kt}{mc}=\left[\left(z\right)\sqrt{z^2+1}\ln \left[z+\sqrt{z^2+1}\right]\right] \end{gathered}$$

Therefore, the value of x as a function of tis .$\frac{2Kt}{mc}=\left[\left(z\right)\sqrt{z^2+1}\ln \left[z+\sqrt{z^2+1}\right]\right]$