Question

Exercise 117 gives the speed u of an object accelerated under a constant force. Show that the distance it travels is given by

$\mathit{x}\mathbf{=}\frac{\mathbf{m}{\mathbf{c}}^{\mathbf{2}}}{{\mathbf{f}}^{\mathbf{2}}}\left[\sqrt{1+{\left(\frac{Ft}{mc}\right)}^2}-1\right]$

Short answer

Distance travelled by an object under constant force is determined by integrating the expression of relativistic speed.

Step-by-step solution

Determine the expression for the differential change in distance dx

Consider an object moving under a constant force, the relativistic speed is given by

$u=\frac{1}{\sqrt{1+{\left(\frac{Ft}{itc}\right)}^2}}\frac{F}{m}t$

Here, velocity $u=\frac{dx}{dt}$and let$k=\frac{F}{mc}$.

$\frac{dx}{dt}-\frac{c}{\sqrt{1+(kt{)}^3}}dt$

$dx=\frac{tki}{\sqrt{1+(kt{)}^2}}dt$

Integrate the above expression

Let’s say

$$\begin{gathered} m=1+k^2{t}^2 \\ \Rightarrow dm=k^2\left(2tdt\right) \\ \Rightarrow tdt=\frac{dm}{2k^2} \end{gathered}$$.

Therefore,

$dx=\frac{c}{2k\sqrt{m}}dm$

$\int dx=\frac{c}{2k}\int \frac{1}{\sqrt{m}}dm$

$x=\frac{c}{2k}\left[\frac{\sqrt{m}}{1/2}\right]+C$

$x=\frac{c}{k}{\left(1+k^2{t}^2\right)}^{1/2}+C$

${\mathrm{At}}^{t=0},\mathrm{x}=0$Therefore, the integration constant

$C=-\frac{c}{k}$

$x=\frac{c}{k}{\left(1+k^2{t}^2\right)}^{1/2}-\frac{\mathcal{c}}{k}$

$x=\frac{c}{k}\left(\sqrt{1+(kt{)}^2}-1\right)$

$x=\frac{m{c}^2}{F}\left[\sqrt{1+{\left(\frac{Ft}{mc}\right)}^2}-1\right]$