Question

Exercise 117 Gives the speed u of an object accelerated under a constant force. Show that the distance it travels is given by$x=\frac{m{c}^2}{F}\left[\sqrt{1+{\left(\frac{Ft}{mc}\right)}^2-1}\right]$.

Short answer

Answer:

Distance travelled by an object under constant force is determined byintegrating 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 $u=\frac{1}{\sqrt{1+{\left({\displaystyle \frac{Ft}{mc}}\right)}^2}}\frac{F}{m}t$

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

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

$dx=\frac{ckt}{\sqrt{1+{\left(kt\right)}^2}}dt$.

Integrate the above expression

Let’s say$m=1+k^2{t}^2\Rightarrow dm=k^2\left(2tdt\right)\Rightarrow tdt=\frac{dm}{2k^2}$ Therefore,

$$\begin{gathered} dx=\frac{c}{2k\sqrt{m}dm} \\ \int dx=\frac{c}{2k}\int \frac{1}{\sqrt{m}}dm \\ x=\frac{c}{2k}\left[\frac{\sqrt{m}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]+C \\ x=\frac{c}{k}{\left(1+k^2{t}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C \end{gathered}$$

At $t=0,x=0.$Therefore, the integration constant $C=-\frac{c}{k}$.

$$\begin{gathered} x=\frac{c}{k}{\left(1+k^2{t}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{c}{k} \\ x=\frac{c}{k}\left(\sqrt{1+{\left(kt\right)}^2}-1\right) \\ x=\frac{m{c}^2}{F}\left[\sqrt{1+{\left(\frac{Ft}{mc}\right)}^2}-1\right]. \end{gathered}$$