Chapter 2: Q 119CE (page 71)

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

Short Answer

Expert verified

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 $u = \frac{1}{\sqrt{1 + {(\frac{F t}{m c})}^{2}}} \frac{F}{m} t$

Here, velocity $u = \frac{d x}{d t}$ and let $k = \frac{F}{m c}$

$\frac{d x}{d t} = \frac{c}{\sqrt{1 + {(k t)}^{2}} k t}$

$d x = \frac{c k t}{\sqrt{1 + {(k t)}^{2}}} d t$ .

Integrate the above expression

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

$d x = \frac{c}{2 k \sqrt{m} d m} \int d x = \frac{c}{2 k} \int \frac{1}{\sqrt{m}} d m x = \frac{c}{2 k} [\frac{\sqrt{m}}{\frac{1}{2}}] + C x = \frac{c}{k} {(1 + k^{2} t^{2})}^{\frac{1}{2}} + C$

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

$x = \frac{c}{k} {(1 + k^{2} t^{2})}^{\frac{1}{2}} - \frac{c}{k} x = \frac{c}{k} (\sqrt{1 + {(k t)}^{2}} - 1) x = \frac{m c^{2}}{F} [\sqrt{1 + {(\frac{F t}{m c})}^{2}} - 1] .$

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

Short Answer

Step by step solution

Determine the expression for the differential change in distance dx.

Integrate the above expression

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Exercise 117 Gives the speed u of an object accelerated under a constant force. Show that the distance it travels is given byx=mc2F1+Ftmc2-1.

Short Answer

Step by step solution

Determine the expression for the differential change in distance dx.

Integrate the above expression

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Rotational Dynamics

Radiation

Kinematics Physics

Magnetism

Turning Points in Physics

Magnetism and Electromagnetic Induction

Study anywhere. Anytime. Across all devices.

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