Answer to Question #183079 in Mechanics | Relativity for mary

Question

Two inertial systems are uniformly separating at a speed of exactly √0.84𝑐. In one system a jogger runs a mile (1609m) 

in 6 min along the axis of relative motion. How far in meters does he run and how long does it take, to observers in the other 

system?

Answer 1

To be given in question

Velocity (v)=√(0.84c)

$L_{0}=1609 meter$

$Z_{0}$ = 6 minutes

To be asked in question

Distance run=?

Times run=?

We know that

$L=L_{0}\sqrt{1-\frac{v^2}{c^2}}\rightarrow(1)$

$Z=\frac{Z_{0}}{\sqrt{1-\frac{v^2}{c^2}}}\rightarrow(2)$

Eqution (1) put values$L=1609\times\sqrt{1-\frac{\sqrt{.84c}^2}{c^2}}$

$L=1609\times\sqrt{.16}$

$L=1609\times0.4$

$L=643.6meter$

Time

Eqution (2)put values

$Z= \frac{360}{\sqrt{1-\frac{v^2}{c^2}}}$

Put v=√(.84c)

$Z=\frac{360}{0.4}$

Z=900 sec

$Z=\frac{900}{60}$ =15 minutes

Z=15minutes after and distance L=643.6meter after observer is other systems $∆d=1609-643.9$

Separation of speed =

$\frac{distance}{time}\rightarrow equation (1)$

Put values

$Speed=\frac{965.1}{900}=1.07mete/sec$

$Speed$ separation of distance $∆d=d_{1}-d_{2}\rightarrow(2)$

Put valuesd₁=1609meter

d₂=643.6meter equation (2) put values

∆d=965.1meter

Separation distance 965.1meter

Separation Speed=1.072meter/sec