Question #183079

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?


Expert's answer

To be given in question

Velocity (v)=√(0.84c)

L0=1609meterL_{0}=1609 meter

Z0Z_{0} = 6 minutes

To be asked in question

Distance run=?

Times run=?

We know that

L=L01−v2c2→(1)L=L_{0}\sqrt{1-\frac{v^2}{c^2}}\rightarrow(1)

Z=Z01−v2c2→(2)Z=\frac{Z_{0}}{\sqrt{1-\frac{v^2}{c^2}}}\rightarrow(2)

Eqution (1) put valuesL=1609×1−.84c2c2L=1609\times\sqrt{1-\frac{\sqrt{.84c}^2}{c^2}}

L=1609×.16L=1609\times\sqrt{.16}

L=1609×0.4L=1609\times0.4

L=643.6meterL=643.6meter

Time

Eqution (2)put values

Z=3601−v2c2Z= \frac{360}{\sqrt{1-\frac{v^2}{c^2}}}

Put v=√(.84c)

Z=3600.4Z=\frac{360}{0.4}

Z=900 sec

Z=90060Z=\frac{900}{60} =15 minutes

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

Separation of speed =

distancetime→equation(1)\frac{distance}{time}\rightarrow equation (1)


Put values

Speed=965.1900=1.07mete/secSpeed=\frac{965.1}{900}=1.07mete/sec

SpeedSpeed separation of distance ∆d=d1−d2→(2)∆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



LATEST TUTORIALS
APPROVED BY CLIENTS