How do measurements of time differ for events in a frame of reference that moves at 50% of the speed of light relative to us? At 99.5% of the speed of light relative to us?

Respuesta :

Answer with explanation:

Relation between Speed , Distance and time

Distance =Speed × Time

It means Speed is inversely Proportional to time.

As distance will remain constant , in that frame of reference

If speed of light in a Medium

                              [tex]=s=3 \times 10^8 \frac{\text{meter}}{\text{second}}[/tex]

Then, time taken to cross the medium = t seconds or hours or another unit of time.

Now, If speed of light is 50% of the speed of light relative to us

    That is Speed of light in another medium

                 [tex]=w=\frac{50}{100} \times 3 \times 10^8\\\\=1.5 \times 10^8 \frac{\text{meter}}{\text{second}}[/tex]

Then, time taken to cross the medium = 2t seconds or hours or another unit of time.

Using Unitary Method

  [tex]\rightarrow v_{1}\times t_{1}=v_{2} \times t_{2}\\\\1.\rightarrow 3 \times 10^8 \times t=\frac{50}{100} \times 3 \times 10^8 \times t_{1}\\\\t_{1}=2t\\\\2.\rightarrow 3 \times 10^8 \times t=\frac{99.5}{100} \times 3 \times 10^8 \times t_{2}\\\\t_{2}=\frac{1000t}{995}\\\\t_{2}=\frac{200t}{199}[/tex]

             

Answer:

1.154 times the proper time

10.013 times the proper time

Step-by-step explanation:

Speed of light = c

At 50% speed of light

v = 0.5c

Time dilation

[tex]\Delta t'=\frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{\Delta t}{\sqrt{1-\frac{0.5^2c^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{\Delta t}{\sqrt{1-0.25}}\\\Rightarrow \Delta t'=\frac{\Delta t}{0.866}\\\Rightarrow \Delta t'=1.154\Delta t[/tex]

Time would differ by 1.154 times the proper time

At 99.5% speed of light

v = 0.995c

Time dilation

[tex]\Delta t'=\frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{\Delta t}{\sqrt{1-\frac{0.995^2c^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{\Delta t}{\sqrt{1-0.990025}}\\\Rightarrow \Delta t'=\frac{\Delta t}{0.09987}}\\\Rightarrow \Delta t'=10.013\Delta t[/tex]

Time would differ by 10.013 times the proper time.

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