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Suppose that, at t = 9.00 × 10 − 4 s, the space coordinates of a particle are x = 135 m, y = 30.0 m, and z = 55.0 m according to coordinate system S . If reference frame S ′ moves at speed 1.50 × 10 5 m/s in the + x - direction relative to frame S , compute the corresponding coordinate values as measured in frame S ′ . The reference frames start together, with their origins coincident at t = 0 .

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Answer:

[tex]x'=134.999983\ m[/tex]

[tex]y'=y=30\ m[/tex]

[tex]z'=z=55\ m[/tex]

Explanation:

Given:

  • time, [tex]t=9\times 10^{-4}\ s[/tex]
  • x coordinates of a particle, [tex]x=135\ m[/tex]
  • y coordinates of a particle, [tex]y=30\ m[/tex]
  • z coordinates of a particle, [tex]x=55\ m[/tex]
  • Relative Speed of frame of reference S' in the +x direction, [tex]v_x=1.5\times 10^{5} m.s^{-1}[/tex]
  • Since the speed of the frame S' is comparable to the speed of light in vacuum therefore the observer from S' frame will observe a contracted length of the dimensions in the direction of motion.

Now from the equation of length contraction:

[tex]x'=x\times \sqrt{1-\frac{v_x^2}{c^2} }[/tex]

[tex]x'=135\times \sqrt{1-\frac{(1.5\times 10^{5})^2}{(3\times 10^8)^2} }[/tex]

[tex]x'=134.999983\ m[/tex]

Rest other values will remain unaffected since they are along the axis of motion. So,

[tex]y'=y=30\ m[/tex]

[tex]z'=z=55\ m[/tex]

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