Question

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 .

Answer 1

`x'=134.999983\ m`

`y'=y=30\ m`

`z'=z=55\ m`

Step-by-step explanation:

Given:

• time,
  `t=9* 10^(-4)\ s`

• x coordinates of a particle,
  `x=135\ m`

• y coordinates of a particle,
  `y=30\ m`

• z coordinates of a particle,
  `x=55\ m`

• Relative Speed of frame of reference S' in the +x direction,
  `v_x=1.5* 10^(5) m.s^(-1)`

• 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:

`x'=x* \sqrt{1-(v_x^2)/(c^2) }`

`x'=135* \sqrt{1-((1.5* 10^(5))^2)/((3* 10^8)^2) }`

`x'=134.999983\ m`

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

`y'=y=30\ m`

`z'=z=55\ m`