Question:

A rocket (S') moves at a speed c2\frac{c}{2}m/s along the positive x-axis, where c is the speed of light. When it crosses the origin, the clocks attached to the rocket and the one with a stationary observer (S) located at x = 0 are both set to zero. If S observes an event at (x, t), the same event occurs in the S' frame at

Updated On: Oct 1, 2024
  • x=23(xct2) and t=23(tx2c)x'=\frac{2}{\sqrt3}(x-\frac{ct}{2})\ \text{and}\ t'=\frac{2}{\sqrt3}(t-\frac{x}{2c})
  • x=23(x+ct2) and t=23(tx2c)x'=\frac{2}{\sqrt3}(x+\frac{ct}{2})\ \text{and}\ t'=\frac{2}{\sqrt3}(t-\frac{x}{2c})
  • x=23(xct2) and t=23(t+x2c)x'=\frac{2}{\sqrt3}(x-\frac{ct}{2})\ \text{and}\ t'=\frac{2}{\sqrt3}(t+\frac{x}{2c})
  • x=23(x+ct2) and t=23(t+x2c)x'=\frac{2}{\sqrt3}(x+\frac{ct}{2})\ \text{and}\ t'=\frac{2}{\sqrt3}(t+\frac{x}{2c})
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The Correct Option is A

Solution and Explanation

The correct answer is (A) : x=23(xct2) and t=23(tx2c)x'=\frac{2}{\sqrt3}(x-\frac{ct}{2})\ \text{and}\ t'=\frac{2}{\sqrt3}(t-\frac{x}{2c})
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