(II) Reference frame S' moves at speed v = 0.88c in the +x direction with respect to reference frame S. The origins of S and S' overlap at t = t' = 0. An object is stationary in S' at position x' = 100m . What is the position of the object in S when the clock in S reads 1.00 μs according to the (a) Galilean and (b) Lorentz transformation equations?

Verified step by step guidance

1. Identify the given values: v = 0.88c, x' = 100 m, t' = 0, and t = 1.00 \u03bcs. Note that c, the speed of light, is approximately 3.00 \u00d7 10^8 m/s.
2. Calculate the time in S' when t = 1.00 \u03bcs in S using the Lorentz transformation for time: t' = \gamma (t - vx/c^2), where \gamma = 1 / \u221a(1 - v^2/c^2). First, calculate \gamma: \gamma = 1 / \u221a(1 - (0.88c)^2/c^2) = 1 / \u221a(1 - 0.7744) = 2.29.
3. Substitute the values into the time transformation equation: t' = 2.29 (1.00 \u00d7 10^{-6} s - (0.88 \u00d7 3.00 \u00d7 10^8 m/s)(1.00 \u00d7 10^{-6} s)/3.00 \u00d7 10^8 m/s) = 2.29 \u00d7 (1.00 \u00d7 10^{-6} s - 0.88 \u00d7 10^{-6} s) = 0.276 \u00d7 10^{-6} s.
4. For the Galilean transformation, the position in S can be calculated using x = x' + vt. Substitute the values: x = 100 m + (0.88 \u00d7 3.00 \u00d7 10^8 m/s)(1.00 \u00d7 10^{-6} s) = 100 m + 264 m = 364 m.
5. For the Lorentz transformation, use the position transformation equation x = \gamma (x' + vt'). Substitute the values: x = 2.29 (100 m + (0.88 \u00d7 3.00 \u00d7 10^8 m/s)(0.276 \u00d7 10^{-6} s)) = 2.29 (100 m + 72.768 m) = 395.44 m.