Question

A meter stick is projected into space at so great a speed that its length appears contracted to only 50 cm. How fast is it going in m/s?

Answer 1

We assume that the meter stick is at rest in S'. As observed by stationary observers in S, the meter stick moves in the positive x-direction with speed v.

$$x'= γ(x − vt)$$

relates the position x' measured in S' with the position x measured in S.

Let ∆x' be the length of the meter stick measured by an observer at rest in S'. (∆x' is the proper length of the meter stick.)

The meter stick is moving with speed v along the x axis in S. To determine its length in S, the positions of the front and back of the meter stick are observed by two stationary observers in S at the same time. The length of the meter stick as measured in S is the distance ∆x between the two stationary observers at ∆t = 0.

Then

$$∆x ' = γ∆x$$

$$\beta=\sqrt{1-\frac{∆x }{∆x ' }}\\\frac{v }{3\cdot10^8}=\sqrt{1-\frac{0.5 }{1}}\\v=2.1\cdot10^8\frac{m}{s}$$