Question

Suppose a particle called a kaon is created by cosmic radiation striking the atmosphere. It moves by you at 0.980c, and it lives 1.24×10^(-8) s when at rest relative to an observer. How long does it live as you observe it?

a) 1.24×10^(-8) s
b) 2.00×10^(-8) s
c) 3.20×10^(-8) s
d) 5.08×10^(-8) s

Answer 1

The kaon particle would live for 5.08 × 10^(-8) s as you observe it.

Step-by-step explanation:

The time dilation formula in special relativity states that the observed time interval Δt' experienced by an observer in a moving frame is related to the time interval Δt in the rest frame by the equation:

Δt' = Δt / √(1 - (v^2 / c^2))

In this case, the kaon particle is moving at a velocity of 0.980c, where c is the speed of light.

The particle lives for 1.24 × 10^(-8) s in its rest frame.

Plugging in the values into the time dilation formula, we can calculate the observed time interval:

Δt' = (1.24 × 10^(-8) s) / √(1 - (0.980^2)) = 5.08 × 10^(-8) s

Therefore, the kaon particle would live for 5.08 × 10^(-8) s as you observe it.