A boy runs a 100meter distance race in 20seconds as measured by an observer on earth.how long does it take according to an observer in space ship moving with speed of 0.98c for the boy to complete the run? What is the length of the track according to the moving observer?
To answer the first part of your question, we need to apply the time dilation formula from special relativity. According to special relativity, the time experienced by an observer in a moving spaceship (relative to an observer at rest on Earth) is different. The formula for time dilation is:
t' = t / √(1 - (v^2 / c^2))
Where:
t' is the time experienced by the observer in the spaceship,
t is the time experienced by the observer on Earth,
v is the velocity of the spaceship relative to Earth, and
c is the speed of light, which is approximately 3.00 x 10^8 meters per second.
In this case, the velocity of the spaceship relative to Earth is given as 0.98c, where c is the speed of light. Plugging in the values:
t' = 20 / √(1 - (0.98c)^2 / c^2)
Simplifying the equation:
t' = 20 / √(1 - 0.98^2)
t' = 20 / √(1 - 0.9604)
t' = 20 / √0.0396
t' = 20 / 0.198
t' ≈ 101.01 seconds
So, according to an observer in the spaceship, it would take approximately 101.01 seconds for the boy to complete the run.
Now, to address the second part of your question, in special relativity, lengths can also appear different to moving observers due to length contraction. The formula for length contraction is:
L' = L * √(1 - (v^2 / c^2))
Where:
L' is the length observed by the moving observer,
L is the length measured by an observer at rest, and
v and c have the same meanings as before.
In this case, we have the length measured by the observer at rest on Earth, which is 100 meters. The velocity of the spaceship relative to Earth is still 0.98c. Plugging in the values:
L' = 100 * √(1 - (0.98c)^2 / c^2)
Simplifying the equation:
L' = 100 * √(1 - 0.96^2)
L' = 100 * √(1 - 0.9216)
L' = 100 * √0.0784
L' = 100 * 0.27926
L' ≈ 27.93 meters
Therefore, according to the moving observer on the spaceship, the length of the track would appear to be approximately 27.93 meters.