A person in a train moving with a speed of 60% that of the speed of light waves to a bystander. If the waving action lasts three seconds as measured by the bystander, how long did the person inside the train wave based on his own clock?
Answer:
Hope This Helps
Explanation:
To solve this, we'll use the concept of time dilation from special relativity. The time dilation formula is:
Δ
′
=
Δ
1
−
2
2
Δt
′
=
1−
c
2
v
2
Δt
Where:
Δ
′
Δt
′
is the time interval observed by the person in the train.
Δ
Δt is the time interval observed by the bystander.
v is the velocity of the train as a fraction of the speed of light.
c is the speed of light.
Given:
=
0.6
v=0.6c
Δ
=
3
Δt=3 seconds
We can plug these values into the formula:
Δ
′
=
3
1
−
0.
6
2
Δt
′
=
1−0.6
2
3
Δ
′
=
3
1
−
0.36
Δt
′
=
1−0.36
3
Δ
′
=
3
0.64
Δt
′
=
0.64
3
Δ
′
=
3
0.8
Δt
′
=
0.8
3
Δ
′
=
3.75
Δt
′
=3.75
So, the person inside the train waves for approximately 3.75 seconds based on their own clock.
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