Question

A young lady of 25 years starts running with velocity 0.8c.After 5 years,her age would appear to her stationary friend?​

Answer 1

the answer is 129 i suppose

i am so not sure

Answer 2

Answer:

The age appearing to the lady's stationary friend is $28$ $years$.

Explanation:

• When a body moves with a very high velocity comparable to the speed of light, time moves slowly. This is the theory of relativity.
• The person's age who is moving at a very high speed compared to the speed of light appears to be less than the actual age.

The formula for the time interval:

$T=\frac{To}{\sqrt{1-\frac{v^{2} }{c^{2} } } }$

where, $To$ is the original time interval.

            $v$ is the object velocity.

            $c$ is the light's speed.

Step 1:

As the time interval becomes larger, we have to calculate the factor by which the time interval increases.

The factor $=\sqrt{1-\frac{(0.8c)^{2} }{c^{2} } }$

                 $=0.6$

Step 2:

Due to an increase in the time interval, the time moves slowly.

The age that appears to the lady's stationary friend is

$=25 + (0.6\times5)$ $years$

$=28$ $years$

So, the age that appears to the lady's stationary friend is $28$ $years$.