Suppose one twin takes a ride in a spaceship traveling at a very high speed to a distant star and back again, while the other twin remains on Earth. When the traveling twin returns, how do the ages of the twins compare

If the twin in the spaceship travels at a speed close to the speed of light, seen from Earth, the time of the traveling twin passes very slowly. Therefore, upon returning to Earth, less time has passed for him, compared to the twin who remains on Earth, that is, he is younger, while the twin who remains on Earth is older.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s