Aspaceship is built to reach a star system 15 light-years away.

(a) how fast must the spaceship move (in units of c) so that the round trip time experienced by the crew is 20 years?

(b) how much time would the round trip take as measured by observers on earth?

a) 0.72c  b) 57.63 years

Explanation:

First you need to know that 1 light year is equivalent to 9.46*10^15m, so 15 light years are 1.374*10^17 m. Now 20 years are equivalent to (20)*(365)*(24)*(60)*(60) seconds = 6.3*10^8 seconds. So dividing the distance by the time you have 2.18*10^8 m/s. And you know that c = 3*10^8, so dividing the velocity by c, you have that the spaceship travels at 0.72c. With the dilation time factor you can calculate (b), the equation is t/sqrt(1-(v^2/c^2)). where t is the time (40 years) and v the velocity (0.72c), so substituting you have 57.63 years, that is the time observed in the earth.

with 5 seconds, no, but over time if the police car stays at 65m/s ad the speeding car stays at 60, then obviously yes, but the answer to your question would be, no.

hope this : )

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