Question

Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship journey that takes T years, as measured by a clock in each spaceship. During the journey they travel at a constant speed v, as measured on earth, except during the relatively short acceleration phases of their journey.
Part A:

Rank these quintuplets on the basis of the year on earth when they return from their journey.
Rank from largest to smallest. To rank items as equivalent, overlap them.

A: T=20years, v=0.4c
B: T= 5years, v=0.2c
C: T=10years, v=0.8c
D: T=10years, v=0.4c
E: T=20years, v=0.8c

Part B:

Rank these quintuplets on the basis of their age when they return from their journey.
Rank from largest to smallest. To rank items as equivalent, overlap them.

A: T=20years, v=0.4c
B: T= 5years, v=0.2c
C: T=10years, v=0.8c
D: T=10years, v=0.4c
E: T=20years, v=0.8c

Answer 1

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = $\frac{t}{\sqrt{1-\frac{v^2}{c^2} } }$

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

$\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

$\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }$

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

$\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }$

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

$\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

$\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }$

=1.67 x 20

t₁= 33.4   years