Question

Romeo departs from Earth in a spaceship to marry Juliet, who lives on the planet Lovelon, ten light-years away from Earth. He accelerates at 15m/s^2 and must decelerate at the same rate, in order to stop at Lovelon. However, Juliet has taken an earlier shuttle back to Earth and she will wait there. Since there is no communication, Romeo will have to travel back again. Romeo left Tuesday, September 24, 2013.

What is the earliest year in which they can be wed?
(Romeo and Juliet are both classical persons, and they ignore any relative effects. 1 light-year is about 9.46x10^15m.)

Answer 1

hi,
this can be solved using a (v-t) time graph..

But before that there r some key concepts to solve this one.
* The question says that the shuttle must accelerate and decelerate at the same rate.
* and the shuttle starts from rest and shuttle must come to rest at planet lovelon
* so the shuttle must be completely at the rest when it arrives lovelon
* Also this must happen in the shortest possible time which means shuttle must accelerate as much as it can and should gain the highest velocity....
but it can't keep accelerating till it reach lovlon 'cause it has to rest at lovelon
* and the return of the shuttle should be exactly the same.. so if we could find the time that the shuttle will take to reach lovelon we just have to multiply it by 2 in order to get the total time that shuttle will take tot travel to lovelon and return earth

The motion of the shuttle from earth to lovelon will be like this...


Now the gradient of the graph is acceleration... which means
tan (alpha) = v/t
15 = v/t
v = maximum velocity
t = time
v = 15t

Now the distance travel = Area of the triangle in the graph
( 9.46x10^15m).*10 =2(vt)/2

let's substitute the value v = 15t in the above expression
94.6x10^15 = (15t^2)
94.6x10^15 = 15t^2
t^2 = (94.6x10^15)/15
= 6.30x10^15
t = 2.51x10^15

so the total time to reach lovelon is 2t(according to the graph)
T = 5.02x10^15
And the total time(go to lovelon and return earth)
2T
=10.04x10^15 s

to turn that in to years
10.04x10^15/60*60*24*365
=10.04x10^15/31536000
=0.032x10^10
= 32x10^7 years

Yep that's it.... but I havent use the cal hear to simplify the answer ... so u have to understand the concept and simplify it by urself if u want answer correct to more than 2 decimal places....

Anywayyy ... Hope this will help ya!!!!