Manned spacecraft to Tau problem

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

The solution for (a) is

And (b) is,

However, I'm confused by this calculation for (b). I get $\Delta x = 11.9~c~years$ and $\Delta ct = \frac{c\Delta x}{v} = \frac{11.9 c^2 years}{0.5 c} = \frac{11.9~c~years}{0.5}$

Does someone please know why they are missing the speed of light factor?

Thanks!

 

[haruspex]

Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 347062
The solution for (a) is
View attachment 347063
And (b) is,
View attachment 347064
However, I'm confused by this calculation for (b). I get $\Delta x = 11.9~c~years$ and $\Delta ct = \frac{c\Delta x}{v} = \frac{11.9 c^2 years}{0.5 c} = \frac{11.9~c~years}{0.5}$

Does someone please know why they are missing the speed of light factor?

Thanks!

$c\Delta x$ is multiplying by c a second time. What units would that give?

 

[member 731016]

$c\Delta x$ is multiplying by c a second time. What units would that give?

Thank you for your reply @haruspex!

Sorry I'm confused. The units for $c\Delta x$ should be m^2/s

Thanks!

 

[Orodruin]

The solution for (b) has the wrong units. $\Delta ct$ has units of length, not time (if one insists on using such unnatural units where $c\neq 1$)

Since c = 1 ly/year, 1 c year = 1 ly.

 

[Orodruin]

I will also add that (a) is not a very well formulated problem. “What spacetime interval passes on Earth?” Between what events?!?!?

 

[berkeman]

Plus, it's a trick question. If the spaceship is launched at v = 0.5c, the crew will not be needing any food on the journey...