Solving for Unknown Variables in Acceleration-Based Time Travel Equations

[DaleSwanson]

I wasn't sure if this should go here, but I guess it falls under independent study.

Homework Statement

Ok, so I'm making a javascript calculator to find the travel time if you accelerate at 1 g. The point of this was to help me learn JS and a bit of physics and math. The possible variables are:
a - acceleration
T - ship time
t - Earth time
d - distance
v - final speed

I wanted to make it so that you could provide whichever variable you knew and it would solve for the rest. So far this has gone well, but I'm having trouble getting the other variables from velocity alone.

Homework Equations

I've been using the equations from this page:
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
In particular I was trying to start with this one and solve for t:
$$v = \frac{at}{\sqrt{1 + (at/c)^{2}}}$$

The Attempt at a Solution

I've spent quite a while trying to solve for t and this is as far as I've gotten:
t² - c^-2t²v² = a^-2v²
I wanted to get rid of all those exponents by finding the square root of both sides, but apparently I can't do this (I've been plugging in known variables and seeing if the equation is still correct). At first I thought this was just because of the minus, which would become i if I square rooted it. However, I then tried x² + y² = 41 where x = 4, y = 5. Simply finding the square root of both sides gave 4 + 5 = 6.4, so I guess you can't do square root when there is addition in the equation at all.

Is there any way to get t by itself here?

 

[Pinu7]

Factor out t-squared.

Remember, when in doubt, factor!

 

[Mentallic]

Just to add to what pinu7 said:

I'm assuming you already know how to solve simple equations, such as multiplying/dividing and adding/subtracting.

Try squaring both sides to rid yourself of that pesky square root. Then manipulate the equation to secondly rid yourself of that pesky denominator. Expand and re-arrange to have all t on one side. Now here is where pinu7's advice comes in. If you made it this far, it should be quite simple to solve for t. Remember to place restrictions on t since you squared the equation earlier on.

 

[DaleSwanson]

Ok, well I started over and did this:

Square to get rid of root:
$$v^{2} = \frac{at}{\sqrt{1 + (at/c)^{2}}}^{2}$$

$$v^{2} = \frac{a^{2}t^{2}}{1 + (a^{2}t^{2}/c^{2})}$$

Multiply to get rid of fraction:
$$v^{2} * \frac{1 + (a^{2}t^{2}/c^{2})}{1} = \frac{a^{2}t^{2}}{1 + (a^{2}t^{2}/c^{2})} * \frac{1 + (a^{2}t^{2}/c^{2})}{1}$$

$$v^{2} * (1 + \frac{a^{2}t^{2}}{c^{2}}) = a^{2}t^{2}$$

$$v^{2} + \frac{v^{2}a^{2}t^{2}}{c^{2}} = a^{2}t^{2}$$

Multiply to get rid of fraction:
$$\frac{v^{2}}{1} * + \frac{v^{2}a^{2}t^{2}}{c^{2}} = a^{2}t^{2}$$

$$v^{2}c^{2} + v^{2}a^{2}t^{2} = a^{2}t^{2}c^{2}$$

Shuffle some things around and:
$$t^{2} = \frac{c^{2}}{a^{2}c^{2}/v^{2} - a^{2}}$$

Thanks for the help.

 

[Mentallic]

No probs. But I just set the directions, you found your way pretty nicely

Just a note: when you take the root of both sides to make t the subject, there are various ways to express the equation, but it would be best if you could manipulate it to look something like your original expression (there should be a 1 in the surd). It isn't necessary though, just... uh... nice? (lol yeah it sounds kinda fruity )