Constant acceleration from accelerated observer's perspective

[Whovian]

First, an unnecessary short introduction of why I'm asking this question in the first place. Due to some doubts about a certain plot device in a science-fiction novel, I'm trying to figure out how long it would take to accelerate someone to, say, 50% of the speed of light under constant acceleration from their own reference frame (for various reasons, I'm assuming about 10*g.) Due to this being a small ameteur project, I'll probably be using Wolfram Alpha.

In any case, I got the outrageously small answer of about 2.5 weeks, which doesn't sound even close to right to me. I think I've isolated my problem to how I formulate this in terms of acceleration as measured from Earth. If this bit's correct, yes, I'll post the rest of my work in a hope of others finding a minor mistake.

So, now for a formulation of the problem. Say Bob's in a spaceship and accelerating with acceleration $100\ \dfrac{\mathrm{m}}{\mathrm{s}^2}\approx 10\cdot g$ in his own reference frame. How long, measured in seconds from Earth, will it take for him to be moving at .5c with respect to Earth?

Now, after a large amount of squandering about with exhaust-based engines in which the exhaust has mass << the spaceship which resulted in very ugly expressions for momentum ($\dfrac{1\ \mathrm{m}^3\cdot\delta\cdot\frac{v-r}{1-\frac{v\cdot r}{c^2}}}{\sqrt{1-\left(\frac{\frac{v-r}{1-\frac{v\cdot r}{c^2}}}{c}\right)^2}}$, for instance,) I decided to stop and just come out with what my intuition told me the acceleration as measured from Earth is, which is $100\ \dfrac{\mathrm{m}}{\mathrm{s}^2}\cdot\sqrt{1-\left(\dfrac vc\right)^2}$. Solving this differential equation yields the nutters result of a periodic function. So, clearly, something's wrong with this expression. But what, exactly?

 

[Nugatory]

Google for "Rindler coordinates" - they were invented for exactly this situation.

 

[Bill_K]

The trajectory of an object undergoing constant proper acceleration is called hyperbolic motion. Because, I guess, it can be expressed in terms of hyperbolic functions. An equivalent but simpler result is

x = (1/a)√(1 + a²t²) - 1
v = dx/dt = at/√(1 + a²t²)

EDIT: Should point out that this is with c = 1, and sure enough, v → 1 as t → ∞.

 

[Whovian]

Okay, thanks, guys!

 

[sardar]

From a scientific perspective, it is important to clarify some of the assumptions and terminology used in this question. First, it is important to note that acceleration is not a constant in an accelerated reference frame, as the name may suggest. In fact, an accelerated reference frame experiences non-uniform motion, where the acceleration changes over time.

In this scenario, Bob is in an accelerated reference frame, meaning that his velocity is constantly changing. This is different from a constant acceleration, which would imply a constant change in velocity over time. Therefore, the concept of "constant acceleration from an accelerated observer's perspective" is not well-defined.

Additionally, the term "acceleration as measured from Earth" is confusing, as acceleration is a relative quantity and can only be measured in a specific reference frame. It is not possible to measure the acceleration of an object from a different reference frame, as it would depend on the observer's perspective.

To determine the time it would take for Bob to reach 50% of the speed of light, we would need to know the initial velocity and the rate of acceleration in his own reference frame. This information would allow us to calculate the time it would take for him to reach 50% of the speed of light, as measured in his own reference frame.

Overall, the problem formulation and assumptions in this question are not well-defined and would require further clarification in order to provide an accurate scientific response.