Here's a simple math problem/complex thought experiment:

If you've learned and retained basic calculus, this will come easily. If not, try to think through it (incorrect answers are better than none).

There are two stations, and two trains. They will be traveling the same distance.

Train A is travelling through station 1 at the same time B is leaving. It does not accelerate, but cruises through at a constant speed to station 2.
( v = c, a=0)

Train B leaves station 1, accelerates uniformly (no jerk or friction to account for), and as it arrives at station 2 its velocity is exactly twice that of Train A.
(v_i = 0, a = ?, v_f =2c)

Who arrives at the station first?

Ah, before the edit train B had .5c, gimme a sec.

Ah, before the edit train B had .5c, gimme a sec.

Yeah I accidentally had it backwards. Good solve.

The hardest math I've done on a regular basis since the eleventh grade is simply converting dimensions and three digit subtraction... once or twice I've had to do long division... so I'm trying my damnest right now to remember the basic physics shit I learned in freshmen year because right now I don't think ballistics and properties of flight are helping me.

But so, if train b's acceleration is constant through the whole time, stations 1 and 2 are endpoints and at station 2 b is traveling at twice A's speed, that means that train B hits A's speed midway through, so really we're trying to figure out how fast each train completes its relative segments and if train b's second half makes up for the slower start enough for it to overtake.

Stop me here if I'm already going about this entirely wrong-like.

Quote from: Azendac on December 07, 2016, 10:08:59 PM

Ah, before the edit train B had .5c, gimme a sec.

Yeah I accidentally had it backwards. Good solve.

Just as an addendum, the kinematic equations can be easily derived using calculus, it's just two integrals. But they can also be derived from simple algebra manipulations of v = d/t, averageV = (Vf + Vi)/2, a = v/t

I know this because I asked my phsycis teachers several times to show me how to derive them and kept forgetting until I sat down and worked it out myself. interestingly, most sources only provide 4 kinematic equations:

 and that's what we had to work with even at university, but a little combinatorics (Binomial coefficient, "5 choose 4") will show that there's a fifth equation:

You don't really need the fifth, but it's nice to know you can use mathematical techniques to improve your knowledge of maths.

It might interest you to see that this problem is actually just a statement of the mean speed theorum.

is this basically suvat? because i've forgotten it.