How to calculate my ETA ,at speed of warp9,distance 5light yrs.,theoretical ?
372609
warp speed is supposed to be a x a x a x speed of light or in otherwords a cubed. so warp 9 would be 9x9x9 x lightspeed or 729 x lightspeed so time would be 5 years divided by 729 = 2.5 days approx
977845
Put on a number plane with B initially at O (0. 0)\nTherefore A is initially at (-150, 0)\n\nNow A's position, t hours after midday, is (-150 + 35t, 0) and B's position is (0, 25t)\n\nSo t hours after midday \n\n|AB| = √[((-150 + 35t) - 0)² + (0 - 25t)²]\n= √[(22500 - 10500t + 1225t² + 625t²)\n= √(22500 - 10500t + 1850t²) \n\nVelocity of change = d(|AB|)/dt\n= 0.5 * (3700t - 10500) / √(22500 - 10500t + 1850t²)\n= 0.5 * (3700 * 4 - 10500) / √(22500 - 10500 * 4 + 1850 * 4²)\n= 0.5 * 4300 / √(10100)\n≈ 21.4 km/h
481716
distance = 160miles\nave. speed = 50mph\n\n160/50 = 3.2 hours or 3hours and 12 minutes\n\nfor the guy to arrive at exactly 11:30am\n\n(11:30) - (3hours12minutes) = 8:18am\n\nANSWER: 8:18am
42921
optimum angle for distance is 45 degrees \nspeed: fucking fast
735157
let x be the distance AP and let Y be the angle <PLA.\n\nThen x=3tanY then differentiate implicitly to time\n\ndx/dt= 3sec^2Y dY/dt\n\nNow dY/dt= 4 rev/min * 2pi radians/rev= 8pi radians/min\n\nWhen x=1, we have tanY= 1/3 sec^2Y=1+tan^2Y= 10/9\n\nThus dx/dt=3*10/9*8pi= 80/3 pi which approximately equals\n\n83.8\n\nThe spot of light is moving along the shoreline at a rate of about \n\n83.8km/min when it is1km from p.\n\n*It is essential to convert revolutions to radians
886238
Remember the third law:  P^2=ka^3, where k simply is telling you to use consistent units (e.g. Earth years and a.u.)  So the square of 5[years]=cube of distance in a.u., or cube root of 25 is the distance you want, expressed in astronomical units.  I'll let you finish this, but as a hint, it falls in the outer part of the asteroid belt (Ceres has a period of 2.72 years).
540450
well, this really isn't a question about astronomy, but imagine the trip were possible, and you could travel as close to lightspeed, c, as you chose.  special relativity shows that you could choose how long the trip would take by choosing your speed.  the distance of the trip will shrink more and more the faster you choose to go.  if the trip were to cross the galaxy (100 000 ly) and you traveled at:\n0.9999c (9 x 10e-4) the distance would look like 1000 ly and take 1000.10001 years.\n0.999999c (9 x 10e-6) the distance would look like 100 ly and take 100.0001 years\n0.99999999c (9 x 10e-8) the distance would look like 10 ly and take 10.0000001 years\n\nthis is actually observed in nature.  high speed cosmic atomic nuclei are continuously raining down on earth, and sometimes one strikes the nucleus of an atom in the upper atmosphere and produces a muon.  muons are very short-lived and shouldn't have time enough to reach us down here before decaying, but they do.  the muons are traveling so fast that they see the distance between the upper atmosphere where they originated to here as much shorter than we do so they have enough time to travel the short distance before decaying.\n\nlook here:\nhttp://en.wikipedia.org/wiki/Special_relativity
725163
Ok, first of all the jet would stop working once you left atmosphere - but ignoring that for a moment and assuming you were in a craft capable of going 500mph:-\n\nWell at about 2.7 billion miles away, approximately 616 years.\n\n\n\n**update** Phew, so pleased to see that someone else concurred (+/-) 1 year, I suddenly thought I might have been out by a factor of 10 :)
797830
Keep in mind these equations\n\nx(t) = x-initial + (v-initial)t + (1/2)(a-gravity)t^2\nv(t) = v-initial + (a-gravity)t\n\n(v-final)^2 = (v-initial)^2 + 2g(x-final - x-initial)\n\na-gravity = -9.8 m/s = -32 ft/s (i think..)\n\nTherefore\n\nx(t) = 0 + (64)t - (16)t^2\nx(t) = -16t^2 + 64t\n\n(v-final)^2 = (v-initial)^2 + 2g(0)\n(v-final)^2 = (v-initial)^2\n\nsince it is going in the opposite direction,\nv-final = -(v-initial)
987748
1.Compute initial velocity V0\nUsing equation S=V0*t-(g*t^2)/2 we find that the initial velocity V0=35/3 + 3g/2\n2.What ever goes up must come down V0*t=-(g*t^2)/2 t=2*V0/g\n3.Total time travel = 2t \nYou to the math\n\nBest\n\nHint:  the arrow will stop midair in about 5.3 seconds
95075
Let's see... \nIf you accept the Nemesis theory, the outer "boundary" is a good 2.5 light years out from the center,\nthat makes the circumference around 20 ly,\nlight travels roughly 386000 miles per second,\n31536000 seconds per year,\nmaking 1 ly about 12172896000000 miles,\nfor a total circumference of 243457920000000 miles.\n\nAnything else?
899910
Speed(S) = distance(D) / time(t) .... the basic equation \n\n45 = D / t \nthus,  D = 45t\n\nthey are already 50 miles away from home \n \nso 50 miles should be added to the d that we have found out above \n\nd = 50 + 45t
535160
what do you want to calculate? gravitation? emission? something else?
283146
you could make it in one second or beyond 12 hours depending on where you are, excluding flight delays at al. it should not actually take long enough to measure in days, unless you are travelling by boat/ship
1033497
Study trignometry and use the inclined plane and measure the radians and then determine the distance from each other by either using measuring instruments or if you want the distance of a planetary or celestial object - then send a laser towards the object and wait for it to bounce back and measure the time and the distance will be there before you - and then knowing the angle of inclination - you will be able to measure the distance - but taking shortcuts like this one will not get you that 98% in the examns - because there is no one to ask there - so open the book and read the trignometry and coordinate geometry and you will be giving answers all your life instead of asking them - &#92;This is written with a smile - and I hated maths and these 'orrible subject when I was in school my self ..but what to do - one has to learn to live with them ... and be sure you will write and tell me the answer in a day or two yourself\nBest of luck
636559
TNG Tech Manual lists it at 656 times the speed of light. This would also be the same for DS9 and Voyager, as these occur in approximately the same time frame. There is some as-yet-indeterminate algorithim to determine the difference in speeds; some say it's logarithmic, but I haven't seen it.\n\nTOS was much easier. According to the orig. ST tech manual, warp speed = (factor)-cubed * speed of light. Warp 1 = 1c (1^3), Warp 2 = 8c (2^3), Warp 3 = 27c (3^3), etc...
29063
http://www.eskimo.com/~jbm/calculations/traj/traj.html click this link it'll calc the answer for you