Time dilation equations?
465096
From memory (so you should double check it):\n\nRa = Rb Sqrt(1-(V/c)^2) \n\n^2 = squared\n\nRa=dilated time\nRb=reference time\nV= velocity\nc=speed of light\n\nSo this is one of the problems with luminal or superluminal travel. If v=c then Ra=0 no matter what Rb is.\n\nAnd if v>c then you get an imaginary number... what could that mean?
177628
Actually, yeah.\n\nAlright, 600mph is 965,606 meters/hr. 965606 meters/hr is 268.2 meters/sec. This is 8.9408e-5% of the speed of light (3e+8 m/s). Following the time dilation formula, for each 12hr period, you'd gain 0.0193104 seconds.\n\nDo this every day for a year and you gain 7.0531 seconds. Do this for 50 years straight and you've just prolonged your life by 5.8776 minutes. Congratulations!\n\nIn case you're wondering, the time dilation formula is:\n\nT = To / sqrt(1-v^2/c^2)\n\nwhere T is the "time that changes" in seconds, To is the "normal" time in seconds, v is the velocity in m/s, and c is the speed of light in a vacuum in m/s.
865054
Let u=ln t, e^u=t, du=dt/t, dt=t du\nThen:\n∫ln² t dt = ∫t u² du = ∫u² e^u du\nIntegrating by parts:\nu² e^u - ∫2u e^u du\nu² e^u - 2u e^u + ∫2e^u du\nu² e^u - 2u e^u + 2e^u + C\nt ln² t - 2t ln t + 2t +C\nAnd we are done.
804155
a) T = 0.5 sec\n\nb) freq = 1/T = 2 Hz\n\nc) w = 2pi*freq = 4 pi\n\nd) w^2 = k/m\nk = m*w^2 = 0.5*16 pi^2 = 8*pi^2\n\ne)  V_max = A*2*pi*f = 0.35*2*pi*2 = 1.4 pi m/sec\n\nf) A_max = A*(2*pi*f)^2 = 0.35*(2*pi*2)^2 = 5.6*pi^2 m/sec^2
957344
t=2.512 seconds\n\n10.3=0.5(0+8.2)t\nTherefore \n\n1 = 4.1t/10.3\n\n1/t = 4.1/10.3\n\nt = 10.3/4.1 = 2.512 seconds
1009103
We have a formula:\nSpeed = Distance / Time\nhere Time= Distance /Speed.==D/S\nD=100km in both cases\nS1= 125km/hr; S2= 80km/hr\n\nT1=D/S1==100/125 hrs === 4/5hrs===4/5*60mins= 48mins\nT2=D/S2==100/80 hrs ===5/4 hrs ===5/4*60 mins= 75mins\n\nTime saved = T2-T1==75-48=27 mins.\n\nIs this clear buddy
478107
It is called the equation of time. It isn't really statistics though.
385849
i was gonna suggest my physics book. it dedicates one chapter to relativity and Lorentz transformation equations are there.\nIt's University Physics by Young and Freedman.
299439
You've confused the statement of the theory. Let two observers be in the same inertial reference frame with two synchronized watches. One observer accelerates in a negligibly small amount of time to some fraction of the speed of light relative to the other observer. The latter observer will be called O1 and the moving observer O2. \nO2 will measure the time in 01's reference frame to be going *slower* than the time in his frame. Likewise, O1 will measure the passage of time in 02's frame to be going *slower* than in 01's frame (the half-life of particles "increases" depending on their speeds). In other words, given two events A and B that always occur t seconds apart in O, they will occur at some separation T > t when measured from O'. The same will occur if O' is switched with O in the sentence above.\nQuestions like this usually come from confusing the relativity of simultaneity. Remember that in relativity, there are no two events that are simultaneous with respect to all inertial frames. \nI recommend you check out the text "Spacetime Physics" by Wheeler et al.
176611
Time standing still at the speed of light is theoretical.  Consider, however, it takes nearly nine minutes for light to travel here from the sun.  Therefore, if something travelling at the speed of light experiences no aging, it's purely relative.
236771
Try this link - it's Relativity in very short words.\n\nhttp://www.muppetlabs.com/~breadbox/txt/al.html
830380
1.time taken=distance/speed=26/43 hours\n 2.time taken=distance/speed=68/54\n=34/27hr\n=1 7/27 hr
1071518
Time And Relative Dimension (or Dimensions) In Space
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
900973
At noon\nA = (0, 0), B = (-90, 0)\n\nAt time t hours\nA = (0, 40t) B = (-90 + 60t, 0)\n\nSo |AB| = √[(40t)² + (-90 + 60t)²]\n= 10√[16t² + 81 - 108t + 36t²]\n= 10√[52t² - 108t + 81]\nd|AB|/dt = 5(104t - 108)/√[52t² - 108t + 81]\n\nWhen t = 1 d|AB|/dt = 5 * (-4)/√[52 - 108 + 81]\n= -20/√25\n= -4 mph ie they are getting closer at a speed of 4 mph
570639
As the speed of light is approached time slows.\nThis is given by the formula Sqrt(1-v^2/c^2) where\nv=velocity\nc=speed of light\n\nAt the speed of light time stops Sqrt(0) = 0\n\nFaster than the speed v exceeds c so v/c >1\n\nsqrt of a negative is imaginary.\n\nThe only thing you can see when traveling faster than a light is a snuffleupagus. (Because their IMAGINARY. or least they used to be back in the good old days.)
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