Explain the equation used to find how much time ahead a clock will read if moving at relativitis speeds?
150092
t=to\n_____\n√1-v2/c2
879627
Time math can be interesting... first, always convert the times to 24 hours:\n\n13:57 - 10:53\n\nNow it's just like math...  13-10 for hours, and 57-53 for minutes\n\n3:04 am or 3 hours and 4 minutes, depending on your frame of reference.\n\nHope this helps!
958698
If t is the time that has been measured in the clock's reference frame, then t/2 is the time that has been measured by the observer.\n\nNow t = (t/2)[1-(v^2)/(c^2)]^(1/2)\n\nthe t's cancel and you are left with 1 = (1/2)[1-(v^2)/(c^2)]^(1/2)\nor [1-(v^2)/(c^2)]^(1/2) = 1/2\n\nsquare both sides to get 1 - (v^2)/(c^2) = 1/4\n\nsubtract one from each side to get -(v^2)/(c^2) = -3/4\n\nor (v^2)/(c^2) = 3/4\n\nthis means that v^2 = (3/4)(c^2)\n\ntaking the square root of both sides gives the final answer:\n\nv = [3^(1/2)]c/2.\n\nNot the most intuitive answer in the world.
168069
assuming your statement is true, then\na=k/t\nwhich gives t=k/a\nso time is a nonlinear function of acceleration.\nGraphically, the plot of a versus t is not a straight line (it is a hyperbola)
997236
10 minutes + 10:50 = 11 o'clock\n \n\n\n\n--- LeeeN
878888
x = speed of train 1\ny = speed of train 2\n\nAt time 2hrs, train 2 has traveled 1hr and reaches the point where train 1 was at time 1:20 (40 mins earlier), so your first equation is:\n                               (y * 1 hour) = (x * 4/3 hours)\n                                              y = 4x/3\n                                  or       x = 3y/4\n(1:20 hrs is one and one-third of an hour, hence the 4/3.)\n\nThe total distance travelled by train 2 is (y * 1 hour) + ((y+0.5) * t) = 540, where t is the time elapsed after the 2-hour point, i.e. both trains arrive at time t+2 hours.  Solving for t:\n                y + (y+0.5)t = 540\n               (y+0.5)t = 540 - y\n               t = (540-y) / (y+0.5)\n\nDoing the same for train 1: x * (t+2) = 540.  Soving for t:\n                                    t+2 = 540/x\n                                    t = 540/x - 2.\n\nSo we have t expressed once in terms of x ,and once in terms of y.  these being equal, we have\n\n                     (540-y) / (y+0.5) = 540/x - 2\n\nSubstituting 3y/4 for x,\n\n                       (540-y) / (y+0.5) = 720/y - 2\n\nFrom here on you should be able to reduce this to a quadratic form, solve for y and all that.
68749
Rabid_Scientist is entirely correct.  Both you and your home see time in the other place as speeding up.  I do have something to add to that though.  If you maintain a constant velocity, it can only be measured relative to some other point of reference, so there is no way to determine which is faster, and consequenty it cannot be determined whose time is actually going faster.  However, if you are accelerating it's a whole different story.  As you speed up to that super fast speed where the time differental starts to be noticable, you will see time on your home planet speed up and they will see your time slow down. (Sorry, I'm not going to try to explain this, I'm not sure I entirely understand it myself).  If you were to go back to your home planet, far more time would have passed for them than for you.  This effect will be greater with greater acceleration and greater speed so in that regard time will pass on your home planet more quickly the faster you go.
1007024
distance equals rate times time\nThat is,\nd = rt\nd = 117 and t = 2.25\nSo we have\n117 = r(2.25) or\nr = 117/2.25\nr = 52 mph
11066
well first, read my other answer to your question about the speed of sound.  You would probably expect the answer to be the same, and the light to reach the other car, but this is a special circumstance because nothing can travel faster than the speed of light, not even light itself.  when a beam of light is shone ahead of a rapidly moving space vehicle, and is expected to be moving at the speed of light plus the speed of the vehicle, it is still only moving at the speed of light, which is puzzling.  Physicists have defined the speed of light to be the ultimate cutoff point for speed.\n\nAnd also, you wouldn't be able to travel at the speed of light anyway because time would probably stop since it does get slower and slower the faster you go.  Velocity warps time, the faster you go, the slower time moves for you relative to a stationary object.  I've read this, and it is definately true, we had to derive the formula in physics from, and calculate the time, lenghth(which decreases with speed), and mass (which increases with speed) distortion as well.  The effects of this are very negligable unless you are moving at speeds very very close to the speed of light, at least over 90% of the speed of light, and we haven't gotten anywhere close yet.  The denominator of this factor that time, length, and mass are distorted by actually is 1 minus the square root of the speed of light(c) over the speed in which the object is traveling (v), so the reason you can't go faster than the speed of light is because if v was greater than c, then that square root would be greater than one, meaning the denominator would be negative, meaning the factor would be negative, and the light would actually be travelling backwards in time...\n\nIf you find any of this interesting, I highly recommend "The Elegant Universe" by Brian Greene.
426864
Roughly 3 ten-thousandths of a second.\n\nIf an object is travelling 60 mph, it moves at 1 mile or 5,280 ft/min.  60 seconds in an minute 5,280/60 = 88 ft per second.  12 inches in a foot = 1,056 inches per second.  2.76/1056 = approx. .00261 seconds.
953778
Let's call the entire track length x, measured in meters.\n\nTo arrive at the time, we have to divide the distance by the speed:\n\nx / 8.5 + x/8.0 = 99\n\nmultiply all sides by 8.5 and by 8.0\n\n8.0 X + 8.5 X = 8.0 * 8.5 *99\n\n16,5 X = 8*8.5*99\n\nX = 8*8.5*99/16.5\n\nTime for the first lap: \nT1 = X / 8.5 s\nT1 = 8*99/16.5 s\n\nT2 = X / 8.0 S\nT2 = 8.5*99/16.5 s\n\n\ncalculator time....\n\nSolution D
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
157544
Well it is a bit technical, but let me try.\n\nThe equations of special relativity include a term that looks like gamma=sqrt[1-(v/c)^2] where v is the speed of the object, and c is the speed of light. The equations for relativistic mass. The equations for relative position. The equations for time in the moving frame. And so on. \n\nSometimes gamma is in the numerator. Sometimes gamma is in the denominator. Sometimes the gamma is squared. But it is in there a lot (you can look them up on wikipedia or any number of web sites).  \n\nThese equations have been verified over and over, in particle accelerators and other experiments. They also explain the Michelson-Morley experiment that provided evidence against the exisence of the ether. \n\nNow what what happens to gamma when v gets close to c?\n\nObviously, gamma gets very small. If you are dividing by gamma in the equations, you get huge numbers. So relativistic mass goes to infinity as v gets close to c, and time slows down and the object moving gets compressed front to back.\n\nWhat happens when v is exactly c? \n\nThen, time stands still, and the object's relativistic mass is infinite and it is infinitely thin. \n\nWhat happens when v is bigger than c?\n\nThen gamma becomes imaginary, and mass becomes complex and time becomes complex and so on. Obviously, a complex mass or a complex time have no meaning in our universe. \n\nThere are postulated objects called tachyons that go faster than c, but no one has ever reliably observed one. Distant galaxies can move away from us faster than c, but that is because no information is being transferred. Just like you can move a light source and make the spot of light very far away appear to move faster than c, but no information is being transferred. Relativity only restricts the speed of information transfer to be c or less. \n\nAre the equations of special relativity true? Well they fit the evidence we have. And until we have evidence that they do not fit, we use them.
202957
How far apart are the marks?\nTime taken to travel between the marks?\nSpeed = distance divided by time.
847424
d= v .t\n> d = 50 km/hr (18 min)\n>d = 50 km/60 min (18min)\nd= 50 km (18)/60\nd =900 km/60\nd = 15 km >>the distance travelled\n\nIf you walk at an average of 3 miles per hour (approximately 4.8 km/hr) at a distance of 15 km:\nt= d/v\n=15 km/4.8 km/hr\nt = 3.125 hrs or (3 hours and 7.5 minutes)>>the time you need to walk on 15 km at the speed of 4.8 km/hr.♥
465629
All light has a velocity of "c."  c is ~ 3*10^8m/s.  With this knowledge you can say that frequency is the inverse of time/distance.  Thus, you can use the formula C(meters/sec) / Freq.(cycles/sec) = Wavelength(meter/cycle) to solve the question.\n\nDo your own work from there.
792607
0 km/hr^2 .... because your speed doesn't change with time due to constant as you mentioned