Here's the story. John Von Neumann is at a party one night and a woman approaches him and says, "Mr. Von Neumann, I hear you're quite the mathematician. I have a problem for you. Two trains are traveling toward each other, each going 30 mi/hr. At the moment they are one mile apart, a fly leaves one train travleing at a constant speed of 60 mi/hr. As soon as the fly reaches the other train, it instantly turns around and keeps repeating this until the trains collide. What is the total distance the fly..." Before she even finishes, Von Neumann replies: "One mile." Stunned, she says, "Ah, you saw the trick. Most mathematicians would model the situation as an infinite series, go off into a corner for an hour and find the limit of the series, but you saw it right away." Von Neumann replies: "What trick? I did the series." My question is what is the trick she was referring to?
203734
The fly has only 1 minute to fly at 60 mph regardless of the back and forth before the trains collide, therefore 1 mile is the distance it will cover.
504421
I like Dr. Greg's solution to disappearing, but let's look at it a bit further, shall we?\n\nAssume the average room is 20 ft X 20 ft.; so the max distance I'd need to whisk out and seemly just disappear would be the diagonal: about 28 ft.  So to travel that in "12.5 milliseconds," like Dr. Greg suggests, I'd need to average v = S/t = 28/12.5 X 10^-3 = 2.24 X 10^3 ft per sec while crossing the room and bugalooing out of there.  That's better than 1,500 mph (there is a reason they say he travels "faster than a speeding bullet").\n\nStarting from a standing stop, our intrepid hero would need to reach 3,000 mph to average that 1,500 because the average would be v = (v1 + v0)/2 = (3000 + 0)/2 = 1,500.  To reach that end velocity, he/she would need to accelerate at a = v/t = (3,000 X 5280)/(12.5 X 10^-3) = 1,267,000,000 ft/sec^2; where 1 mile = 5,280 ft.  This is almost 1.3 X 10^9 (that's billion) ft/sec^2\n\nAcceleration due to gravity at Earth's surface is g = 32.2 ft/sec^2; so Superboy is accelerating at n = a/g = (1.3 X 10^9)/32.2 = .04 X 10^9 = 40 X 10^6 g's.  In other words, he would be pulling 40 million g's during that escape from the room.  But, hey, he's a superlad; so he can take it.  But could you...?\n\nConsider the amount of energy the caped crusader expends while getting away in less than the blink of an eye.  J = Fd; where J is work in foot-lbs, F is force (ma) in pounds, and d = 28 ft, the distance the big S moves across the room.  Thus, we have J = Fd = m(ng) d = mg (n) d = W n d = 220 X 40 X 10^6 X 28 = 246,400 X 10^6 = 246 billion ft-pounds of energy or work.  W = mg = Superkid's weight (I'd judge about 220 lbs), n = the number of g's pulled while accelerating (a = ng) up to 3,000 mph.\n\n246 billion ft-pounds of energy (work) is equivalent to 80 tons of TNT; or 160,000 pounds of TNT going off in that room when the super dude splits in less than the blink of an eye. [See source.]\n\nIf I were you, I'd not wait for Clark to split...could be hazardous to your health.
659408
Wow.\n\nA few people here know what they're talking about.  A few.  Very few.\n\nMost of these responses are absolutely hilarious.  Most of them are from people trying to sound smart, in which case they do NOT.  Other cases are from kids with big imaginations, with "Anything is possible!"\n\nEinstein was brilliant, no question.  But all great scientists are capable of mistakes, and Einstein was no exception.\n\nHis theory on time travel was based on the speed of light and the "paradox" in which light travels at a fixed rate of speed.  This is true; light moves at 186,000 mps no matter what conditions are in place.\n\nHis theory's flaw is accounted to the fact that light has NO mass.  Light, in it's fixed movement, cannot be compared to any living organism (which is made up of mass) or any other object.  Any object made up of mass (for example, a human being, or any kind of lightspeed-capable rocket ship) does NOT have a fixed speed; therefore the velocity equation paradox that Einstein based his theory on does not apply to them.\n\nThis is all I want to say tonight.  Anyone who understands his theories will understand what I've said here, so think about my statements.  Let me know what you think (again, only those of you who understand what Einstein's theories were all about), as I'm no expert on relativity.  Like everyone else posting on a Yahoo answers forum about Albert Einstein and his theories on time travel, I'm just a fan.  Don't act like any of us are any more.\nEveryone else:  Please don't worry about it.\n\nAlthough I am skeptical by nature, like any smart person, it is NOT my goal to bash the idea of time travel.  After I spotted the flaw in Einstein's theory I presented it to many professionals in the field (or the few that I know, anyway) and have so far received very positive feedback.  I searched online as well and have discovered that others seem to have come up with the same idea, or similar.  Let me know what you think about the whole deal.\n\nBeing proven wrong only makes you smarter.
530084
Conservation of angular momentum must be considered as PhysicsDude mentioned. The thing to keep in mind here is the speed of the train given is "with respect to the track". The train is moving in one direction and the wheel/track is the other. If the wheel were not moving the speed of the train would be the 0.15 m/s, however since the wheel is moving counter to the train, and observer standing by the track (that is moving on the wheel along with the track) would see the speed of the train as faster than the actual train speed by an amount equal to the speed of the track. \n\nVr = Vtt + Vtr\n\nwhere Vr (=0.15 m/s) is the relative speed, Vtt is the speed of the toy train, and Vtr is the speed of the track.\n\nNow, considering conservation of angular momentum\n\nLtt = Itt*wtt = Itr*wtr = Ltr\n\nwhere Itt is the moment of inertia of the train, wtt is the angular speed of the train, Itr is the moment of inertia of the track, and wtr is the speed of the track. The two equate since the intial angular momentum is zero so that the two must cancel each other out after the train has started moving. The moments of inertia are:\n\nItt = m*r^2\nItr = 1.1*m*r^2\n\nwhere m is the mass of the train (1.1*m is the mass of the wheel/track), and r is the radius of the wheel. Recall that the angular speed is the linear speed divided by the radius. This means that the above conservation equation is\n\nm*r^2*wtt = m*r^2*(Vr/r - wtr) = 1.1*m*r^2*wtr\n\nSolving for the angular speed of the wheel/track (wtr)\n\nwtr = Vr/(2.1*r)\n\nPlugging in the values given\n\nwtr = (0.15 m/s)/(2.1*0.43 m) = (0.15 m/s)/(0.903 m)\n\nwtr = 0.166 rad/s
495723
OK... This kind of problem is easiest if you break it down into smaller problems. Also, we need to have everything in the same unit of measurement to make life easier... \n\nFirstly, we need to know how far apart in miles they will be when they are 4 minutes apart. We need to convert the mph figures to mpm (miles per minute) by dividing by the number of minutes in an hour (60). That gives us:\n\nAirplane 1: 240mph/60 = 4mpm\nAirplane 2: 360mph/60 = 6mpm\n\nNext, we know the 2 airplanes are moving toward each other. If we add the new figures together, we get the "closing speed" (how fast they are moving toward each other):\n\n4mpm+6mpm = 10mpm\n\nWe know that there is 440 miles between them at the start of the problem, and now we know that the airplanes are closing at 10mpm. We can use that to work out how far apart they are in time (we need to know this, because the next part of the problem gives us a seperation in minutes, not miles):\n\n440 miles/10mpm = 44 minutes\n\nSo now we know how fast each one is going, and how far apart they are in minutes... Next, we need to know ho long each has traveled in minutes when they have only 4 minutes remaining to go:\n\n44 minutes - 4 minutes = 40 minutes.\n\nSo... \n\nAirplane one has flown for 40 minutes at 4mpm\nAirplane two has flown for 40 minutes at 6mpm\n\nThe distance travelled by each airplane is:\n\nAirplane 1: 40 minutes x 4mpm = 160 miles\nAirplane 2: 40 minutes x 6mpm = 240 miles\n\nSo, the total distance covered by BOTH airplanes (in the first 40 minutes) is:\n\n160 miles + 240 miles = 400 miles\n\nAnd we know that there was originaly 440 miles between them, so, the distance between airplane 1 & 2 when they are 4 minutes apart is:\n\n440 miles - 400 miles = 40 miles.\n\n\nAnd there you have it!
154830
When V(t)=0 the particle stops.\nIt can have a(t)=0, but still moving (constant speed)
110076
Depends on how far away the wall is, don't you think, and the limits on your methods of measuring "half way"?  Eventually the difference between the distance from the center of the room to the wall and the sum of your moves will be less than your ability to measure.  That's the role of the little epsilon in the theory of limits.  Let the distance, center to wall, be D.  Then\n\nD - [(1/2)D+(1/4)D+...+(1/2^n)D]=(1/2^n)D\n\nWhen you cannot measure (1/2^n)D, you're there, for all practical purposes.\n\nTell a mathematician and an engineer they can approach the naked girl tied to the tree in this fashion [sorry, ladies, but in ancient times engineers were all male].  The mathematician will throw up his hands {so were mathematicians, with a few notable exceptions), saying, "I'll never get there."  The engineer will grin and say, "I'll get close enough."
607518
i am no genius but i did try. here's how i went about it.\nspeed = distance/time.\n(1/30)s = 525 lines, therefore time for 1 line = ((1/30)/525)\n=(1/15750)s\n20in = 1 line.\nresweeping the line, therefore distance = 40in.\n40in = 3.33ft (1ft = 12in)\nagain, speed = distance/time, so\nspeed = (3.33ft)/(1/15750s) = 52500 (ft/sec)\nhopefully, this is the right answer. please let me know. thanks.
144883
As in many other pseudo-demonstrations of this kind, the error is to divide by zero (2-2) both sides of the identity. You can't do this because the division by zero is an undefined operation.\n\nFrom a well-known logic theorem, starting from a false statement, you can consistently derive any other false statement. This is the common source of many apparent paradoxes.
124650
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.
466786
Great Question! I have the answer!\n\nImagine you are moving an object from left to right. You will see there is a displacement and a definite time taken to move from one place to the other. \nImagine you are moving faster. Notice that the time reduces.\nNow you are moving in ultrafast speed such that the time taken is nearly zero. \nWhat does this mean? There has been a displacement with no time. \nThis is what is called Yoga! Ancient Saints use to travel in time space exactly by controlling their displacement at no interval of time.\nSo do not confuse yourself with the things which will not happen with our life.\n\nTimelessness would be felt in time machines, where in the the machine travels at a speed equal to infinity. Or atleast close to infinity!\n\nOk! is it possible to have a timelessness?\n\nThis looks like an imaginery thing. Just in fictions!
410160
Wow. Your classmate just asked this question. But here's your answer anyways...\n\nThere are a few things to keep in mind here: all of the energy in the initial system is rotational, and the initial energy is equal to the energy lost to friction. Also note that the energy is lost to friction at a constant rate.\n\n(a) The initial energy of the system is:\n\nE = (1/2) * I * w²\n\nWhere I is the moment of inertia and w is the angular velocity. The power transferred from the system due to friction is equal to the time derivative of the energy loss:\n\nP = dE/dt = I * w * a\n\nwhere a is the angular decceleration.\n\nSince the power is transferred at a constant rate and removes all of the energy from the system, we can take this as:\n\nP = -E/t (negative sign denotes power loss)\n\nCombining the power equations yields:\n\nI * w * a = -E/t\n\nSolving for a:\n\na = -E / (I * w * t)\n\nFinally, substitute the expression for E to obtain:\n\na = -I * w² / (I * w * t)\n\na = -w/(t)\n\nSubstituting the values yields:\n\na = -(150 rev/min * 2*pi rad/rev * 1/60 min/sec)/(2.2 hr * 3600 sec/hr)\n\na = -(15.71 rad/s)/(7920 s)\n\na = -.001983 rad/s²\n\n(b) The rotational behavior of the fly wheel can be modeled as simply:\n\nR(t) = -1/2*a*t² + w*t + Ri\n\nwhere R(t) is measured in radians and Ri is the initial angular displacement. Assuming that Ri = 0 (we'll measure the number of rotations from its initial position), then the flywheel will revolve:\n\nR(t) = -1/2*a*t² + w*t (radians)\n\nR(t) = (-1/2*a*t² + w*t)/(2 * pi) (revolutions)\n\nSubtituting the values:\n\nR(2.2 hr) = R(7920 s) = (-.5*.001983*(7920 s)² + (15.71 rad/s)(7920 s))/(2*pi)\n\nR = 9900 revolutions\n\n(c) The tangential component of the linear acceleration of a point on a rotating body is given as:\n\nA-tangent = a * r\n\nwhere A = linear acceleration. The angular acceleration is constant, so for all w (including w=75 rev/min), the linear acceleration will also be constant. Thus:\n\nA-tangent = (-.001983 rad/s²)(.5 m)\n\nA-tangent = -9.915 x 10^(-4) m/s²\n\nA-tangent = -0.9915 mm/s²\n\n(d) The magnitude of the net linear acceleration on the particle is composed of a tangential and a normal component. The equation is of the form:\n\nAnet = [ (A-tangent)² + (A-normal)² ]^(1/2)\n\nOr:\n\nAnet = [ (a*r)² + (v-tangent²/r)² ]^(1/2)\n\nNote that the tangential velocity is given by:\n\nv-tangent = w * r\n\nThus, the magnitude of net linear acceleration is:\n\nAnet = [ (a * r)² + (w² * r)² ]^(1/2)\n\nSubstituting the known values yields (note that for this case, the angular velocity of 75 rev/min is equal to 7.854 rad/s):\n\nAnet = [ (-.001983 * .5)² + (7.85² * .5)² ]^(1/2)\n\nAnet = 30.81 m/s²\n\nFINAL ANSWERS:\n\n(a) a = -.001983 rad/s²\n(b) R = 9900 revolutions\n(c) A-tangent = -0.9915 mm/s²\n(d) Anet = 30.81 m/s²
381150
I'll give the calculations for the principle given by the first answerer above because I agree with what he said.\n\nInitial momentum of 3-car train=3m*2\nInitial momentum of 4th car=m*4\n\nMomentum of 4-car train=4m*v\n\nBased on conservation of momentum:\n3m*2+m*4=4m*v          m cancels out.\n6+4=4v\nv=10/4\n  =2.5m/s   This is the speed of the 4 car train.\n\nTherefore, initial momentum of 4-car train=4m*2.5\ninitial momentum of 5th car=m*0\n\nMomentum of 5-car train=5mv'\n\nAgain using the same principle:\n\n4m*2.5+m*0=5mv'\n10=5v\nv=2m/s.
514565
1. d=t, any form of d=t+b would be parallel. b = any number. That is a line.\n\n2. d=t would also work here.\n\n3. Well if you go 60 MPH and you wnat to know how many miles a minute you can go convert using (60miles/hour)*(1hour/60minutes) and you get 1mile/minute. If you need feet, 1 mile is 5280 feet, so 5280 feet/minute
266719
All the answer I saw is wrong so far. Come on people.Anyway here is your answer buddy trust me on this. 100% sure\n\nA travels from Earth to Jupiter 6 hours before B starts from opposite side, which means 180000 km/h extra. The total distance is 980000 and when you substrack 980000-180000= 800000 you find the total distance of the meeting point of A and B and since they are traveling towards each other you add they speed which is 80000 and you divide 800000/80000=10hr this is when they are going meet. But since A already traveled extra 6 hours it makes 16 hours and C coming from same point as A but 12 hours later. So in order to C meet with A and B at the same time C has to travel 480000/4= 120000km/hr this means. It is divided by 4 because 16-12=4
403122
I solved the problem by stating that P would travel distance d in time t\nd=20*t+.5*.3*t*t\n\nand Q would travel d+20 in time t\nt is the time to collision\n\nd+20=30*t\nd=30*t-20\nSubtract the equations and solve for t:\n\n0=.15*t*t-10t+20\nt=2.064 seconds\n(the second root is superfluous)\nd=41.9m\n\nWhich proves collision\n\nj
113105
Let's assume time measurement was started when the front of the train just entered the tunnel.\n\nThe train travels 1 mile per minute.\n\nIn 1 minute:\nThe front of the train just emrges out of the tunnel.\nThe rear end of the train had just entered the tunnel.\n[ie. The whole train is inside the tunnel now; the front is just about to emerge out]\n\nIn 2 minutes:\nThe rear end of the train emerges out (ie. leaves) the tunnel. Now the train has completely emerged out.\nThe front end of the train is 1 mile away from the tunnel.