A train goes around a circle, the first time with an average velocity of 100 km/h. How fast should the train be when going around the circle again a second time, so that the average velocity over the two loops is 200 km/h?\nHint\n•The obvious answer 300 km/h is wrong. \n\nFirst right answer with a mathematical proof will be chosen as the BEST ANSWER
124553
Let the time taken to go round twice be T1 +T2.\n100=D/T1 where D is the circumference of the loop.\nWhat we want is, 200=2D/(T1+T2)\nor D= 100(T1 + T2)\n   D= 100(D/100 + T2)\n   D=D+100T2\n   T2=0\n   which means it should travel with infinite velocity.
365942
Let us assume that the speed of the train was x km/h and let us also assume that the length of the tuneel to be "d" km.\nNow the man has already travelled one-fourth of the tunnel.So he has two options open for him either to run back a distance of d/4 back to his entry-point or to run forward to cover a distance of 3d/4 miles.\nAgain, as the man has two options,the train may also come from two direction.So in one case the relative speed of the man and the train would be x-12 mph and in another case it would be x+12 mph and as x>12 (ioh! otherwise the man could have saved himself and we need not had to toil so hard for solving this problem),x+12 would be greatr than x-12\nNow we can say that the shorter distance traelled by the man was covered by less relative speed and the loner distance coverd by the greater measure of relative speed and as the man got knocked down just at the end of tunnel,the time taken was the same\nTherefore,(d/4)/(x-12) =(3d/4)/(x+12)\n1/(x-12)=3(x+12)\nBy solving the above equation,we get x=24 miles\nTherefore the speed of the train was 24 miles per hour.\nNow it may be noted that the ration of the distance the man ran \nand the ratio of the relative speed in two cases are the same It is\n1:3.So in eithr cases the train was at the mouth of the tunnel\n(You may ask how I solved the problem.Well I started my service-life as an enginer in the Railwys and used to look aftr railway track and yes u r right tunnels)\n\n.I
50817
If you start with the times from A to B is t1 and the time from B to C is t2, then the distance from A to C is t1*10 + t2*50, and the total time is t1+t2, so the average speed is:\n\n    (t1*10 + t2*50)/(t1+t2)\n\nThat's the weighted mean of 10 and 50, with weights t1 and t2.\n\nIn general, to average a bunch of averages, you need to compute a weighted average.  For example, if you have ten rocks averaging one pound apiece, and I have five rocks averaging four pounds apiece, then to determine the average of all the rocks, you would take the weighted average of 1 (with weight ten) and 4 (with weight 5.)  That is:\n\n    (1*10 + 4*5) / (10+5) = 30/15 = 2
396208
Now there's some strange questions. I'll help as best I can. I hope I've understood the questions properly.\n\nQuestion 1) Bob will always win, assuming they play perfectly.\nIf Ann takes 1 matchstick, Bob should take 2. That leaves 3 left. Now whether Ann takes 1 or 2 from the remaining 3, Bob is left with 1 or 2, so he can take what's left and wins.\nIf Ann takes 2 matchsticks,  Bob should take 1. That also leaves 3 left, so again regardless of how many Ann takes from the 3 remaining ones, Bob can take what's left and win again.\n\nQuestion 2) The answer is 5.52pm.\nLets say the slow train moves at speed X. That means the fast train moves at speed 2.75X. This is because the fast one moves 2.75 times faster than the slow one (since it travels the same distance in 8 hours compared to the slow time of 22 hours, so it must be going 22/8 times faster).\nIn the time taken for the trains to meet, between them they have travelled the full distance. Therefore you can add their speeds together to calculate the relative approaching speed of the trains, which would be 2.75X + X = 3.75X.\nNow, the slow train travels at speed X and takes 22 hours to go the same distance. Therefore we can say that the time taken for the two trains to meet is 22X/3.75X, or 22/3.75, which is 5.8666.\nTo convert 0.8666 of an hour to minutes is simple - just multiply by 60. This gives 52.\n\nSo the total time taken is 5 hours 52 mins. \n\nTo put it another way:\nSpeed of train 1 = 2.75 times faster than train 2. So the speeds of the two trains together = 3.75 times faster than the slow train.\nThe slow train takes 22 hours to go the distance. Therefore the two trains together take 22/3.75 to do the same distance, which is 5.8666.\n\nI hope this helps :)
301552
distance = rate times time or\nd = rt\nLet x = rate of the freight train, then\nx+30 = rate of the express train.\nt = 3 hours for both trains\nFor the freight train, d1 = rt\nd1 = x(3) or d1 = 3x\nFor the express train, d2 = rt \nd2 = (x+30)3 or d2 = 3x + 90\nThe sum of the two distances must equal 390\nd1 + d2 = 390\n3x + 3x + 90 = 390\n6x = 300\nx = 50 km per hour for the freight train\n50+30 = 80 km per hour for the express train
231295
In theory it will never get there as every half-distance will take an hour, and the number of half-distances is infinite. This is Xeno's Paradox in disguise, I think.\n\nIn practice the train cannot change speed with  a few metres left to a crawl that would take an hour, and it would grind to a halt or be unable to maintain the regime.\n\nBut the series 1 + 1/2 + 1/4 + 1/8 + 1/16 ... converges on 2 but never actually makes it.
351135
its right.......... average speed= distance/time\n\nso average speed= 10*3/1\n=>30km/hr
559574
I assume it's starting from zero speed?  If so,\n\nd = v1t + 1/2at^2\nafter 2 seconds (i.e. 4 half-seconds), it's travelled\nd = 0(2) + 1/2(2)(2)^2\n   = 0 + 1/2 * 2 * 4\n   = 4m\n\nafter 2.5 seconds (i.e. the fifth half-second), it's travelled\nd = 0(2.5) + 1/2(2)(2.5)^2\n   = 0 + 1/2 * 2 * 6.25\n   = 6.25 m\n\nSo in the last half-second, it's travelled 6.25 - 4 = 2.25m.
171655
you can try somthing like...if one train starts in los angeles and heading to new york traveling at 55mph and a train leaving new york is heading to los angeles traveling at 67mph. if both trains leave at 12pm from thier starting point, at what time will they pass each other? (assuming the trains are on two different parellel traacks next to each other and the distance between them is 3000 miles).
410420
Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.
387648
LivingDownSouth's answer is not right.\nHe's tried to simplify it too much by simplifying the curve to a quarter of a circle.\n\nIf I have time later, I'll try to determine the actual curve ... and provide the answer for you.\n\nAdditional information entered later:\nI don't have a complete solution to the problem, but the following reduces it to a differential equation.\n\nFor simplicity, assume that the initial distance between the boats is 1 unit, that A is at (0,0) and B is at (-1,0), and that both travel at a speed of 1 (= 1 distance unit per time unit).\n\nThen A's position at any time t is (0,t).\nB's velocity consists of an x (eastward) component and a y (northward) component. these two velocities can be expressed as derivatives in the following way:\ndx/dt = -x/sqrt(x^2 + t^2)\n[Remember that x is a negative number between -1 and 0, so -x in the numerator is a positive number, indicating movement in the positive (eastward) direction.]\ndy/dt = t/sqrt(x^2 + t^2)\n\nIf you can solve these two equations (i.e., manipulate them and take the integrals in order to get rid of the derivative), then you will have expressions for x and y in terms of t. There will be a constant of integration, but you eliminate that by the initial condition (position at time 0 is (-1,0).\n\nTo find the eventual distance between the boats, take the expression t - y and determine what its value will be for a very large value of t. In this expression, t represents boat A's distance from the shore, and y (a function of t) represents boat B's distance from the shore. The difference is the distance separating the two boats.\n\nHope you can use this info.
445673
vA = 35 m/h\nvB = 45 m/h\ndA + dB = 320 m (dA, dB are the distances traveled by each train)\n\ntA = tB = t (the 2 trains will travel for the same amount of time before they meet)\n\nvA (t) = dA\nvB (t) = dB = 320 - dA\n\n35(t) = dA\n45(t) = 320 - dA\ndA = 35t = 320 - 45t\n80t = 320\nt = 4 hours
471064
First one's pretty easy.  Since you have a frictionless track (and we're assuming there is no air resistance here, either), the speed is due solely to the change in potential and kinetic energies.  At the start, the cube has a potential energy of mgh, or 0.530 kgX0.387mX9,8 m/s^2 = 2.01 J.  By the time it reaches the bottom of the loop, it has converted all of that potential energy into kinetic energy (since we're calling the bottom of the track to be zero altitude).  As it travels up the loop, it gains potential energy again as it fights against gravity, slowing it down.  At the top of the track, therefore, it has a potential energy of 0.30 J, meaning the kinetic energy must be the difference, or 1.71 J.  Solving for velocity (KE = 1/2 mv^2), you get 2.54 m/s.\n\nThe second case is a little more tricky, in that it involves rolling friction.  However, friction only plays a role in starting and stopping motion -- if the cylinder achieves pure rolling motion, the frictional component goes to zero because there will be no relative motion between the points of contact on the ramp and the cylinder (no sliding).  If we assume that there is never any slipping along its path, then it should behave just like the cube sliding along a frictionless track, and the speed will still be 2.54 m/s.  I could be wrong, but I think that's the answer.
1442
lambda =velocity/freq\n       =360/10\n       =36 m.\n\nb)1/10 secs.
243532
The circumference of the circle with radius of 1 m:\nd = 2*pi*r = 6.28 meters.\n\nThe time t = 1.35 seconds\n\nYou get the velocity by dividing the travelled distance by the time interval:\nv = d/t = 4.65 m/s
514166
Southbound: 150 km/h\nNorthbound: 200 km/h\n\n\nThey're moving apart at 850 - 500 = 350 kph.\n\nSo at 9 am they were 500 - 350 = 150 km apart.\n\nWhich means that South was going at 150 kph;\n\nSo North was moving at 200 kph.
410940
0 km/hr^2 .... because your speed doesn't change with time due to constant as you mentioned