if a train leaves town at 7:25 pm and another train leaves a city at 5:00 pm How many passengers do they have?
401576
0.  They are freight trains.
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
917540
right now i am in Buffalo,me & a freind are on our way to New York ,its 5:25 pm
818855
You are correct that if the bell and passenger are both on the train then the freq would be the same.  But, I think the intent of this question is to say that the bell is on the station.  It says 'train station bell'.  Since the bell is on the station and the person is on the train, there is a relative motion between them.
983724
Current revenue is $2.00 * 1,000.\n\nFor every 25 cent increase, there will be a 50 person decrease in ridership.\n\nSo, in that case revenue would be ($2.00 + $0.25x) * (1,000 - 50x).\n\nSo, I would use R = (2+0.25x)(1000-50x) = -12.5x² +150x + 2000\n\nAre you in calc or pre-calc?  Do you know if you need to determine what the maximum fare increase is before revenue starts going down?  (That's a typical type of question that's usually asked with these types of questions.)\n\n~ ♥ ~
72826
To be specific, what cities are you talking about?? The time will vary depending on whether or not you are connecting. The time enroute could be upwards of 18 hours of you must connect a few times.\n\nCheck http://www.expedia.com for service between your city pair.
416112
that same day... just depends who the last kid on the bus is \n\n;)
103992
You need to know the speed of each car to answer the question. Assuming car 1 has a speed of s1 miles/hour and car 2 has a speed of s2 miles/hour. The time they will take to meet each other will be 8.5/(s1+s2). You can now substitute the values to arrive at the answer!
669522
It is easiest to think of this in 2 steps.  First, how long does it take the east bound train going 65 mph to travel the 175 miles to the station?\n\nanswer: 175/65=2.6923 hours\n\nStep 2:  How fast does the westbound train need to go to travel the 63 miles in 2.6923 hours?\n\nanswer 63/2.6923 = 23.4 mph
182504
It can help school busses get more use out of less busses, because if one school starts at 7:30 and the other at 8:30, it can do two routes.
839270
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
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.
935120
for the first hour:\nJose travels 5 miles\nMario travels 7 miles\nMelanie travels 9 miles\n\nfor the second hour:\nJose travels 10 miles\nMario travels 14 miles\nMelanie travels 18 miles\n\nJose   (1,5) and (2, 10) are your points\n(10 - 5)/(2-1) = 5 mph\nMario  (1, 7) and (2, 14)\n(14-7)/(2-1) = 7 mph\nMelanie (1,9) and (2, 18)\n(18-9)/(2-1) = 9 mph\n\nANSWERS\nJose\n    5 mph\nMario\n     7 mph\nMelanie\n    9 mph
116496
Any number of places.  You leave from 1 hour east of a time zone, and arrive west at the same 'time'.
247974
2 with another on the way
702157
Let\nx = speed of Sachiko\nx+5 = speed of Jose\n\nSince they travel a distance of 476 miles in opposite direction\nthen,\n\nd(Sachiko) + d(Jose) = 476\ns(Sachiko)t + s(Jose)t = 476\n(x)(4) + (X+5)(4) = 476\n4x + 4x + 20 = 476\n8x = 456\nx = 57 \nx+5= 62\n\nTherefore,\nspeed of Sachiko = 57 miles per hour\nspeed of Jose = 62 miles per hour
951909
Depending on where you are in New Jersey, your best bet is going to be the NJ Transit train system.\n\nTrying to get into the city by car at 5:00AM might not be too bad, but parking will be really difficult and getting out of the city and back to Jersey will be even worse.\n\nThe train will be running on a holiday schedule, so check and make sure you have the correct time.\n\nwww.njtransit.com