2 trains left the same station at the same time. The first train traveled north with a speed of 10km/h ...?
543583
Use the pythagorean theorum.  Train One is the left leg, train 2 is the right leg, and the distance between them is the hypotenuse.  Since they only traveled an hour, their speed is irrelevant.  That triangle gives the formula of (x+10)^2 + x^2 = 71^2\n\nWork it out from there.  X will equal the distance that the second train traveled, and since they traveled for an hour, that is the magnitude of its speed as well.
1054333
Hey, mammalboy. Get a clue. Trains are still used today but tracks don't run everywhere that roads go. Genius.
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
694191
S = southbound speed\nN = northbound speed\nd = distance\nx = time from 8am\nS(x) + N(x-1) = d\n\nS*2 + N*1 = 500  => N = 500 - 2S\nS*3 + N*2 = 850 = 3S + (1000 -4S) = 1000 - S\nS = 150, N = 200
691525
Well, when you jump out of a moving train, you will be carried forward in the direction of its movement, this is because when the train is travelling, the train is moving with a velocity of v towards the x-component. Therefore, as you are on the train, you are also moving with the velocity of v towards the x-component. Hence, during your jump, you are still 'travelling at the x-component' with the velocity of v. So, you will be carried forward. Hopefully I answered your question and make you understand better.
838947
I used to get tripped up with these too!\n\nSo, it's all about speed, distance, time (s = d/t) and finding the 2 things you know, and getting an equation for the third unknown and solving it. Here, the speeds are given, and the difference in times are given.\n\nSay d is the distance Matt has to go, and t1 & t2 are the times he took to get there the first and second time.\n\nt1 = d/75\nt2 = d/45\n\nalso, he arrived one hour early the first time, and one hour late the second ==> the difference between these two times is 2 hours.\n\nt2 - t1 = 2\n=> d/45 - d/75 = 2\n=> (75-45)d/45*75 = 2\n=> d = 2*45*75/30\n=> d = 225 miles.
754249
The first car has been traveling for 7.5 hours, so it has covered 330 miles; therefore the second car has been traveling at 330 miles/5.5 hours=60 miles per hour.
1051502
Trick question! They aren't even on the same road!
685470
the speed = distance/time
237365
Say that x is 10 miles.\n\nIt takes 10/10 hours on the first trip (1 hour)\nIt takes 10/8 hours on the return trip (1.25 hours)\n\nHe has therefore travelled 20 miles in 2.25 hours.\n\n20/2.25 = 8.89 miles per hour.
1039781
Depends on how long the train is.....right?
951097
the one who answered first... plzzzzzzzzzz\n\nexcellent answer but.. trains moving towards each other can also be said to move in opp directions.
104070
Heather leaves a town at 2 p.m., going 56 mph, traveling east. Erin leaves the same town at the same time going 64 mph, traveling in the same direction. What time of day will it be when Erin is 72 miles ahead of Heather?\n============================================================\n\nHeather's distance = 56x\nErin's distance = 64x\nx = hours traveled\n\nThe distance between them at any time can be expressed as a simple equation:\n\nd = 64x - 56x\nd = 8x\n\nYou want to find "x" when d = 72 miles:\n\n8x = 72\nx = 9 hours\n\n2:00 p.m. + 9 hours = 11:00 p.m.
965437
okay when does the train leave?
735262
1.  The key here is to understand how far the two trains have traveled in total during the first 3 hours.  We are told that the two trains started 500 miles apart, and 3 hours later are now 80 miles apart.  Thus they traveled 420 miles in total. \n\nIf you let x represent the rate of speed that the Freight train is traveling, and x+20 represent the rate of speed that the Passenger train is traveling, you can set the following algebric equation:\n\n3 hrs * x mi/hr  + 3 hrs * ( 20 + x ) mi/hr = 420\n 3x + 60 + 3x = 420\n6x = 360\nx =60 mi/hr.\nx+20 = 80 mi/hr.\n\n2.  The key is too figure out the total number of employees needed to break even.  Regardless of how many employees you have, we are told that it costs $2000 per day to run the factory.  In addition to that  we are told that for each employee, we must pay him or her $300.  These are the total costs for the operation.\n\nOn the other hand, we are told that each employee can produce $500 in revenue each day.\n\nThus, we need to create an algebric equation where the total costs = the total revenue\n\nLet x represent the number of workers, and set up the following algebric equation:\n\nCosts = Revenue\n\n$2000 (Fixed Costs) + $300 * x (Variable Costs) = $500 * x\n\n2000 + 300x = 500x\n2000 = 200x\n10 = x\n\nYou will need 10 workers to break even, and the total gross value of ball bearings produced will be 10 * $500 or $5000.\n\nHope that I have cleared things up for you!
981312
310 times faster\n\nif A is going 10mph, and B is going 5mph, the 10/5=2 or twice as fast. so 9.3X10^10 / 3.0X10^8 = 310 times faster.\n\nbut nothing can't go faster than 3X10^8m/s!!!! The normal speed of light means it's going though a vacum (nothing), so if it's going though a gas, it will slow down!! Not speed up! Since the energy will hit atoms and slow down. \n\nIf an object is going 3E8 (short way of 3X10^8m/s) releative to the ground and an object passes it at 3E8, then the object passing the first objects speed was 3E8 releative to the ground... so both were going the same speed, even if some was twice as fast as it!
312163
the first after collision moves in an opposite direction and the second ball moves towards the direction of collision\nfirst ball moves with a velocity of  16.66cm/s\nsecond ball with 66.66cm/s