If a car enters a parking every 3 sec and a car leaves the parking every 7 sec, how much time to fill parking?
401527
7 seconds or 3 seconds
979936
Let x=east, y=north, time=t in hours\nThe distance of the first car from the intersection as a function of time \n= 5-40t\nThe distance of the second car from the intersectionas a function of time t\n= 6+30t\n\nAs the two cars are travelling at 90 degrees of each other, they form a right triangle with the intersection, therefore using Pythagoras theorem,\nD=sqrt(x^2+y^2)\n=sqrt((5-40t)^2+(-6+30t)^2)\n=sqrt(2500t^2-760t+61)\n\nNote that the answer corresponds to the question you asked.\n\nYour given answer corresponds to distances from the intersection of 3 miles (instead of 5) and 2 miles (instead of 6) from the intersection.
393461
First put everything in the same units.\n\n36 km/h * 1000m/1km * 1hr/3600s = 10m/s\n\nOk...the equation for on objects position as a function of time is.\n\nx(t) = xo + v*t + a*t^2\n\nwhere \nxo = initial position (0 for both car and truck)\nv  = velocity (10 m/s for the truck)\na  = acceleration (4 m/s2 for the car)\nt  = time\nx(t) = position at any given time t\n\nIn question 1 you are looking for the time at which the car and the truck are at the same position.  Or the time at which x(t) for the car equals x(t) for the truck.  So set the position equations equal to each other and solve for t.\n\ncar equation:  x(t)car = 0 + 0 * t + 4 * t^2\ntruck equation: x(t)truck = 0 + 10 * t + 0 * t^2\n\ntherefore when they are at the same position\n\n4t^2 = 10t\n\nor\n\nt = 10/4s or 2.5s (note: the answer you have given is incorrect.  At 5 seconds the car would be 100 meters from the light and the truck would be 50 meters from the light)\n\nFor question 2 we need to take the derivative or the position equation for the car and solve for the velocity at time t = 2.5\n\nx(t)car = 4t^2\n\nd(x(t)car)/dt = v(t)car = 8t\n\nv(t)car (at t = 2.5) = 8*2.5 = 20 m/s  your answer for this question proves to be correct
724020
5 hour trip times 3 students standing at all time = 900 standing-minutes. 900 min divided by 19 students = 37 min 22 s.\nAt departure student 1 2 and 3 is standing. When a third of student 1's standing-time is over he/she sits down on student 4's seat, who just stood up. When two thirds of student 2's standing-time is over he/she sits down on student 5's seat, who just stood up. Student 3 stands for his fair share of time, and switch place with student 6. All students from #3 to #19 do their standingtime in a single streach. When student 17 is done he/she sits down on student 1's current seat. When student 18 is done he/she sits down on student 2's current seat, and when student 19 is done, they are at their destination!\n\nThe math teacher only has to shout "switch!" every 12min 42sec.\n\nI know you said no trickery, but she could also force the students to all stand for 5 straight hours...
822312
wish I could solve em for you but ill try and help still.\n\n1) The acceleration should be the derivative of the equation for the car's velocity.  The original distance/time equation would be the anti-derivative of the velocity.  \n\n2 The equation to find out how long it will take to get there is h+ax-16x^2 in feet.  (a is the original speed and x is number of seconds and h can be the height of the building.)\n\nHope this helps
14241
Because 33 sec < 43 sec, add 60 seconds and then subtract one minute from the first time (notice how that like adding nothing).\n3:15:33 -> 3:14:93\n3:14:93 - 3:14:43 = 0:0:50
108141
2.4\n\n75 miles/60 minutes   =   3 miles/X minutes\n\nCross multiply and solve for x:\n\n75X = (60 * 3)\nX = 2.4\n\nThe weight of the car is not relevant.
702945
this is gonna be impossible to prove , the fact is your car was there around that time , and you arguement that it was only a matter of minutes will work both ways , you say 6;59 she says 7;04 , if you go in saying you were there at 6;59 that is like admiting guilt ......... you are gona lose time from work going to court  and a letter is a laughable defense ....... just pay the cash and move the car at 6;30 next time
1009103
We have a formula:\nSpeed = Distance / Time\nhere Time= Distance /Speed.==D/S\nD=100km in both cases\nS1= 125km/hr; S2= 80km/hr\n\nT1=D/S1==100/125 hrs === 4/5hrs===4/5*60mins= 48mins\nT2=D/S2==100/80 hrs ===5/4 hrs ===5/4*60 mins= 75mins\n\nTime saved = T2-T1==75-48=27 mins.\n\nIs this clear buddy
1136388
a = gallons used by car 1\nb = gallons used by car 2\n\nYou can write two equations:\n1) 15a  + 40b = 2050\n2) a + b = 70\n\nMultiply #2 by 15 on both sides:\n3) 15a + 15b = 1050\n\nSubtract #3 from #1:\n4) 25b = 1000\n\nSolve for b:\n5) b = 40\n\nSolve for a (from #2)\n6) a = 30\n\nDouble check your answers:\nCar 1 goes 15 x 30 miles (450 miles)\nCar 2 goes 40 x 40 miles (1600 miles)\nTogether they go 2050 miles\n\nSo the answer is:\nFirst car 30 gallons\nSecond car 40 gallons
966246
d=r*t\nhighway= 60*highwaytime\ncity = 30*citytime\ncity = highway -  100\ncitytime =highway time - 5/6  50 min = 5/6 hour.\nsub in\n\nhighway - 100 = 30 (highwaytime- 5/6)\n60 * highwaytime - 100 = 30 * highwaytime - 30*5/6\nsimplify\n30* highwaytime = 75\nhighwaytime = 2 1/2 hours\nh  =2.5*60 = 150 miles\ncity =30* 1 2/3 = 50 miles\ntotal 200 miles
741402
a) F = ma \nYou're given F = -6070 N (negative because it's in the opposite direction) and m = 2600 so a = -2.33 ms^-2. Then over 1.76s, the change in velocity is (-2.33)(1.76) = -4.10 m/s. Car velocity is then 34.9 - 4.10 = 30.8 m/s.\n\nb) v^2 = u^2 + 2as\nv = final velocity = 30.8 m/s (from above)\nu = initial velocity = 34.9 m/s\na = acceleration = -2.33 ms^2 (from above)\nRearrange to get s = (v^2 - u^2)/2a = 155.61 m\n\nc) v = u + at\nYou want v = 0. So if the force is constant then acceleration is constant at 2.33ms^-2. \nu = initial velocity = 34.9 m/s\nSo rearrange to solve for time: (v-u)/a = t = 14.98 seconds.\n\nHope that helps. And learn your equations!
959780
Well, the car travels 20 m in 1 sec. There are 60 seconds in one minute, so multiply 20 m by 60 to find out how far the car travels in one minute.\n\n20 x 60 = 1200\n\nSo the car travels 1200 m in one minute. Multiply that by 15 to find out how far it travels in 15 minutes.\n\n1200 x 15 = 18000 m
109042
13 and a third hours is the number i came up with...\n\nJust plug some fake numbers in, say, for example, the tank that these two pipes are filling is 500 gallans.  then you can figure out how much each pipe put into it, take that five hundred, divide by 8, then multiply by 3, that's how many gallens the smaller pipe put into the tank in 5 hours...   now, take that number and divide by the 5 hours, that's how much that smaller pipe can pour out into the tank in one hour...  to take that 500 gallens and divide by the amount that can be poured in one hour and it'll give you how many hours it'd take to fill at that slower rate...\n\nThis is not a standard way of figuring out this problem, but it seemed like it worked to me, it should have taken more then double the 5 hours, but less then 3 times...   does this help at all??
791778
20meters/5 minutes = x meters/60 minutes\n\nx = 1200/5\nx = 240m / hour\nx = 0.240 kph\n\nIt's a really slow car only going less than a 1/4 km per hour.  I can *walk* faster than that!
139303
the physics formula that can be applied here is:\n\nvt = x, where v = speed tub is being filled, t = minutes, x = number of tubs filled.\n\nsince there are three valves, each contributes to the filling/emptying of the tub.  thus, v = valve1 + valve2+ valve3.\n\nt is unknown, the minutes we are solving for.\n\nx = 1, since we want the time that it takes to fill 1 tub, no more.  putting these all into the equation:\n\n[valve1 + valve2 + valve3]t = 1 tub\n\n[(1/3)tub/min. + (1/5)tub/min. - (2/15)tub/min.]t = 1 tub\n\n[(2/5)tub/min.]t = 1 tub\n\nt = (5/2)min. = 2.5 minutes\n\nchecking the answer, in 2.5 minutes, valve 1 will fill the tub 5/6 of the way, valve 2 will fill it 1/2 of the way, and valve 3 will empty it 1/3 of the way.  (5/6) + (1/2) - (1/3) = 1, therefore our answer works!\n\nt = 2.5min.
544642
3 feet per hour\n3 1/3 city blocks per hour\n62 miles per hour