Distance, Rate, Time equations!?
816384
Distance = Speed * Time.\nWe don't know the "distance" variable for either. We need to find the "time" at which the "distance" will be the same. Let distance = d\n\nWe want to find Thomas's time, so let Thomas's time = t.\nAlexis left 3 hours earlier, so her time is t + 3.\n\nThomas speed is 30, Alexis is 20. So lets fill this in.\nThomas:\nD = 30t\nAlexis:\nD = 20(t+3)\n\nThe distance will be the same, so the D in both equations is the same variable. A basic principle is that if a = b and b = c, then a = c.  Therefore, we can combine the equations for the two people as follows:\n30t = 20(t+3)\n30t = 20t + 60\n10t = 60\nt = 6 hrs\n\n\nThe same applies to the second.\nTimothy:\nD = 58t\nSavanah:\nD = 47(t+2)\n\n58t = 47(t+2)\n58t = 47t + 94\n11t = 94\nt = 8 6/11 OR 8.5454545454545\nt = 8 hours, 32 minutes, 43 seconds\n\n\n\nThe last one is slightly different.\nJet 1 Distance = 331t\nJet 2 Distance = 333t\nThe two distances aren't equal this time, but the two distances added up will equal 1,705, SO:\n331t + 333t = 1705\n664t = 1705\nt = 1705/664 = 2 377/664 = 2.56777108\nt = 2 hrs, 34 minutes, 4 seconds
751006
speed = distance / time\n---rearange this----\n\nyou get ----->  time = distance / speed\n\ntime = (1.8 * 800) / 180\n\ntime = 1440 / 180\n\ntime = 8 hours\n\na looonnnng time for a bunch of fat men in a bus
832829
Speed = distance/time\n\nlet x be the time it takes to get to school on time\n\n3=d/(x+15/60)                                      4=d/(x-15/60)\nd=3x+3/4                                         d=4x-1\n\ntherefore 3x+3/4=4x-1\nx=1  3/4 = 7/4\nHence distance=d=3(7/4) + 3/4 = 6\nTherefore the school is 6 km away
953778
Let's call the entire track length x, measured in meters.\n\nTo arrive at the time, we have to divide the distance by the speed:\n\nx / 8.5 + x/8.0 = 99\n\nmultiply all sides by 8.5 and by 8.0\n\n8.0 X + 8.5 X = 8.0 * 8.5 *99\n\n16,5 X = 8*8.5*99\n\nX = 8*8.5*99/16.5\n\nTime for the first lap: \nT1 = X / 8.5 s\nT1 = 8*99/16.5 s\n\nT2 = X / 8.0 S\nT2 = 8.5*99/16.5 s\n\n\ncalculator time....\n\nSolution D
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.
160821
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s
806819
A.\n\nF=ma, so, a= F/m\n\nIn this case, a=7/5ms^2\n\nB.\n\nv=u+at, u=0 (initial velovity), t = 2 sec, so\n\nv=0+7/5 x 2 metres per second = 14/5 m/s\n\nC.\n\nThe time to reach Sasha will be:\n\nt = d/v = 11.2/(14/5) seconds\n\nI will let you use your calculator!
325281
OK, this is a simple unit conversion problem, buddy. First, get your measurements into similar units: in this case, change your minutes into hours (1 hr/60 min). You'll end up with.\n\n(40 km/h * .5 h) + (60 km/h * .25 h)\n\nThe hours will cancel out, and you'll be left with the kilometers.\n\n20 km + 15 km = 35 km\n\nIn this instance, the direction of travel doesn't matter: the question asks the distance traveled, not how far away from the point of origin you are.
232038
Distance = rate x time\n\nif you're going 8 feet per second for 4 seconds, you've gone 32 feet...
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
297241
Average rate per hour = total distance / total time\n=( x + x )/ total time          (1)\nTime required to travel x distance at the speed of 10 miles\nper hour = x / 10 hours.\nTime required to travel x distance at the speed of 8 miles\nper hour = x / 8 hours.\nSo the total time is\n=(x / 10) + (x / 8)\n=18x/80 \n=\nPutiing this value in the denominator of (1)\nWe get\nAverage rate per hour = total distance / total time\n= 2x/[18x/80]\n=80/9\n=8.88\n=8.9 miles/hr\n\n( In general if the speeds are 'a' & 'b', the the average \nis given by their harmonic mean viz. 2*a*b/(a+b) )
923403
Start off by drawing the picture. You have a right triangle, with a hypotenuse of 25, a base (b), and height (h). \n\nDon't get jumbled down by numbers yet, ignore them. Visualize what you are calculating.\n\nVelocity / speed is the derivitive of position. (v = delta x / delta t). So all you need to do is find an eqation relating the geometry using known and unknown quantities and then differentiate with respect to time. Sounds confusing but all you have to do is....\n\nYou know, a^2+b^2 = c^2 for a right triangle. In our problem,\n\nb^2+h^2 = 25^2\n\nThis is our position equation. Now differentiate with respect to time.  DO NOT forget the CHAIN RULE! <------ VERY important\n\n2*b*db/dt +2*h*dh/dt = 0.\n\nFrom here just use algebra; where,\n\nb=7 ft, and db/dt = 2 ft/sc.\n\nWhich leaves two equations and two unknowns.\n\nThe important thing to remember is to take this step by step and try to visualize what you are calculating. Also, looking at the units of the given quatitities can give you some idea of what you will be required to do. \n\nAlso, just remeber that speed / velocity is the derivative of position with respect to time (the units reflect this).
399954
SPEED=dy/dt=distance/time.its not exactly something you invent.it was an axiom,a basis of understanding.
830380
1.time taken=distance/speed=26/43 hours\n 2.time taken=distance/speed=68/54\n=34/27hr\n=1 7/27 hr
1006215
Use the formula d = rt (distance = rate * time)\n\n1) Train A: d = 40t\n\n2) Train B: d = 60(t - 1)  [left an hour later]\n\nthe answer is when the two distances are the same, so we set the two equations equal to each other\n\n40t = 60(t - 1)\n40t = 60t - 60\n60 = 20t\n3 = t   [3 hours]\n\nPut this result back into equation 1\n\nd = 40 * 3 = 120\n\nAnd we expect the same result in equation 2\n\nd = 60(3 - 1) = 60 * 2 = 120\n\nI provide this level of detail in case you need this level of help.
712175
Distance = Rate * Time\n\nYour first question does not make sense.  you are telling us that they will streak around a 1.5 mile track at 15 mph for 15 minutes and then asking us how long it would take.  You just told us 15 minutes.\n\nif you want to know "how far would they have traveled in 15 minutes at a rate of 15 mph on a 1.5 mile track" you can find out by:\n\nthe answer is 3.75 miles.\n\nD=RT\nD = 15 miles per hour * 15 minutes\nfirst you have make sure you are using the same units for time, either use minutes or hour for both the rate and time.\n\nI chose to change the 15 minutes into hours which = 1/4 of an hour.  use a ratio:  of minutes to hours and reduce 15/60 = 1/4\n\nthen multiply  15 m/h * 1/4h and you end up with 3.75 miles.\n\nwhen you say 15 miles per hour it = 15miles/1 hour in fraction form.\n\nwhen you multiply 15 miles / 1 hour * 1 hour /4  you cancel out the hours units and you are left with only miles.\n\nIf you want to know how long it would take them to travel one time around the track:\n\nD = RT\nT= D/R\n\nTime = 1.5 miles / (15 miles / 1 hour)\n\nRemember when you divide fractions you invert the second one and them multiply them:\n\nTime = 1.5 miles * 1 hour / 15 miles\nmultiply and cnacel out the miles unit and you get:\n(1.5 * 1) / 15\n1.5/15\n= .1 hours\n\nchange .1 hours into minutes:\n1 hour = 60 minutes so:\n.1 * 60 = 6 minutes\n\nit takes them 6 minutes to do a lap, (which is crazy for someone that size)
867482
Let t be the time, the same time for both then because distance is rate times time the equation is\n\n90t + 80t = 425\n170 t = 425\nso t = 2.5 hours