Find the average speed?
544642
3 feet per hour\n3 1/3 city blocks per hour\n62 miles per hour
900973
At noon\nA = (0, 0), B = (-90, 0)\n\nAt time t hours\nA = (0, 40t) B = (-90 + 60t, 0)\n\nSo |AB| = √[(40t)² + (-90 + 60t)²]\n= 10√[16t² + 81 - 108t + 36t²]\n= 10√[52t² - 108t + 81]\nd|AB|/dt = 5(104t - 108)/√[52t² - 108t + 81]\n\nWhen t = 1 d|AB|/dt = 5 * (-4)/√[52 - 108 + 81]\n= -20/√25\n= -4 mph ie they are getting closer at a speed of 4 mph
899910
Speed(S) = distance(D) / time(t) .... the basic equation \n\n45 = D / t \nthus,  D = 45t\n\nthey are already 50 miles away from home \n \nso 50 miles should be added to the d that we have found out above \n\nd = 50 + 45t
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
757576
To solve this problem, you need to find the time the bus needed to travel the 200 miles.  You should divide this problem into 3 intervals:\n\n1) 150 MPH for first 50 miles\n2) 125 MPH for 2 minutes\n3) 150 MPH for reaminder of trip\n\nFor interval 1, we need to caclulate the time taken to travel 50 miles:\n\n(50 miles)(1 hour / 150 miles) = 1/3 hour = 20 min\n\nFor interval 2, we needed to calculate the distance traveled in order to solve for interval 3:\n\n(125 miles / 1 hour)(2/60 hour) = 25/6 miles\n\nNow, we can find the # of miles for interval 3:\n\n# miles = 200 - (50 + 25/6) = 875/6 miles \n\nNow we can find the time needed to travel this distance:\n\n(875/6 miles)(1 hour / 150 miles) = 35/36 hour (around58 min)\n\n---------\n\nTo find average speed, you need to take [total distance/ time].  therefore,\n\navg. speed = [200 miles]/[(1/3) + (1/30) + (35/36) hr] ------->\n\n149.38 miles per hour\n\n------------\n\nHope this helps
1073159
90 mph. 90+30= 120 speed / 2 LAPS = 60 mph average
995779
To all those who answered 90 mph - BE MORE CAREFUL!\n\nMath Teacher
448267
the average orbital velocity of earth is 29.783 km/s.\n\nhttp://en.wikipedia.org/wiki/Earth
932092
Something's wrong with the writeup for the first question.  I'm guessing it should have said THR = RHR + 0.65(MHR-RHR), because otherwise you're gonna get a THR of, like, 10000 beats per minute.\n\nProceeding from that assumption, the THR is RHR + 0.65(MHR-RHR).  You're told that the RHR is 54 bpm.  The MHR is 220 - the age(43), so it's 177.  Therefore the THR is 54 + 0.65(177-54) = 54 + 0.65(123) = 54 + 79.95 = 133.95 bpm.\n\nIn the second question, you didn't provide the "following equations," but think of it this way.  The airplane flew for a total of 15 hours at an average speed of 350 miles per hour.  That means it went 5250 miles.  In the final 7 hours it went 2275 miles.  So in the first 8 hours it had to go 2975 miles.  2975 miles/8 hours = 371.875 mph.  I don't exactly know how to write that in an equation, but now that you understand the reasoning, hopefully you'll be able to recognize the correct equation.  If all else fails, plug the data into each equation and see which one returns 371.875.
696298
1st way :-\ndistance = speed x time = 55 mph x 3 h = 165 m\n\n2nd way :-\ngiven that d = 55t , so on putting t = 3hrs , we get d = 165 m
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) )
754657
The time taken is (200 miles divided by 185 mph) plus 5 minutes.  Divide 200 miles by that time to get the average speed.  Obviously, it's going to be less than 185 mph by around 1/12th
960583
plz find car position and after \nincrement in car position and car is moving
128700
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
847424
d= v .t\n> d = 50 km/hr (18 min)\n>d = 50 km/60 min (18min)\nd= 50 km (18)/60\nd =900 km/60\nd = 15 km >>the distance travelled\n\nIf you walk at an average of 3 miles per hour (approximately 4.8 km/hr) at a distance of 15 km:\nt= d/v\n=15 km/4.8 km/hr\nt = 3.125 hrs or (3 hours and 7.5 minutes)>>the time you need to walk on 15 km at the speed of 4.8 km/hr.♥
1178980
Fastest ------- 6.11 km/sec (13,667 mph)\nSlowest ------ 3.67 km/sec  (8,209 mph)\nAverage ------ 4.74 km/sec  (10,603 mph)
741279
Total distance traveled divided by total time in transit.\n\nr = d/t.