a man ran to a phone and back in 15 minutes the speed to the phone is 5m/s and back is 4m/s find the distance?
477042
Assuming that the initial distance between the man and the phone is x:\nTime to get to the phone = distance/speed = x/5\nTime to get back = distance/ speed = x/4\n=> 15 * 60 (time in seconds) = x/4 + x/5\n900 = 9x / 20\n=> x = 2000 m
836862
distancetravelled by the bird=2000m\ntime taken=187 1/2 sec\n                 =375/2 sec\nspeed is in the ratio of 1:2\ndividing the time in the ratio of 1:2\ntime to=375/2(2/3)=125 sec\nspeed=distane/time=1000/125=8 m/sec\nspeed in thefro direction=16m/sec\ncheck\n1000/8+1000/16=3000/16\n=187.5 sec checks
437647
Let's think about how to solve these and maybe that will help you with others,\n\n1. Christian makes the trip twice as fast as Eloise.  Therefore his trip would take half the time.  If his trip was three hours shorter, then 3 hours is half the time.  His trip was 3 hours, Eloises was 6 hours.  So if Christian traveled at 30 mph for 3 hours, and Eloise traveled at 15mph for 6 hours .. how far is that?\n\n[modified below with details for additional answers]\n2. Basically \n4mph x SomeTime = SomeDistance\n5mph x (Sometime-1 hour) = SomeDistance\n\nor, \nif 4x=y and 5(x-1)=y\nthen,\n4x=5x-5\n4x-5x=-5\n-x=-5\nx=5  (this is the SomeTime)\n\ngoing back to the first equation\n4mph x SomeTime = SomeDistance\n4mph x 5hours = 20 Miles\n\ndoublecheck with the second equation,\n5mph x (Sometime-1hour) = SomeDistance\n5mph x (5hours-1hour) = 5mph x 4hours = 20 miles\n\n\n3. If the fast freight was going 60mph for 6 hours (noon to 6pm), then the whole trip is how far? .. 360 miles.  If the slow freight takes 8 hours to go 360 miles, then what is the speed of the slow freight?  360mph/8m = 45 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
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
334516
Well, 60 minutes = 3600 seconds\nIf he moves at 5 meters per second, he'll travel 18000 meters (5*3600) or 18 kilometers.
2556
You're really close! The answer is actually 1:03:11. You were off one minute because you forgot to borrow a minute from the 34 so you could subtract the seconds.
884244
a. 1.10 MJ\n\nEnergy = mass * acceleration * distance = m * a * s\n\nm = 1600 kg\nAssume that the car is accelerating at 1 m/s^2 =>  a = 1\n\nIt would take 25 seconds for a car accelerating at 1 m/s^2 to go from 15 m/s to 40 m/s.\n\nThe average velocity during the acceleration is (15 + 40)/2 = 27.5 m/s.\n\nThe distance, s, is 27.5 m/s * 25 seconds = 687.5 m\n\nm*s*a = 1600 kg * 1 m/s^2 * 687.5 m = 1.1 e6 J = 1.1 MJ\n\nIf you assume another acceleration (let's say 5 m/s^2) you will get the same because the same amount of work is being done.
557602
This is how you solve it. Use dimmensional canellation, but you will have to do it twice and subtract the difference\n\nA. (5 miles/1)  *   (1 hr/55 mi)  * (60 min/1 hr)\n   =5.45 minutes\n \n   \nB. (5 miles/1)*  (1 hr/65 miles) * (60 min/1 hr)\n   =4.6 minutes\n    \nYou saved 1.2 minutes. Not much at all. Which is what they say, speeding only saved 2 or so minutes in a 15-20 minute trip. BUt it tears the bejesus out of your engine
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
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
963172
average speed =total distance/total time\n1.\ntotal distance=2km\ntotal time=1/2hr\naverage speed =4km/hr\n2.\ntotal distance=360km\ntotal time=5hr\naverage speed =360/5=72km/hr\nthanks & good luck
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.♥
661447
The difference in rates is 1 km/hr. So if they are moving in the same direction, the faster walker will be 5 km ahead at the end of 5 hours.\n\nThe sum of the distances traveled per hour is 11 km. So at the end of 5 hours of travel in opposite directions, they will be 55 km apart.
253105
well... man is jogging at 10 km/h... and yhe distance between A and B is 25 km.... then the man is jogging for\nt = (25 km)/(10 km/h) = 2.5 h\n-------\nNow the distance that bird makes in that time....\nd = (45 km/h)*(2.5 h) = 112.5 km
976772
15 / 3.5 = 4 with remainder of 1\n\nTherefore the leftover piece is 1 meter long
732600
SRY!! Matt is going to a basketball game out of town. If he could travel 75mph, He would arrive one hour early. If he could travel 45mph he arrives one hour too Late. How far is Matt's trip\n\ndis=75*(time-1)\ndis=45*(time+1)\n\nthen\ndis= 225mile,    time=4hour