A man ran to a telephone and back in 15 minutes. His speed on the way to the telephone is 5m/s and his speed on the way back is 4m/s. Find the distance to the telephone?\n\nnote:( please state each step you did in your calculations get your answer).
216632
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
515038
Ok. 45 minutes is equal to 3/4 of an hour. To find the distance Travis has travelled at that point in time, you will need to multiply 55 (his rate) by 3/4. That is equal to 41.25 miles. Now we need to find if Roy will ever catch up to him. We are trying to find when their distances are equal, so time can be our variable. \n\n41.25 + 55t = 65 t \n41.25 = 10t\nt = 4.125 hours\nNow, if we plug that in to 41.25 + 55 t and it is greater than 450, then Roy can't reach Travis before Travis reaches Pecos. \n41.25 + 55 (4.125) = 268.125\n4.125 hours after Roy leaves Dallas, he will catch up to Travis 268.125 miles from Dallas.
352035
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)
501337
Horizontal component of distance is x=cos(45)(30 m/s)(time)\n\nVertical component of distance is y=sin(45)(30 m/s)(time) - (1/2)(9.81 m/s^2)(time)^2\n\nFirst solve for how long it will be in the air. It start 2 meters off the ground so we must solve for time when y = -2 (that's when it will hit the ground):\n\n-2=sin(45)(30 m/s)(t) - (1/2)(9.81 m/s^2)(t)^2\nt= about 4.417 sec\n\nNow we find horizontal distance when the guy hits the ground:\n\nx=cos(45)(30 m/s)(4.417 sec)\nx= about 93.699 meters\n\nThat's about one football field!\n\nNow we must use the pythagorean theorem to add the distances. He want 2 meters down and 93.699 meters out.\n\nsqrt(2^2+93.699^2) = 93.720 meters is the total distance.
120528
This was the best question I have seen on this site.  (That isn't saying much, so let me say it this way:  This is an excellent question!)\n\nThe answer is 100 steps. (Not considering the hidden steps since it's constantly revolving).\n\nFirst we have to consider the ACTUAL speed of each man since the escalator is moving, too.\n\n(Fast Speed) + (Escalator Speed) = (Fast Adj. Speed)\n(Slow Speed) + (Escalator Speed) = (Slow Adj. Speed)\n\nAnalyzing that if the fast man had been able to climb the entire time that the slow man did, he would have gone 150 steps, we can see that he finished in half the time it took the slow man (because 75/150 = 1/2).  Therefore we also have:\n\n2 (Fast Time) = (Slow Time)\n\nWe also use the simple equation defining speed (s=d/t) to determine that:\n\n(Fast Adjusted Speed) = (Distance) / (Fast Time)\n(Slow Adjusted Speed) = (Distance) / (Slow Time)\n\nwhere (Distance) is equal to the actual distance traveled (eg. the distance from the top of the escalator to the bottom of the escalator).\n\nInserting the few givens and abbreviating the variables, we have, respectively:\n\nEq 1) 3 + ES = FAS\nEq 2) 1 + ES = SAS\nEq 3) 2 FT = ST\nEq 4) FAS = D/FT\nEq 5) SAS = D/ST\n\nThis looks like 6 variables with only 5 equations, but it is still actually solvable.\nUsing Eq.3 to simplify Eq.4 and Eq.5:\n\nEq 4) FAS = D/FT\nEq 5) SAS = D/2FT\n\nand we can reduce our number of variables by simply saying that X = D/FT, hence,\n\nEq 4) FAS = X\nEq 5) SAS = X/2\n\nSo now we use these to analyze Eq.1 and Eq.2:\n\nEq 1) 3 + E = X\nEq 2) 1 + E = X/2\n\nTwo variables, two equations.  E turns out to be 1.  This means that the escalator moves at the same speed as the slow man (remember that we assigned his abitrary, but proportional speed to be 1 in the equations).  If he and the escalator are both the same speed, and he contributes 50 steps to the journey, then the escalator must also have contributed 50 steps, totalling 100 steps.\n\nOr guess and check works pretty well, too.
344633
Let\nx = speed of Sachiko\nx+5 = speed of Jose\n\nSince they travel a distance of 476 miles in opposite direction\nthen,\n\nd(Sachiko) + d(Jose) = 476\ns(Sachiko)t + s(Jose)t = 476\n(x)(4) + (X+5)(4) = 476\n4x + 4x + 20 = 476\n8x = 456\nx = 57 \nx+5= 62\n\nTherefore,\nspeed of Sachiko = 57 miles per hour\nspeed of Jose = 62 miles per hour
440053
Let x be the distance he walked.\n\ntime = distance/speed\n\nso time walked + time in truck = 10 hours\n\nx/5 + x/20 = 10\n\nmultiply by 20\n\n4x + x = 200\n5x = 200\nx= 40 km\n\nCheck - he spent 40/5 = 8 hrs walking out, and 40/20 = 2 hours walking back for a total of 10 hrs.
18520
distance = rate x time, \n\nLet x be the distance she can go out on the lake, and t the time it takes to get there.\nGoing out she paddles a 3km/h so d = 3 x t\nComing back, the distance is the same, the rate is 2 km/h\nand the time she has is whatever is left of the 2.5 hrs.\nShe use t amount of time going out so (2.5 -t)\n\nGoing out   d = 3 x t\nComing back d = 2 x ( 2.5 - t)\nThe distance is the same both ways, so \n3 x t = 2 x (2.5 - t), now solve for t and then figure out d.
112295
3.84*10^5 km ( 1000m /  1km ) =  384000000m\n\n\n  distance / rate = time:\n\n   384000000m/3.0*10^8ms = 1.28sec
10839
pier_haringsma is correct.  You do not need any additional information.  Let \nD = total distance of the trip  \nD/2 = the distance the driver has gone so far\nT1 = amount of time he took to travel the first half of the trip\nT2 = the complete length of time of the whole trip (both halves).\n\n   Now, he wants the average ( D / T2 ) to equal 30.  His average so far is ( (D/2) / T1 ) = 15.\n\nD/T2 = 30, so D = 30*T2\n(D/2) / T1 = 15, so D = 30*T1\n\nD = 30*T1 = 30*T2, so T1 = T2.\n\n  So the driver wants the total time for the trip (BOTH halves), T2, to equal the time for the first half of the trip, T1.  This means he would have to travel the second half of the trip in no time, which is impossible.
537905
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
195869
The train has to travel 0.17 km for the head to pass the train and another 0.11 km to let the whole train pass the other train completely. So the train has to travel a total of 0.28 km. By this we can already solve the problem.\n\n(0.28 km)/(60 kph)\n\nThe kilometer unit will be cancelled giving us the total time. The answer is 0.00467 h or 0.28 min or 16. 8 sec approximately.
518923
My answer to the first part, by conservation of momentum, is that the speed equals 1.41m/s
297281
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.
175947
Part 1: Mike\nWork=Force x Distance(moved) x cos (theta) (angle between work and distance)\nF=215\nD=64.7\nTheta=35 degrees\n\nW=215*64.7*cos35\nW=11,394.815 Joules\n\nPart 2: Chris\nF=8N\nD=3.9\nTheta=90 degrees\nW=8*3.9*cos90\nW=0\n\nThats all you get...good luck
514565
1. d=t, any form of d=t+b would be parallel. b = any number. That is a line.\n\n2. d=t would also work here.\n\n3. Well if you go 60 MPH and you wnat to know how many miles a minute you can go convert using (60miles/hour)*(1hour/60minutes) and you get 1mile/minute. If you need feet, 1 mile is 5280 feet, so 5280 feet/minute
433053
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