I just through up this question, answer if and show your working. (I know the answer BTW, but i just wanna see how clever people are, here)\n\nA man leaves home at 8:00 a.m. and travels towards southampton which is 146 miles away. At 10:30 a.m. he stops for a cup of coffee and at that point his average speed for the journey so far is 48 mph. How far is he from southampton?
204384
Speed = Distance/Time\nDistance = Speed x Time\nDistance traversed = 48 x 2.5 = 120 miles\nDistance Remaining - 146 - 120 = 26 miles.
538387
1) two hours from 9 to 11. so distance 45*2 = 90 km\n\n2) The car has done 100+108= 208 km in 5 hours\nthe speed is 208/5 = 41.6 kph
259698
As you can tell, there's not really enough info within the problem itself to solve it.  You'll need to get a map, atlas, or some internet source to determine how much of the 481 miles will be driven in Oregon, and how much in California.  I won't look that up for you, but let's pretend that it's 121 miles in Oregon, and the other 481-121=360 miles in Cali.\n\nAssuming you constantly maintain the speeds indicated, you can calculate the time it takes you for each leg of the trip by using the formula:\n\n   distance = speed x time\n\nOr, equivalently,\n\n   time = distance / speed\n\nSo, for the 121 miles in Oregon, over which you will be traveling 65 mph,\n\n   time = 121 miles / 65 mph = (approximately) 1.86 hours\n\nand for the 360 miles in California, over which you will be traveling 70 mph,\n\n   time = 360 miles / 70 mph = (approx.) 5.14 hours\n\nAltogether, then, you'd be traveling 1.86 + 5.14 = (approx.) 7.00 hours.\n\nIf you want to write that out as one long problem, you could write it this way:\n\nTotal Time = Time1 + Time 2\n           = Dist1/Speed1 + Dist2/Speed2\n           = 121 mi/65 mph + 360 mi/70 mph\n           ~ 1.86 hr + 5.14 hr\n           ~ 7.00 hr\n\nNow, don't forget, I picked the distance values, 121 and 360 miles, out of thin air.  You'll need to look them up and do the problem yourself.  But hopefully that should give you an idea of how.  Good luck!
27559
distance is speed multiplied by time.\nexample: you travelled for one hour (t=1) and your driving speed was 70mph (s=70)\nThen, D=70m/h x 1h (the units "h" get cancelled out)\nANSWER: 70m\nYou went a distance of 70 miles
504455
Let's assume your range of vision was 0.2 miles infront of you, and 0.1 miles behind with the mirror.  (I'm pulling these numbers out of my behind.)\n\nAt t=0, the car was at position x = -0.1 relative to you.\nAt t=5 seconds, it was at x = 0.2.\n\nSo the speed (relative to you) is 0.3 miles / 5 seconds = 0.06 miles per second = 216 miles per hour.  Since you were doing 85 mph, it real speed was 301 mph.\n\nAre you sure this took only 5 seconds?  I picked some pretty small numbers for the range of vision and this still results in some wicked speed.
658118
The width of the river is 1 mile =  1760yards\n\nYes, let A be the faster boat leaving from shore A, and B the slower boat leaving from shore B. A will a travel a distance (720+x) yards from shore A to where they first meet while B travels 720 yards from shore B. A reaches shore B first and waits10min and then starts back, while B reaches shore A later and does the same thing and they meet 400 yards from shore A. Since they wait the same time and keep their speed constant we must have 720-400 = 320 = x. So the river = 720+720+320 = 1760 yards.\n\nIt's easy to see if you eliminate the wait time and imagine A turns around as soon as he reaches shore B.  A will be distance x from shore B on the return trip when B is distance x from reaching shore A. So they must still travel the width of the river to meet 400 yards from shore A. So we must have x+400 = 720 as the distance B travels to reach shore A and then return to the 2nd meeting point. So x = 320 and the river is a mile wide,
39489
A would take 22 minutes to get to 15 miles and 32 minutes to get to 20 miles. Unfortunately, there's insufficient information to be precise on B's journey. Did he spin off before the start?.. or at which point en route? Guessing it, I would say A arrived at 15 miles and at the 20 miles point first. NB. I have had to make certain assumptions in your question to arrive at this answer (eg I asumed A slows down to 40mph after 4 miles?)
403122
I solved the problem by stating that P would travel distance d in time t\nd=20*t+.5*.3*t*t\n\nand Q would travel d+20 in time t\nt is the time to collision\n\nd+20=30*t\nd=30*t-20\nSubtract the equations and solve for t:\n\n0=.15*t*t-10t+20\nt=2.064 seconds\n(the second root is superfluous)\nd=41.9m\n\nWhich proves collision\n\nj
117916
ok well his initial velocity is 0 and the distance is 402m and time is 4.913. you have to find the acceleration. well the only way i can think of doing this  without using these basic equations that are derived from newtons law (i forgot the derivation) is this.     you find the average velocity needed to go 402m in 4.913s. which is just 402m/4.913s = 81.823....\nthis is just the average velocity and because acceleration is constant, this means you can just find the average by doing  Vi + Vf / 2. you have Vi =0    so Vf/2 = 81.823........ so Vf = 163.6474.........m/s\nacceleration is  change in velocity /time. you have change in velocity  (Vf-Vi)   so divide it by time.   \n163.6474-0/4.913   =  33.3 m/s^2\n\n2. so the acceleration due to gravity is 9.8m/s^2. you know the distance is 10m and initial velocity is 0. so Vf = a x t\nso Vf = 9.8t. and Vf+Vi/2 = d/t    same things basically. so Vf/2 = 10/t  which means  Vf x t = 20\nVf = 9.8t\nVf x t = 20\nso then that means that (9.8t) x t = 20\n 9.8t^2 = 20\nt^2 = 2.04....\nt = 1.4285.....  so it takes her that long\nand now you just find Vf by putting this time in the equation  Vf = 9.8t = 14m/s. thats final velocity\n\nb. ok so this question confused me at first but it snot hard. once she hits the water (with speed of 14m/s), she travels for 2m before coming to a stop. so distance is 2m. Vf = 0. Vi = 14m/s and you have to find acceleration. \nso a = (Vf-Vi) x t = -14t\nso you need time now. you do average velocity  again\nVi + Vf / 2 = d/t\n14/2 = 2/t\n7 = 2/t\nt = 0.2857.... s\nthen put it in the other equation and you get a = -14t\nso a = -4m/s^2.\nforce = mass x acceleration\nmass is 61kg so you do 61kg x -4 = -244 N\nthe water exerted 244N against the movement (thts why its negative).\n\nso hope this helped. let me know if you need more help or whatever. laterz
461787
Rate * Time = Distance\n\n2x = Rate (x) * Time (2) for first two hours\nx-10 = Rate (x-10) * Time (1) for third hour\n\n2x + (x-10) = 152\n3x-10 = 152\n3x =162\nx = 54\n\nThe correct answer is B. 54mph
445431
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.
171655
you can try somthing like...if one train starts in los angeles and heading to new york traveling at 55mph and a train leaving new york is heading to los angeles traveling at 67mph. if both trains leave at 12pm from thier starting point, at what time will they pass each other? (assuming the trains are on two different parellel traacks next to each other and the distance between them is 3000 miles).
390581
D = distance in feet\nt = time in seconds\nS = speed in miles per hour\n\n[D feet x (1 mile/5280 feet)] / [t seconds x (1 hour/3600 seconds)]\n= S miles per hour\n\nDoing the arithmetic, we get:\n0.6818 D / t = S mph\n\nIf D is 150 feet and t is 4 seconds, \n0.6818 (150)/ 4 = 25.56 mph\n\nIf the time is 8 seconds, the speed would be half as much, or \n12.78 mph
402576
try: \n\nF=m*a =>     0.3*M*g = M*a    =>    a=0.3*g\n\na*t = 60 miles/hr   =>    \nt = 60miles/hr / (0.3 g) = (60 miles per hour) / (0.3 * 9.8 ((m / s) / s)) = 9.12 seconds\n\ndistance\nD=1/2*a*t^2 = ((0.5 * 0.3 * 9.8 * (m / s)) / s) * ((9.12 sec)^2) = 122.266368 meters
469570
Let D = distance between home and school.\n\nUsing the formula : distance = speed * time, we have :\n\nFrom home to school : D = x * T1\nFrom school to home : D = kx * T2\n\nTherefore, x * T1 = kx * T2\nDividing through by x gives :\nT1 = k * T2\n\nWe know that T1 + T2 = 1 (total time travelled)\nTherefore, T2 = 1 - T1.\n\nSubstituting this into T1 = k * T2 gives :\nT1 = k (1 - T1)\n\nFrom this, we find that :\nT1 = k / (k + 1)\n\nSubstituting into 'home to school equation' gives :\nD = x * k / (k + 1)\n\nTherefore total distance covered\n= 2D\n= 2 * x * k / (k + 1) miles.
440882
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
421542
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!