2 Math questions! =)(Finding average speeds)?
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
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!
115023
Imagine a `speed` x `time` chart. Plotting the data, we will have parallelogram whose area will the length the car walked between the two speeds.\n\nWe'll have, then, 30*((5.56+13.89)/2) = 291.75m
966205
(a) too trivial to answer\n(b) Take the average speed times the number of seconds over which it had that average speed.
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.
742241
The second problem is easier than the first problem, so I'll tackle it first.\n\nLet's imagine that the trip is 120 miles. The distance doesn't matter if the average rates are as given above.\n\nIt takes 3 hours to travel 120 miles at 40 miles per hour.\nIt takes 2 hours to travel 120 miles at 60 miles per hour.\n\nSo it took 5 hours to travel a total of 240 miles.\n\nThis is an average rate of 48 miles per hour.\n\nI'll work on the first problem and hopefully I'll get back to you.\n\nUPDATE\n\nI think I have the first problem solved as well. It involves an infinite series! Call the people involved A and B.\n\nThe probability that A wins on the first toss is 1/2.\n\nIf A doesn't win on the first toss, then the probability that B wins on the second toss is also 1/2. But this only happens in 1/2 of all games. (In the other half, A won on the first toss.)\n\nIf B doesn't win on the second toss, the the probability that A wins on the third toss is also 1/2. But this only happens in 1/4 of all games (In half the other games, A won on the first toss. In half the remaining games - 1/4 of all games, B won on the second toss.\n\nFollowing this logic, A wins:\n\n1/2 + 1/4(1/2) + 1/16(1/2) + 1/64(1/2) + ...\n\nof all games.\n\nB wins:\n\n1/2(1/2) + 1/8(1/2) + 1/32(1/2) + ...\n\nof all games.\n\nLooking again at A, we have:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...)\n\nThe term in parentheses is an infinite series having initial term 1 and ratio 1/4.\n\nThe sum of an infinite series having ratio whose absolute value is less than 1 is equal to:\n\na/(1 - r)\n\nWith a = 1, and r = 1/4, the sum = 4/3. So:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3\n\nLooking again at B, we have:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...)\n\nThis sum should be half of the sum of the infinite series in A.\n\nWith a = 1/2 and r = 1/4, a/(1 -r) = 2/3\n\nSo:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...) = 1/2 * 2/3 = 1/3.\n\nA should win 2/3 of the time, and B should win 1/3 of the time. There is an advantage to going first!
1003883
distance = rate times time or\nd = rt\nr = (x+100) mph\nt = (2x+3) hours\n\nd = rt\nd = (x+100)(2x+3)\nUse FOIL\nd = (x)(2x) + (x)(3) + (100)(2x) + (100)(3)\nd = 2x^2 + 3x + 200x + 300\nCombine like terms\nd = 2x^2 + 203x + 300
810450
1.\nConvert 80 mph to feet per second:  80 * 5280 / 3600 = 117.333\n\na = (Vf - Vo) / t  = (117.333 - 0) / 6 = 19.556 ft /sec^2\n\n2.  \n\nConvert 70 mph to feet per second: 70 * 5280 / 3600 = 102.667\n\nVf^2 = V0^2 - 2*a*d\n\n0 = 102.667^2 - 2*a*157\n\n10540.44 = 314 a\n\na = 33.57 ft / sec^2\n\n\n3.  a = (Vf - Vo) / t = 80 / 5 = 16 ft/sec^2\n\nV = 80 - 16t\n\nIntegrate that equation from 0 to 5\n\n80t - 8t^2 = 80 * 5 - 8*25 = 200 ft
815192
d = 1/2 a * t^2\n98m =(1/2) * 9.8m/s/s * t^2\n20 sec^2 = t^2\nt = 4.47 sec\n\nd = v * t\nd = 27m/s * 3.16sec\nd = 120.7 m
958697
10.4 m/sec\n\nV(t) = (30+a*t) m/sec\n\nwhere a = - 9.8m/sec^2\n\nV(2) = (30 - 09.8*2) m/sec\n\nV(2) = 30-19.6 = 10.4 m/sec
1009103
We have a formula:\nSpeed = Distance / Time\nhere Time= Distance /Speed.==D/S\nD=100km in both cases\nS1= 125km/hr; S2= 80km/hr\n\nT1=D/S1==100/125 hrs === 4/5hrs===4/5*60mins= 48mins\nT2=D/S2==100/80 hrs ===5/4 hrs ===5/4*60 mins= 75mins\n\nTime saved = T2-T1==75-48=27 mins.\n\nIs this clear buddy
701707
The Average Linear Speed is {total distance / total time taken} = 65/ 5.7 = 11.4 m/s. \n\nDivide this value by the radius of the wheel to get the ‘The Average Angular Speed’. \n\nThe Average Angular Speed is = 11.4 / 0.02 = 570 radian per second.\n\nSince the motion is uniform,\n\n The Average Angular Speed = {Initial Angular Speed + Final Angular Speed} / 2   \n\nBut the Initial Angular Speed is zero.\n\nTherefore, the Final Angular Speed   is twice the Average Angular Speed\n= 2 x 570 \n=1140 radian/ second. (The answer for the second question)\n\nSince the Initial Angular Speed is zero,\n\nThe Angular acceleration is, “Final Angular Speed/ total time”.\n= 1140/5.7 = 200 [radian / second] / second. The answer for the first question)\n\n\nThe number of revolutions is found by “total distance / (pi x diameter of the wheel)” = 65 / pi 0.04 = 517.25 revolutions.
918850
If initial speed is V0 and the speed after a time t second is V,\n\nAcceleration is change in speed per unit time.\n\nAcceleration a = (V - VO) / t.\n\nThe average speed during this time t is \n\nV (average) = (V + V0) / 2.\n\nNote; if any quantity is changing uniformly, the average is found by adding the initial and final value and dividing the sum by 2.\n\nThe average speed can also be determined by dividing the total distance by total time of travel.\n \nV (average) = S / t.\n\nTherefore S/t = (V + V0) / 2.\n\nS = (V + V0)* t / 2.\n\nFrom a = (V - V0) / t we have V = V0 +at.\n \nTherefore S = {V0 +at + VO} * t /2 \n--------------- = {2 V0 t + at t } /2\n\n-------------- S = V0t + 0.5 a tt
124298
Answer to #2\nt=1.565 seg\nV= 2.556 m/seg\nt=1.565 seg, the same.
446591
1)X=V1t+1/2gt^2\nat the interval between getting to the highest point and being dropped back to the first place is half of the full time = 2 seconds\nV1 at the highest point = 0\nX=displacement\nX=0+1/2*9.8*4=19.6 m \n2)V2=V1+gt\nV2 = velocity of landing\nV1 = velocity at highest point\nV2 = 0 + 9.8 * 2 = 19.6 m/sec
782101
East bound speed=x, in 5 hours 5x\nWest                     =x+6,             5x+30\n10x+30=240\nx=21 miles per hour.\nEast=21\nwest=27\nVR
577594
12.66 mph\n\nor 12.65558195 to be (more) exact!