Anyone know the equation for speed, thinking distances, and braking distances?
916308
Breaking distance is related to speed ^2.  From the examples given, breaking distance = speed^2/168.  For example, at a speed of 32, breaking distance = 32^2/168 = 6.1 and at 112, bd = 112^2/168 = 74.7.
947787
In this equation, you have a right triangle.  After 3 hours, the first car has traveled 150 km south and the 2nd car ahs traveled 210 km west.  You can find the between these two cars at 3 hours by using the Pythagorean Theorem since you have a right triangle.\n\nDistance = sqrt(150^2+210^2) = 258.07 km\n\nThe general form of the Pythagorean Theorem is:\n\na^2 + b^2 = c^2, where:\n\na = distance of car 1\nb = distance of car 2\nc = distance between cars.\n\nTake the derivative to find how the rates are related to one another.  Take the derivative with respect to time:\n\n2a(da/dt) + 2b(db/dt) = 2c(dc/dt)\n\nIn this problem, we are looking for dc/dt.  These are our known values:\n\na = 150 km\nb = 210 km\nc = 258.07 km\nda/dt = 50 km/hr\ndb/dt = 70km/hr\n\nPlugging these values for the equation above, solve for dc/dt:\n\ndc/dt = [2a(da/dt) + 2b(db/dt)]/[2c] ------->\n\n[(2*150*50) + (2*210*70)]/(2*258.07)\n\nTherefore, dc/dt = 86.02 km/hr.  Now change to km/sec:\n\n(86.02 km / 1 hr)(1 hr / 60 min)(1 min / 60 sec) = 0.024 km/sec\n\n-----------\n\nHope this helps
841789
Assume s=275, v=43, r=0.21 & t is time from start to stop; then v=a*t & s=a*t^2/2, ‘a’ being deceleration. Therefore t=v/a, s=a*(v/a)^2/2; thus 2*s*a=v^2 or a=v^2/(2*s).\nAngular deceleration w’=a/r.\nNow I=0.175, so frictional torque T=I*w’. plug numbers in.
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
532726
CW= PW / (.5)(A)(Gr)(vcubed)\n\nCW=drag coefficient\nPW= aerodynamic power (kilowats)\n.5= constant\nA= total front area of the vehicle in meters squared.\nGr= gradient of the terrain.\nv3= velocity cubed.\n\nGr can be found with the equation:\nGR=(mass)(gravity)(velocity)(sin of the angle of the gradient)\n\nyour total answer should be  no more than 1
108141
2.4\n\n75 miles/60 minutes   =   3 miles/X minutes\n\nCross multiply and solve for x:\n\n75X = (60 * 3)\nX = 2.4\n\nThe weight of the car is not relevant.
966246
d=r*t\nhighway= 60*highwaytime\ncity = 30*citytime\ncity = highway -  100\ncitytime =highway time - 5/6  50 min = 5/6 hour.\nsub in\n\nhighway - 100 = 30 (highwaytime- 5/6)\n60 * highwaytime - 100 = 30 * highwaytime - 30*5/6\nsimplify\n30* highwaytime = 75\nhighwaytime = 2 1/2 hours\nh  =2.5*60 = 150 miles\ncity =30* 1 2/3 = 50 miles\ntotal 200 miles
399954
SPEED=dy/dt=distance/time.its not exactly something you invent.it was an axiom,a basis of understanding.
311411
Math model of problem:\n\nD2S/Dt^2 = 2.1  \n\nIntegrating:\n\nDS/Dt = 2.1*t + c1\n\nIntegrating again:\n\nS = 2.1*t^2/2 + c1*t + c2\n\nFrom the statement of the problem:\nWhen t=0 DS/Dt=0 and S=0\nTherefore c1 and c2 both are zero\n\nSo S = 1.05*t^2\nSo t = square root of S/1.05\nThat is t = square root of 70/1.05\nthus t = 8.165 secs.\n\nSo boat accelerates for 8.165 secs in travelling from rest over a distance of 70 metres
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
160821
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s
845679
You have mixed units. To solve, units must be consistent. Speed is in miles per hour so time must be in hours.\n\n210=(70/60)(y-x)\n210=(50/60)(y+x)\n\n1260=7y - 7x\n1260=5y + 5x\n\n6300=35y - 35x\n8820=35y + 35x\n\n2520=70y\n\ny = 36
69843
work = force times distance\nW=Fs
1007024
distance equals rate times time\nThat is,\nd = rt\nd = 117 and t = 2.25\nSo we have\n117 = r(2.25) or\nr = 117/2.25\nr = 52 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
702361
ok biker bike, here we go.\n\nacc.= g = 9.8 m/s2, (any object moving in air will have this acc. or retardation)\n\nb) we apply formula\ns=ut + 1/2at^2\nu=20\nt=4\ns=20*4 + 1/2 *9.8 *4*4\n  =158.4 m\n\nc) v=u + at\n  therefore, t= v-u/a\n                   =50-20/9.8\n                   =3 seconds (approx)
799226
They travelled a total of 476 miles in 4 hours.  So together they are travelling (476/4) mph or 119 mph.\n\nSo you have two equations:\nJ + S = 119\nJ - S = 5\n\n2J = 124\nJ = 62 mph\nS = 57 mph