Uniform Motion Problems - Equal Distance?
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
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I think you can use a simpler equation than the one for distance.\nWe don't need to find distance, just time.\n\nv = u + a.t\nv= 0, u= 64 f/s, a= -32f/s^2\nsolve for t. \nt = 2 s\nThats the time it takes to go up just before it starts to fall again.\nThe time taken to fall is going to be the same: another 2 s\n\ntotal time: 4 s
946504
Intensity varies as the square of the distance.\nThat should help
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
438060
Without regard to air resistance you solve the problem as such.\n\nv(f) = final velocity\nv(i) = initial velocity\na = acceleration\nt = time\nx = displacement (distance)\n\nWhen you throw the ball up vertically, you know several things:\n1) v(i) = 30 m/s\n2) v(f) = 0 m/s\n3) a = -9.8 m/s^2 (this is because acceleration due to gravity goes in the opposite direction of the object)\n\nSolve for x with this equation:\nv(f)^2 = v(i)^2 + 2ax\n0^2 = (30 m/s)^2 + 2(-9.8 m/s^2)x\n0 = 900 m^2/s^2 - (19.6 m/s^2)x\n(19.6 m/s^2)x = 900 m^2/s^2\nx = (900 m^2/s^2)/(19.6 m/s^2)\nx = 45.9 m (rounded off)\n\nNow you know how high the ball went.  The time it takes to reach this is equal to the time it takes to get back down (assuming that it is the same distance up and down).\n\nTo solve for t, use this equation:\n\nv(f) = v(i) + at\n\n0 = 30 + (-9.8)t\n9.8t = 30\nt = 30/9.8\nt = 3.06 s  this is going up to max height.\n\nto check if it will be the same time going down, use this equation:\n\nx = v(i)t + (1/2)at^2\nv(i) = 0 (because when you reached max height, your velocity will be zero)\na = 9.8 because your ball is travelling in the direction of the acceleration.\n45.9 = 0 + (1/2)(9.8)t^2\nt^2 = 45.9/4.9 = 9.367\nt = 3.06 s\n\ndone
83305
I suppose the problem was already due so I'll finish it for you now.\n\nF=m*a\na=F/m= 2*10^6N/1.2*10^6kg = 1.67m/s^2 * g/9.81m/s^2 = 0.17g
927736
Everyone above assumed a constant speed.  the more general formula is \n\nD = ∫v dt
835037
Let the two case be identified as having  moment-of-inertia I1 and I2. \nTorque T on to the each of the two bodies produces angular accelerations (alfa1 and alfa2) given  by the relation T= I x alpha \n\nalfa1 = T/ I1 and alfa2 = T/I2\n\nStarting form rest(wi =0), let the time taken to reach final angular velocity wf be t1\nusing:  wf =- wi + alfa x t  \n \nwf/alfa1 = t1 and wf/alfa2 = t2 \nleading to \n\nt2  = alfa1x t1/alfa2 \n\n     = I2 xt1/I1\n     = (I2/I1) x t1\n     =. ......Ans \n(Hoping you will put the values and get the answer you want)
125809
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
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
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
1053820
Stationary in reference to what fixed point?
782037
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
798759
1- yes, 2-yes, 3-no Vx=the same 24.5m/s, 4-yes, 5- no Dy=21, 6-yes Ax=0 throughout the whole parabola. 7- yes coz the only force (constant force) is gravity. Now (8) – its final Vy=26, and Vx=  still the same 24.5, thus Vtotal=sqrt(Vx*Vx+Vy*Vy)
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That's the trouble with answers that give no explanation. Who can you believe?\n\nSince we go in each direction once, the ORDER of the moves doesn't matter. \n\nDraw a grid. Start at 0,0\n10 North means 0,10 You are due north 10 miles. Then we come back south 6 miles, so we're 4 miles north of where we started. \n4 miles east and then 7 west means we're 3 miles west of the starting point. \n\nDraw a right triangle 4 miles up and then 3 miles to the left of the starting point. We want the hypotenuse of the triangle whose sides are 3 and 4. \nTime for Pythagoras: a^2 + b^2 = c^2 \n3^2 + 4^2 = c^2 \n9 + 16 = 25 Take the square root of that for your final distance. Direction: NW
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I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
125621
B) Increase\n\nThe difference in the two objects' velocities would be constant since that are both accelerating at the same rate, but the first object already had a higher initial velocity (from Object B' point of view).\n\nThe equation relating the distance traveled with respect to initial velocity, accelertion and time is,\nd = v_i * t + 1/2 a*t^2\nSo for both objects, after B is released the 1/2 a*t^2 part of the equation is the same, so we can ignore that part and focus on the v_i * t part.\nIf we start t=0 at the moment object B is released, object A's velocity is some non-zero number while B has no initial velocity.  So as time goes on, the distance between the 2 objects inceases.