An object has an initial velocity "v0" and an acceleration "a". How far ("x") does it travel in a time t?
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
46701
v=u+at\nv^2-u^2=2as\ns=ut+1/2at^2\ns=disance traveled\nu=initial vel.\nv=final vel.\na=accelaration\nt=time
448264
Given v_i = initial velocity = 0; v_f = final velocity = 44 m/s; a = 9.8 m/s^2:\n\ndv/dt = a = 9.8 m/s^2\ndv = 9.8 dt\nint(dv, v_0, v_f) = int(9.8 dt, 0, t)\nv_f - v_0 = 9.8t\n44 - 0 = 9.8t\nt = 44/9.8 = 440/98 = 220/49\n\nint(dv, v_0, v) = int(9.8dt, 0, t)\nds/dt = v = 9.8t\nds = 9.8tdt\nint(ds, 0, h) = int(9.8tdt, 0, 220/49)\nh = 4.9(220/49)^2/2 = (49*22*220)/(49*49*2) = (11*220)/49 = (2200 + 220)/49 = 2420/49 ~ 49.3878\n\nAlternatively, since accelration was constant, you could always get it more quickly with the formulas that were derived with integrals as above:\n\nv_f = v_0 + at\ns = v_0t + 1/2at^2\n\nEither way you should get 2420/49 m if the initial conditions are correct.  I don't know where 122/5 comes from although it seems to be approximately a half of this answer.
834761
Lets call the distance of the first part S1 and the total distance S2, and the corresponding times T1 and T2.  We know:\nS1 = 40\nS2 = 100\nT2 = 9.85\n\nIf V is the maximum velocity we know S1 = 0.5VT1 (const acc.)\nFor the second part we know S2-S1 = V(T2-T1) = VT2 - VT1\n\nVT1 = 2S1 so S2-S1 = VT2 - 2S1 or V = (S2+S1)/T2 = 140/9.85\n\nSo a) - V = 14.21 m/s\n\nWe know V^2 = 2AS1 where A is acceleration so A = V^2./2S1\nor A = 14.21^2 /(2x40)\n\nSo b) A = 2.525 m/s^2\n\nWe know V = AT1 so T1 = V/A = 14.21/2.525 = 5.62 s\n\nSo c) the two parts are 5.62s and 4.23s
808209
Using one of the equations of motion\n\n(v*v) - (u*u) = 2*a*s\n\nwhere \nv is final velocity\nu is initial velocity\na is accleration\ns is displacement\n\n\nOR \nuse \n\nv * (dv / ds) = a
546224
Energy = force x distance\n=>  Energy = (mass x acceleration) x Distance\n=>  Energy = mass x acceleration x Distance\n=>  Energy = mass x (speed / time) x Distance\n=>  Energy = mass x speed x (Distance/time)\n=>  Energy = mass x speed x speed\n=>  Energy = mass x (speed)²
987378
150m.\nThe equation is:-\nd=ut + (1/2).a.t^2\nWhere u= velocity=40m/s\nt= time =6sec.\na=acceleration/retardation= -5m/s^2 \n\nSource(s):
716765
The potential energy at the top is mgR = 2 x10 x R = 20 R.\n\nKinetic energy at the bottom of the curve = 0.5 m U ^2 = 0.5 x 2 x U^2. = U^2.\n\nAs there is no friction  K.E = P.E\n\nU^2 = 20 R.\n\nOn the horizontal path, \n\nInitial speed is U.\n\nFinal speed is V = 0.\n\nDistance traveled = 8 m.\n\nAcceleration = Fricttional Force / mass = -8/ 2 = - 4 m/s^2. ( Negative sign since the force acts opposite to the motion.\n\nUsing V^2 - U^2 = 2as, and also noting that V=0,\n\nU^2 = 2 x 4 x 8 = 64.\n\nU = 8m/s.\n\nThe time of travel is found using V= U+ at.\n\nt = -8/ (-4) = 2 second.\n\nUsing U^2 = 20 R.\n\nR = 64/ 20 = 3.2 m.
332219
An egg is pretty aerodynamic, so we can neglect friction. Thus, a falling object satisfies the equation\n\ns(t) = -0.5*g*t^2+v0*t+s0\n\nwhere g is 9.82 m/s^2, v0 is the start velocity (in this case, 0) and s0 is the start height (in this case, 3 metres).\n\nSo, letting s(t)=0 and solving for t, we get\n\n0 = -0.5*g*t^2 + 3\n\nt = sqrt(6/g) = 0.78 seconds.
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc
767928
1. its final velocity depends on the direction it was heading to start off with. If it was flying in the x direction (no components in y or z) then after 4.3 seconds the x component has increased by 8.92*4.3  and the y component has increased (from 0 this time) by 1.21*4.3.\nso now you can work out the new direction (in the x direction but slightly downwards by whatever angle has a tangent of 1.21*4.3/(152+8.922*4.3) \nand using pythagoras to determine the speed.\n\n2. You can ignore the gravity as it is << the precision with which the acceleration due to the electric field is stated (I'm assuming the 106 is 10^6 = 1000000) \nso the final velocity is 3.80*2.45*1000000 m/s upwards.
716223
The change in velocity of an object undergoing uniform acceleration, a, for a given time, t, is given by:\nv = at\nwhere v is the change in velocity of the object.\n\nIf the object is accelerating at a constant rate of 10 m/s^2 for 5 seconds, the change in velocity the object experiences is 50 m/s.\nSince (we assume) the object started from rest, the objects final velocity after 5 seconds is 50 m/s.\n\nTo find the average velocity, divide the total displacement of the object by the total time.\nIn this case, the object has traveled 125 meters in the 5 seconds it has been accelerating (d = 1/2 at^2), so dividing this by 5 seconds and we get an average velocity of 25 m/s.
1186433
Well, I think that for the first oneyou can get the acceleration by deriving it twice, then you get the integral from 1 to 5, let's see...\n\nds/dt= 6t^2+3=v\ndv/dt= 12t=a\n\nfor a\n\nI think you should do it this way\n\nintegral of vdt/ integral of dt from 1 to 5\n\nso you will get\n\n(2t^3+3t)/t from 1 to 5\n\nso it will be\n\n2t^2+3 from 1 to 5\n\nevaluating you'll get\n\n2*(25)+3-2-3\n\n=50-2\n\naverage speed = 48\n\nfor b\n\nyou just substitute t=4 to the first derivative\n\nV=6*(4)^2+3=99ft/min
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.
927736
Everyone above assumed a constant speed.  the more general formula is \n\nD = ∫v dt
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.
75568
There are four (at least that I know of) to choose from.  You have to first analyze your problem and find out what you know and then solve for your unknown.  However, these equations assume that the acceleration is constant.\n\nv = v0 + a(t)\nHere:  v=final velocity; v0=initial velocity; a=constant acceleration; t=time\n\ns = s0 + (1/2)(v0 + v)(t)\nHere:  s=final position; s0=initial position; v0=initial velocity; v=final velocity; t=time\n\ns = s0 + v0(t) + (1/2)(a)(t^2)\n\nv^2 = v0^2 + 2*a(s-s0)\n\nI didn't say what each of the variables in the last two equations stood for because they are the same as what you see in the first two equations.  One good way to check if you are on the right track or not is to make sure your units work out right.  For example, if you are looking for velocity, chances are you'll be looking for something like meters per second (m/s) or miles per hour (mi/hr or mph).  Hope this helps!