Velocity and Acceleration for SHM?
922827
They tell you what the period is. The frequency is 1/period angular frequency is 2*pi*frequency\n\nThen you make a lot of assumptions : air resistance and frictional forces in the spring are very small, the spring mass can be neglected. The spring response is linear.\n\nThen look up expression for frequency :\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/shm.html#c2\n\nso k is about 79 N/m \n\nmax speed is about 4.4 m/s\n\nmax acceleration is about 55.3m/s/s and max force is therefore about 27.65N
844762
There are jthree  formulae cncerning velocity.These are v=u+at , V^2=u^2+2as ,and s=ut +(1/2) at^2 ,where u= initial velocity,v=final velocity ,a=accelaration.,s= distance travelled and t= time.Now how to decide which formula is to be used?in the problem,just see what they want you to find out and note down what is are already given in the problem, if they give you u,a and t and ask you to find out v,apply formula no.1.But if they gdonot give you time but give the distance or displacement and asks you to find v,donot hesitate to apply formula no,2 and so on.if you have any difficulty to understand.please do not hesitate to write to me through yahoo answersand i shall be glad to assist you
908729
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.
1001657
V= V(i) + AT Initial velocity plus acceleratin x Time\n\n\n   = V(i) + (F/m)T     Acceleration = Force/ Mass\n\nNote  use same system of units
822312
wish I could solve em for you but ill try and help still.\n\n1) The acceleration should be the derivative of the equation for the car's velocity.  The original distance/time equation would be the anti-derivative of the velocity.  \n\n2 The equation to find out how long it will take to get there is h+ax-16x^2 in feet.  (a is the original speed and x is number of seconds and h can be the height of the building.)\n\nHope this helps
700119
Velocity:\nv=d/t\n\nd=distance\nt=time\n\nAcceleration:\na=(vf^2-vi^2)/2d\n\nvf=final velocity\nvi=initial velocity
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
98111
Code:\nx'' - 2nd derivative of x with respect to time (t) (acceleration in x)\n\nx' - 1st derivative of x wrt time (velocity in x)\n\nAnd the same system for the y coordinate.\n\nx''=1.2 then by integration wrt time gives x'=1.2t+constant. Boundary conditions are x'=4750 m/s at t=0, so the constant = 4750 m/s. Equation is now Vx = x' = 1.2t + 4750. Substitute in t=743s, answer is Vx=5641 m/s. \n\nSame method for y'' with acceleration of 8.5 m/s^2, and boundary conditions of y'=0 at t=0 gives Vy=6315.5 m/s.\n\nFinal answers:\nVx=5641.6 m/s\nVy=6315.5 m/s
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
712458
Simple harmonic motion can be described by a sinusoidal function\n\nX(t) = X0*sin(w*t)\n\nwhere X0 is the amplitude of the motion, w is the angular frequency, and t is time. Given this expression for the time dependence of the displacement, the velocity is expressed as\n\nV(t) = X0*w*cos(w*t)\n\nand the acceleration can be written as\n\nA(t) = -X0*w^2*sin(w*t).\n\nBased on these expressions, the maximum velocity is\n\nVm = X0*w\n\nand the maximum acceleration is\n\nAm = X0*w^2\n\nThe angular frequency w is\n\nw = 2*pi*f\n\nwhere the linear frequency for this problem is f = 4940 /min. Be sure to convert to the proper units to get the answers in the form you need.
863271
mass doesn't matter here, just the acceleration\nv=u+at\na=(v-u)/t\na=(9-0)/0.15=60 m/s^2
423229
Go to this site http://www.bksv.com/pdf/Measuring_Vibration.pdf it should help you out with your question.
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
912968
Your formula give the displacement, just put in the value of t.  The velocity is the derivative of the displacement,  Take the derivative dx/dt and put in the value of t.  The acceleration is the second derivative of displacement, or the first derivative of velocity.  Same method.  The period is given by ∆t when the argument of cosine goes from 0 to 2π, and the frequency is 1/period.
769161
Its a pain to do this with X and Y components, but there is an easier way....\n\nRange Formula - used when start and end heights are the same\nQ = Theta\nd = ((Vo^2)(sin 2Q))/g\n\n301.5 = ((Vo^2)(sin 2(25)))/9.81\n\nVo = 62.13 m/s\n\nNow that u have initial velocity, find the y component\n\nVo Sin Q\n\n62.13 sin 25 = 26.26 m/s\n\nThen No Time Formula\n\nVf^2 = Vo^2 + 2gd\n\n0 = (26.26^2) + 2(-9.81)d\n\nd = 35.15 m\n\nHope this helps
728641
acceleration can be found using\na=(m1-m2)g/(m1+m2) = 3*9.8/5 = 5.88 m/s^2
418402
Using one of the equations of linear motion:\n\nv² = u² + 2as\n\nInitial speed u = 0\nAcceleration a = 32 ft/s\nDistance s = 9 ft\n\nSolve for final velocity v = sqrt( 2 * 32 * 9) = 6.2 ft/s\n\nThis assumes negligible air friction.