How do you find acceleration???
123108
The car accelerated from 0mph to a speed of 100mph.\nIt took 50s to do this.\nThe rate it changed its speed is its acceleration - how long it took to reach that speed.\nSo, 100mph / 50s = 2mph per second.\n\nA car that could accelerate faster reached 100mph in only 25s.\nSo, 100mph / 25s = 4mph per second.
454773
final speed=initial speed+(acceleration*time)\n=10+(30*30/60)\n=10+15\n=25km/h\nhope that helps..\ngood luck
362151
Remember Newton's second law:  F = ma.  You can rearrange that to read a = F/m.  So the original acceleration was equal to the original force divided by the original mass.  After the adjustments, you have a2 = 3.3F/2.5m.  Therefore, a2 = (3.3/2.5)a, or 1.32a.  The original acceleration was 6 m/s^2, so the new acceleration is (6*1.32) or 7.92 m/s^2.
30699
hey, didnt i answer something very similar to this for you? you need to do F=ma so 43000000N/2100000kg then you get an acceleration, but then you have to subtract gravity from this because thats the acceleration produced by the rocket but there is still gravity pulling with 9.8 m/s^2 in the opposite direction. so just do (43000000/2100000) - 9.8, and that's your answer :)
449781
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
81300
Well um...\n\na = r''\n\na = (tangential acceleration * unit tangent) + (normal acceleration * unit normal) = (at * T) + (an * N)\n\nT and N are your unit tangent and unit normal of the function that defines your position.\n\nat = (r' dotted with r'') / (|r'|)\n\nThose are the tangential acceleration components but for this problem i think that you are going to have to reparamiterize your curve.
233617
I'll get you started...\n\nFirst, determine how long the ball was in the air.  From that, determine the velocity at which the ball was thrown (the rate necessary, factoring in gravity's acceleration, to keep it airborne for that period of time).  Construct a vector using the velocity of the truck vs. the velocity of the ball to determine the angle.\n\nGood luck!
422571
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
421860
here, you basically use the kinematics equations:\nx=x0+v0cos(theta)t+.5axt^2\ny=y0+v0sin(theta)t+.5ayt^2\nhere, you can set x0 to be 0 since you're only interested in the change in x.  also, there's no acceleration in the x direction, so ax=0.  you start out at y=0, and after the jump, you end up at y=0 again:\nx=v0cos(theta)t\n0=v0sin(theta)t-.5gt^2\nv0sin(theta)t=.5gt^2\nv0sin(theta)=.5gt\nt=2v0sin(theta)/g\nx=v0cos(theta)*2v0sin(theta)/g\nnow, you must use the trig identity 2cos(theta) *sin(theta) =sin(2*theta):\nx=v0^2*sin(2*theta)/g\nthis formula is known as the range equation.\ng=v0^2*sin(2*theta)/x\nhere, you have 2 unknowns (g and theta), but you do know that for g to be at a maximum, sin(2*theta) must also be at a maximum.  since sin(2*theta) has a maximum value of 1, which occurs at theta=45 degrees, you have:\ng=v0^2/x\n  =(5.90 m/s)^2/7.52 m\n  =4.63 m/s/s
374549
Hi,\n\nThe constant acceleration formula is\n\nx(t) = at^2 + vot + xo\n\nwhere \nx(t) = position at time t\na = acceleration\nvo = initial velocity\nxo = initial position \n\nuse the acceleration to solve for t when x(t) = 30.\n\nHope that helps, \nMatt
459303
1) I think your 2nd c.f. should say c. acceleration; that's the units you're using------ the c force of a mass m with acceleration a is f = ma, so youhave 1.5 kg*18 m/s² = 27 N\n\n2) First find the velocity: v = d/t = 2πr/t = 2π*.75m/.8sec = 5.89 m/s\nThen, acceleration: a = v²/r = 5.89²/.75 = 46.256 m/s²\nThen, the tension: T = ma = .1kg*26.256m/s² = 2.626 N\n\n3)  v = 2πr/t = 2π*90/12.3sec = 45.974m/s\na = v²/r = 45.97²/90 = 23.485 m/s²\nF = ma = 900*23.485 = 21,136.5 N\nThe lateral friction of the tires on the track
33714
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
374334
Velocity(at time t) = Velocity (at time = 0) + Acceleration x Time\n\nV(t) = Vi + At\nV(t) = 32 ft/sec + (-32 ft/sec/sec)(t)\nV(1 sec) = 32 - 32 = 0 ft/sec\n\ny(t) = y(initial) + (V x t) + 1/2(A)(t^2)\ny(1 sec) = 0 + (32) +  (-16) = 16 feet\n\nTo hit ground, y = 0, so (V x t) = 1/2(A)(t^2)\n32t = 16t^2\n2 = t, t = 2 seconds\n\ny(max) occurs at zero velocity, which is at t = 1 sec, x = 16 feet
297281
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
131355
We have to find the resultant acceleration of acceleration in East direction and acceleration in the North direction, since there is change in direction.\n\nSince it moves with same speed in both directions it is enough if we find in any one direction.\n\nFor the first 2.5 s there is no acceleration, similarly there is no acceleration in the second 2.5 s. \n\nAt the end of first 2.5 s it has to come to stop and then turn to north all with zero time. \n\n In the Eastern direction the change in speed is 5- 0 = 5m/s. To find acceleration divide this by zero.  It is not defined. \n\nIf we assume that it takes a very small time, then the acceleration in this direction is nearly   minus infinity toward west (as the speed is reduced).\n\nSimilarly in the north direction it is plus infinity toward west.\n\nThe resultant of these two is inclined toward North West with a magnitude of infinity.
31731
Yeah i know:\nDo this:\n\na.)Find the work done by the applied force.\n\nb.)Find the work done by friction\n\nc.)Find the work done by gravity\n\nd.)Find the work done by the normal force.\n\ne.) IF the object starts at rest, how fast is it going after the 5.0m.\n\nhope this helped\n-----------\n\nNO I WILL NOT HLEP YOU. DO YOUR OWN WORK! your asking someone to do your work for you.
403039
dC/dt = 4 m/s\nC = 2*PI*R\ndC/dt = 2*PI*dR/dt = 4\nSo dR/dt = 2/PI\n\ndA/dt = d(PI*R^2)/dt = 2*PI*R*dR/dt = C*dR/dt = 32 * 2/PI = 64/PI m/s