Aceleration,velocity,position(physics,calculus base)?
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
693415
You take the avaerage of the instantaneous velocity over an interval of time.\n\ntake say five instantaneous velocity measurements over about 40 seconds. then take the average of these five measurements to get a linear velocity. to make your answer more accurate just take more than 5 mearsurements and average them in the same matter.
250190
This is really simple stuff. Newtons is force, seconds is time, meters is length, and meters per second is length per time, or velocity.
201825
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.
985855
velocity = df(t)/dt = 4π + 4πcos(4πt)\nacceleration = dv(t)/dt = -16π^2sin(4πt)\n\nParticle is at rest when velocity = 0, or 4π+4πcos(4πt) = 0\n1+cos(4πt) = 0\ncos(4πt) = -1   The values for which cosø = -1 are ø=n(odd)π, so when\n4πt=n(odd)*π, t = 1/4*n(odd); thus for t = 1/4, 3/4, 5/4, 7/4, 9/4, 11/4 the velocity is zero\n\nThe maximum velocity occurs when cosø = 1, so the max velocity is 8π
874949
c. Velocity will have size and direction.
614422
Depends what information you already have. The simplest one I can think of is:\n\nVelocity = Acceleration x Time\n\n-Therefore-\n\nAcceleration = Velocity / Time\n\nBut you might have other variables involved here.
379523
There is a great website on this kind of stuff, check out:\n http://scienceworld.wolfram.com/physics/NewtonsLawofCooling.html\n\nThere is also an equivalent one for mathematics:\n\nhttp://mathworld.wolfram.com/\n\nExcellent resources for basic to advanced theory.
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.
889055
Power:\nP=dE/dt = dW/dt\n\nForce:\nf=ma\n\nEnergy:\ne=mc^2
912927
1D\n2D (acceleration can be negative in which case you mihgt call it deceleration)\n3B
224170
please (2 pdf documents)\nhttp://physics.angelo.edu/~dbixler/phys3444lab4new.pdf\nhttp://physics.angelo.edu/~dbixler/phys3444lab8.pdf
779861
Ok, here's how i think you do it\n\ncar displacement = 66*t where t is the time in seconds\n\nbaloon displacement = 15*t + 200\n\nso according pitagoras\n\nc^2 = 4356*t^2 + (15*t+200)^2\n\noperating you get\n\nc^2 = 4581*t^2 + 6000*t + 40000\n\nnow, deriving dc/dt you get\n\nVc = 3*(1527*t+1000)/(4581*t^2+6000*t+40000)^(1/2)\n\nso now you just subtitute t\n\nVc = 33.7 ft/s\n\nI hope this helps.
863361
x = -11 + 12t + 3t^2 + 4t^3 - 4t^4\n\nTaking the 1st derivative gives the velocity.\nv = dx/dt = 12 + 6t + 12t^2 - 16t^3\n\nTaking the 2nd derivative gives the acceleration.\na = dv/dt = 6 + 24t - 48t^2\n\nTo find when the acceleration is maximum, we have to take the derivative of the acceleration equation and set it equal to zero.\n\nSo, da/dt = 24 - 96t = 0.\nThus, t = 1/4.\n\nThe location of the particle at this time is then found by substituting t = 1/4 into the location equation.\n\nx = -11 + 12(1/4) + 3(1/4)^2 + 4(1/4)^3 - 4(1/4)^4\n= -11 + 3 + 3/16 + 1/16 - 1/64\n= -497 / 64\n\nThe velocity at this time is found by substituting t = 1/4 into the velocity equation.\n\nv = 12 + 6(1/4) + 12(1/4)^2 - 16(1/4)^3\n= 12 + 3/2 + 3/4 - 1/4\n= 14
756774
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
879746
Ek = 1/2.m.v2   (1/2 mass times velocity squared) - \n\nsorry can't do superscript on this site
611013
Oh good grief.  If it's anything like what I've just done it'll include velocity, centrepetal forces, magnetic fields, friction, gravity, forces generally, working out the mass of something and the speed at which it will fall, balances with pivot points, static electricity, AC & DC electricity, transformers, vectors and so on and so on.  It's quite hard but it's maths based so if you like maths, you should be ok.  I wish you luck.