calculus word problem with velocity?
813109
this is a simple dervitive problem....we recall that given a position the first derivitive would give a velocity at any time t, and a second derivitive would give an accelaration at any time t\n\nso we need to take the derivitive of h(t)\n\nso h(t) = -6.5t^2 +40t+6\nh'(t)= -13 t +40\n\nnow you find when h'(t) is zero...this will tell you when the slope of a tangent line of position is zero...in otherwords...a max or min\n\nso you find h'(t)=0=-13t+40\nso t= 40/13\n\nnow we need the second derivitive to  apply the second derivitive test (recall this tells us that if the h''(t) is negative at a specific t value we have a max, and if its positive at that t value its a min)\n\nso h''(t)=-13, so for all values of t h''(t) is negative so at t=40 h'(t) is zero and h''(t) is negative so there is a max height at t= 40/13\n\nnow to find the height you plug 40/13 into h(t)\n\nso h(40/13) i will leave to you cause this is teh easy part\n\nhope this helps\n\nmatttlocke
931347
x =70m, t = 7s, Vf = 15m/s\na.)\nx = 1/2(Vi + Vf)t\nVi = initial velocity\nVf = final velocity\n\n70m = 1/2(Vi + 15m/s)(7s)\n10m/s = 1/2Vi + 7.5m/s\n2.5m/s = 1/2Vi\nVi = 5m/s\nb.)\nVf = Vo + at\n15m/s = 5m/s + 7s(a)\n10m/s = 7s(a)\na = 1.43m/s/s\nbasic kinematics\nyou can't use distance = rate x time because there is acceleration and the speed is not constant.
723707
s'(t) =(t-3)(-2t^{-3}) + t^{-2}\n=( -2t+6 +t)/ t^{-3}\n=(6-t) / t^{-3}\nso the velocity is 0 when the top =0\ni.e 6-t-0\nt=6\ns
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
1016838
Use the quotient rule for differentiation.  Taking the case where the firetruck is approaching:\n\nThe numerator of the derivative has the denominator of the original function times the derivative of the numerator minus the numerator of the original function times the derivative of the denominator.  \n\nThis is pretty easy, since the numerator of the original function is a constant.  The derivative of a constant is 0.  When approaching, you only have to worry about the original numerator times the derivative of (331 - v).  The 331 is a constant with a derivative of 0 leaving you the derivative of -v.  That's -1.  \n\nTherefore, the numerator of your derivative is -132400 * -1, or 132400 when approaching.\n\nThe denominator of the derivative is just the square of the original denominator, or (331 - v)^2 when approaching.  When approaching, that would give you 301^2 (301 squared), or 90601. \n\nWhen approaching, the rate of change would be 132,400/(301^2), or about +1.46.\n\nFinding the rate of change when moving away works the same, but with a +v instead of a -v.
547930
Velocity is speed in a particular direction, anyone changes means the velocity is not constant.
497479
When calculating with error, remember to MAXIMIZE error wherever possible. For example, since V1 and V2 are in the denominator of their respective terms, error should be subtracted to maximize the overall terms. Check me on this, but I found an average velocity of (19.5 ± 0.5) m/s.
784184
θ =0.3t² rad\nω = dθ/dt = 0.6t rad/s\nα = d²θ/dt² = 0.6 rad/s²\n\nWhen t = 5\na. ω = 0.6t\n= 0.6*5 = 3 rad/s\n\nb. v = rω\n= 10*3 = 30m/s\n\nc. α = 0.6 rad/s²\n\nd a = v²/r (= rω² = 10*3² = 90 m/s²)\n= 30²/10\n= 90 m/s²
901648
put x=2 directly into the given equation to get value of x at t=2...the answer u get is the displacement\n\ndifferentiate the equation wrt time to get velocity equation and then differentiate the velocity equation to get acceleration equation\n\nin both cases put t=2 to get velocity and acceleration\n\nconvert the given equation in terms of sin\ncomparing the new equation with x=asin(wt+@)\nwhere a=amplitude\n          w=angular frequency\n          @=phase\n\nalso use equations w=2 x pi x f and T=1/f\nhere x denotes multiplication \nf=frequency \nT=period
1158387
as you can see that velocity vector as a function of time is given \nv(t) = 5 t^2 - 200 t - 150;\nthe acceleration is given,\nso using the vector calculus (vector differentiation)\nwe differentiate the velocity vector,we have,\na(t) = 10 t - 200;\nso in the given problem the acceleration of the particle is given to be 100 m/s/s;\nthere fore equating both we have,\n10 t - 200 = 100;\nwhich gives us t = 30 s.\nto know the distance through which the particle traveled we have to find the displacement vector first.\nto get the displacement vector again we take the help of vector calculus this time vector integration\nv(t) = 5 t^2 - 200 t + 150;\nint [v(t)] = s(t) = 5 t^3 / 2  - 200 t^2 / 2 + 150t;\nand substituting t = 30s will give the answer as -40,500m;\nthe answer is negative indicating that the particle has travelled along the negative direction of the axis taken into consideration.
762251
If 2 billiard balls of mass m (ma and mb) are perfectly elastic and ball A approaches from the left with velocity 2 m/s and hits ball B which has velocity 1 m/s, and is also going to the right, what's the status after the collision. \n\nA is going to the right with velocity 1 m/s and B is going to the right with velocity 2 m/s.\n\nma*2 m/s + mb*1 m/s = ma*1m/s + mb*2m/s\nmomentum is conserved
987748
1.Compute initial velocity V0\nUsing equation S=V0*t-(g*t^2)/2 we find that the initial velocity V0=35/3 + 3g/2\n2.What ever goes up must come down V0*t=-(g*t^2)/2 t=2*V0/g\n3.Total time travel = 2t \nYou to the math\n\nBest\n\nHint:  the arrow will stop midair in about 5.3 seconds
107382
v^2-u^2 = 2gh\nHere, v=1, g=-10m/s/s, h=0.6m\nHence, u^2 = 1+12=13\n=> u = 3.606 m/s
703031
Well the first thing to do in a problem like this is extract information from it.\n1.) Weight of Block = 10kg\n2.) Velocity of block(Vi) = 6m/s\n3.) Final velocity of block(Vf) = 0m/s\n4.) Friction acting on block = 20N The friction is always opposite of the direction of the block.\n\nSecond:\nWhat are we calculating? Distance traveled by block! DISTANCE.  We need a formula that will give us distance as an answer! So...we have final velocity, initial velocity, and distance....lets see...how about this one...\n vf^(2) = vi^(2) + 2*a*d \n\nThis equation has Vf, Vi, and D.  \nVf = 0 m/s\nVi = 6 m/s\nD = What we are trying to find.\nThe only thing missing in this equation that we have to find is (a) or acceleration. Hmm.\n\nThere is actually acceleration on  the block.  Friction is acting on the block in the opposite direction; therefore the block is decelerating.  We can use F=ma to find the force of deceleration.\nFriction Force = 20N\nMass = 10kg\nAcceleration = ?\n\nUse the following equation:\nF=ma\n1.) -20N = 10kg * a\n2.) a = -2 m/s (Yes it's negative b/c we are decelerating!)\n\nNow we go back to the original problem!\n vf^(2) = vi^(2) + 2*a*d \n\nThus\n(0^2) = (6^2) + (2)*(-2)*(d)\n0 = 36 - 4d\n-36 = -4d.\nDistance = 9meters! (Make sure to get the units correct!)\nI think I got this right...if not tell me what I did wrong pls\n\n-----------------------------------------------------------------------------------\nThe guy who did it using Energy equations forgot to add one thing....\nmass x (velocity)^2 = Force x distance\nso 10kg * (6m/s)^2 = 20N * x\nso (10 * 36)/20 = 18 meters\n\nThis is another way to do it.  He is using the energy equations to calculate energy in the block. When the block is in motion there is Kinetic Energy in it.  When the block has no motion, there is zero Kinetic Energy in it.  However, the individual that did this messed up the equation as follows:\nKe = (1/2) * Mass * Velocity^2\n\nThus the answer is not 18 but 9 meters!!!!!\nHere is the correction:\n1.) .5*mass x (velocity)^2 = Force x distance\n2.) .5*10kg*(6m/s)^2 = 20N * x distance\n3.) (5 * 36)/20 = distance&#92;\n4.) Distance =9 meters!
569320
v=vo + at\nv=0 + (9.8)(8)\nv=78.4\n\ny=1/2(vo + v)t\ny=1/2(78.4)(8)\ny=1/2(627.2)\ny=313.6m\n\nB
109422
This is pretty close to an question that was asked previously.\n\nI presume you mean Velocity= Distance/time, and you want the derivatives with respect to time t (for velocity in a 1-dimensional space; velocity strictly speaking is a vector and needs magnitude and direction). \n\nThe first derivative of velocity with respect to time is called acceleration. The second derivative of velocity with respect to time is called the boost or jerk. \n\nYou have not stated if distance D is a function of time or not. Assume that D is a constant. \n\nThen dV/dt=-D/t^2 is the acceleration\n\nAnd d/dt(dV/dt)=2D/t^3 is the boost.
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