what is the formula for calculating velocity?
111448
velocity is a vector which means it must have a magnitude and a direction.  speed is not a vector so it only has a magnitude with no direction.  \n\nthe "formula" for speed is distance/time.  so if you were traveling 50 miles and you did that in 1 hour you would be travling at a speed of 50miles/hour.   \n\nnow to answer your question velocity must have a magnitude so it is the speed with a specified direction. therefore if you were travling at a speed of 50miles/hour and you said you were going north. then it would be a velocity of 50miles/hour north.\n\nvelcoity = distance/time with a direction.
879746
Ek = 1/2.m.v2   (1/2 mass times velocity squared) - \n\nsorry can't do superscript on this site
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.
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.
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
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
715989
Calculate the number N of seconds in a year (365*24*60*6). Time t in s = N*Ta. Velocity in m/s = 2*pi*632,000,000*1000/t.
991014
Ballistic Coefficient 1 \nHere, you enter a known velocity for your bullet, and then either a second velocity or a time value followed by the distance travelled by the bullet for that second velocity or time value. Use the combo box to tell the program which value you wish to use. When you click on the calculate button, the other value (velocity or time) will be calculated and displayed. The resultant Ballistic Coefficient will be that coefficient for the drag function selected in the Shooting Parameters window. For example, if G1 is the currently selected drag function, enter 2600 ft./sec. for Velocity 1, 2500 ft./sec. for Velocity 2, leave time blank, Specify 'Use Velocity 2' for Select, and 20 yards for distance. Click on the calculate button and a G1 Ballistic Coefficient of 0.178 will be displayed. You will also see that the time to travel those 20 yards is 0.02353 seconds. The computation also incorporates the weather parameters currently defined in the Shooting Parameters window. The formula for this is as follows: \n\nbc = [Cd(mach)](pi)(rho)(Vavg)(Vavg) / [(8)(a)] \n\nwhere, \n\nbc = ballistic coefficient \nCd = drag function \nmach = mach number (bullet velocity / speed of sound) \nrho = atmospheric density \nVavg = average bullet velocity, (V1 + V2) / 2 \na = constant bullet acceleration over the interval \n\n\nif you know V2 then: \n\ntime = distance / Vavg \n\nif you know time then, \n\nV2 = (2)(distance / time) - V1 \n\nand, \n\na = (v1-V2) / time \n\nNote that the speed of sound as well as atmospheric density is a function of weather parameters. \n\nBallistic Coefficient 2 \nThis calculation computes the ballistic coefficient when the bullet diameter, bullet weight, and form factor are known. Some early ballistics techniques allowed the determination of a form factor from charts where you could match the shape of your bullet to known shapes and extract a form factor for a particular drag function (see Hatcher's Notebook). The formula used here is: \n\nbc= weight / [(form factor)(diameter)(diameter)]
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
162313
Let V = speed (the magnitude of the initial velocity)\na = angle, t = time, \nVx = horizontal (x) component of velocity\nVy = vertical (y) component of velocity\nAx = acceleration in x direction\nAy = acceleration in y direction\nDx = net distance traveled in x direction\nDy = net distance traveled in y direction\n\nVx = Vcos(a)\nVy = Vsin(a)\n\n  Dx = 1/2 Ax*t^2 + Vx*t\nbut Ax = 0, so Dx = Vx*t, t = Dx/Vx\n\n  Dy = 1/2 Ay*t^2 + Vy*t\n(Ay = -g, the acceleration of gravity, negative for downward)\n -1/2*g*Dx^2/(Vcos(a))^2 + Vsin(a)*Dx/(Vcos(a)) = Dy\nMultiply by V^2, and collect terms with V^2 on the left:\n(Dx*tan(a)-Dy) V^2 = 1/2*g*(Dx)^2/(cos(a))^2\nV = sqrt(1/2*g*(Dx)^2/(cos(a))^2/(Dx*tan(a)-Dy))\nsubstitute a = 30 degrees, Dx = 6.05 m, Dy = 3.05-2.2 = 0.85 m, g = 9.8 m/s^2\nV = 9.51 m/s\nFor a = 60 degrees,\nV = 8.63 m/s
888452
f=ma\n\na=change in velocity/change in time = (5.53-2.25)/1 \n\nm= 57\n\nf = 57 (5.53-2.25)\n\npunch the numbers in to your calculator, and there it is!
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.
792153
‘impulse’ or momentum\n=m*(v2-v1)=0.42*(-17-4.1)=-8.862 kg*m/s
864633
He's right. You take the derivative of the equation and plug in the values for t (just consult any book on differential calculus for details on this)\n\nSo h'(t)= -3.92t+10 (the derivative)\n\nWhen t=2.3, the velocity will then be (-3.92)(2.3)+10, or .984 m/s\n\nWhen t=2.7, the velocity will then be (-3.92)(2.7)+10, or -.584 m/s (.584 m/s in the opposite direction)\n\nNot bad for a high school sophomore who taught himself differential calculus, eh?
49799
In polar the velocity vector is v = [ dr/dt, r * d(theta)/dt ]\n\nSo r(theta) = .1 * theta and theta(t) = 2t so:\n\nr(t) = r(theta(t)) = .2*t\n\nSo now:\n\ndr/dt = d/dt(.2 * t) = .2\nd(theta)/dt = d/dt(2 * t) = 2\n\nSo then the velocity vector becomes:\n\nv = [ .2 , 2 * .2 * t ]\nv = [ .2 , .4 * t ]\n\nAnd at time t = 1 we have\n\nv( t = 1 ) = [.2 , .4]\n\nSo finding the magnitude of the velocity vector:\n\n|v| = sqrt( .2^2 + .4^2 )\n\n----------------------------\nNow for acceleration:\n\na = [d^2(a)/dt^2 - r * ( d(theta)/dt )^2 )\n, 1/r * d( r^2 * d(theta)/dt )/dt ]\n\nAnyway, we note that\n\ndr^2/dt^2 = 0\nd(theta)/dt = 2\n\na = [ - .2 * t * 2^2, 1/(.2*t), .8 ]\na = [ -.8, .8 ]\n\nSo then |a| = sqrt( (-.8)^2 + (.8)^2 )
67284
well the formula is the square root of 3kT/m  where k is the boltzman constant(1.38066 x 10^-23 J/K) T is in Kelvin so you would have to convert(just add 273 to the temp in Celcius) so your T is 290K and m is the mass. the mass can be found using the density of water(1000 kg/m3) the formula for density is d=m/V and V can be found by using the volume of a sphere, 4/3(pi)r^3 oh, and don't forget to convert the nanometers into meters, then cut the diameter in half for radius.\nusing this information to get you started, you should be able to do it yourself. so you should try it first before reading further :) im sure you were familiar with all of those formulas, you just need to remember to think of things all at once so you can put them together.\n\nok, so we start with the rms velocity formula. first problem is that we need mass. ok, lets solve for mass using density. oh right, we need volume, ok, lets solve for volume then:\nwe need to make the diameter into meters, so thats 1.0 x 10^-8 meters. now we cut this in half to get the radius(r) so that's 5.0 x 10^-9.\nV=(4/3)(pi)r^3\nV=(4/3)(3.14159)(5 x 10^-9)^3\nV=5.236 x 10^-25 m^3\nnow we have volume, so using the density of water(1000 kg/m3), we can solve for mass:\nd=m/V\n1000=m/(5.236 x 10^-25)\nm=5.236 x 10^-22 kg\nok, now we have mass, so we can use our original rms velocity formula and solve for velocity. *sigh* finally.\n\nv(rms)= sqrt[3kT/m]\nv(rms)= sqrt[3(1.38066 x 10^-23)(290)/(5.236 x 10^-22)]\nv(rms)= 22.94 m/s\n\ni would check my math if i were you though, i tend to make stupid calculator errors. the concepts are there though :p
797399
velocity=107,300 km/h (or if you prefer 67,062 miles per hour)