This seems easy but I still need help?!! Also how do you find the acceleration?
956088
x = -11 + 12t + 3t² + 4t³ - t^4\nv = 12 + 6t + 12t² - 4t³\na = 6 + 24t - 12t²\nfor max or min a, \n24 - 24t = 0\nt = 1 (and this yields a miximum)\na = 6 + 24 - 12 = 18 units/sec²\nv = 12 + 6 + 12 - 4 = 26 units/sec\nx = -11 + 12 + 3 + 4 - 1 = 7 units
456780
There is no one answer to this question. You need to know either the rate of acceleration or the time taken to get to 30m/s.\n\nYou will then be able to solve with the formula:\nv = u + 2as\n\nWhere v is the final velocity, u is the initial.\na is the acceleration.\ns is the time.\n\nTherefore:\n30 = 2as\n15 = as\n\nSo as long as acceleration and time equal 15 when multiplied together then that is a valid answer.
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
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
750537
acceleration = v^2/r or v*v/r (this is m/s^2 so divide by 9.81 to get g's)\n\nwhere v is the velocity\n\nr is the radius\n\nBe careful, think about the one g the pilot feels from the pull of the earth. ;)
77860
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 :)
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
210676
Use free body diagram to balance the forces and u can derive the equation of motion \nIf u want i can give u full derivation.\n"::
284176
ok well his initial velocity is 0 and the distance is 402m and time is 4.913. you have to find the acceleration. well the only way i can think of doing this  without using these basic equations that are derived from newtons law (i forgot the derivation) is this.     you find the average velocity needed to go 402m in 4.913s. which is just 402m/4.913s = 81.823....\nthis is just the average velocity and because acceleration is constant, this means you can just find the average by doing  Vi + Vf / 2. you have Vi =0    so Vf/2 = 81.823........ so Vf = 163.6474.........m/s\nacceleration is  change in velocity /time. you have change in velocity  (Vf-Vi)   so divide it by time.   \n163.6474-0/4.913   =  33.3 m/s^2\n\n2. so the acceleration due to gravity is 9.8m/s^2. you know the distance is 10m and initial velocity is 0. so Vf = a x t\nso Vf = 9.8t. and Vf+Vi/2 = d/t    same things basically. so Vf/2 = 10/t  which means  Vf x t = 20\nVf = 9.8t\nVf x t = 20\nso then that means that (9.8t) x t = 20\n 9.8t^2 = 20\nt^2 = 2.04....\nt = 1.4285.....  so it takes her that long\nand now you just find Vf by putting this time in the equation  Vf = 9.8t = 14m/s. thats final velocity\n\nb. ok so this question confused me at first but it snot hard. once she hits the water (with speed of 14m/s), she travels for 2m before coming to a stop. so distance is 2m. Vf = 0. Vi = 14m/s and you have to find acceleration. \nso a = (Vf-Vi) x t = -14t\nso you need time now. you do average velocity  again\nVi + Vf / 2 = d/t\n14/2 = 2/t\n7 = 2/t\nt = 0.2857.... s\nthen put it in the other equation and you get a = -14t\nso a = -4m/s^2.\nforce = mass x acceleration\nmass is 61kg so you do 61kg x -4 = -244 N\nthe water exerted 244N against the movement (thts why its negative).\n\nso hope this helped. let me know if you need more help or whatever. laterz
30757
just draw a free body diagram. you're solving for F parallel. that's just FgsinX which is just mgsin(30_ then you take this and divide it by mass to get the acceleration. so actually, you can cancel out mass and its just (2.15)sin(30) which is 1.075\n\nfor the frictional force, either you need more info, or it's zero.
59053
actually... although the first answerer did the correct calculation, he left out a step. that is the acceleration produced by the rocket, however you must take gravity into account. \nF=ma, a=F/m= 23.3 is the acceleration produced by the thrust of the rocket, but the rocket is not actually going to accelerate that much because an acceleration of 9.8(due to gravity) is pulling it down in the opposite direction. so 23.3-9.8= 13.53\nso the initial acceleration of the rocket is:\n13.53m/s^2
806029
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
716372
1 and 2 are easy. I'll help with #3.\n\n3. You need I, the rotational inertia. From a table in my physics book, I = MR^2 for the reel. You need to convert the force to torque, T = F*R. \n\na. Then you can get the angular acceleration, alpha, from \nT = I*alpha\n\nb. In t = 0.25 s, omega will reach alpha*t. The angular displacement, theta, will be (1/2)*omega*t. From that and the circumference you can get the line drawn from the reel.
111044
I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?
728641
acceleration can be found using\na=(m1-m2)g/(m1+m2) = 3*9.8/5 = 5.88 m/s^2
125809
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
1150360
The acceleration due to gravity is 32.2 ft. per second per second, which is pretty near 32.  Now we need two fundamental equations of physics:\n\nv = a / t\n\nwhich we can re-write\n\nt = a / v\n\nSince a (acceleration) is 32 ft/sec^2, and v (velocity) is 32 ft/sec, time (t) is 32/32 = 1 second until the ball's velocity equals zero.\n\nSo how much distance in 1 second? The second equation gives us the answer:\n\nd = 1/2 a t^2\n\nwe know a = 32, and since we now know t = 1, we can solve the equation. \n\nt^2 = t * t = 1 * 1 = 1\n\nso\nd = 1/2 * 32 * 1\nd = 16 feet