Can you help me with acceleration?
964035
You need to consider vectors ....\n\nAcceleration has a direction. In the case of gravity its down towards the ground. \n\nNow what accelerationis doing is changing the instantaneous velocity of the particle or object or whatever.\n\nVelocity is also a vector.\n\nSo if the velocity is in the opposite direction to the acceleration then the object will motion will slow down and eventually increase in the same direction as the acceleration.\n\nIf the velocity is in the same direction then it will increase in that direction.\n\n So if UP is defined as positive then down is defined as negative. This goes for:\n- position (above (+) and below (-) ground if ground level is defined as zero);\n- velocity is positive if the object is moving up and negaitve oif it is moving downward; and\n- acceleration is positive if applied upward or negative if applied downward.\n\nNote acceleration is in the same direction as an applied force as by Newton's Second Law, F = ma and mass is a scalar quantity .. it has no directionality. It is the same wherever the object is. (Not to be confused with weight which is a force attracting us to the ground and is dependent on the strength of gravity. So for example we weigh on earth and are weightless in space and are 1/6 of our weight on the moon etc ... In all locations our mass is the same but the forces acting on us are different)
826560
3.0m/s is not an acceleration, it is a speed.\nIs the acceleration constant?
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.
351440
well ok. for part a, the acceleration is in the horizontal direction, because its moving horizontally. so the force is just mass x acceleration which is 70*6=420N. but thatz only in the horizontal direction.\n\nfor part b, you have to account for gravity. so gravity is pulling down at 9.45m/s^2, but youre accelerating upwards at 6m/s^2. so that means that youre overcoming the gravity and accelerating at 6m/s^2 MORE than the gravitational acceleration.\nso 9.45+6 = 15.45. this is your actual acceleration.\nso now you just do F=ma again and you get 70*15.45=1081.5N. \n\nhope this helps. let me know if you need more help. and this isnt thermodynamics.
1186433
Well, I think that for the first oneyou can get the acceleration by deriving it twice, then you get the integral from 1 to 5, let's see...\n\nds/dt= 6t^2+3=v\ndv/dt= 12t=a\n\nfor a\n\nI think you should do it this way\n\nintegral of vdt/ integral of dt from 1 to 5\n\nso you will get\n\n(2t^3+3t)/t from 1 to 5\n\nso it will be\n\n2t^2+3 from 1 to 5\n\nevaluating you'll get\n\n2*(25)+3-2-3\n\n=50-2\n\naverage speed = 48\n\nfor b\n\nyou just substitute t=4 to the first derivative\n\nV=6*(4)^2+3=99ft/min
783790
F=-kx and F=ma\n\nfind mass from period     T = 2pi(m/k)^(1/2)\n\nhmmmm,  you seem to be missing k\n\nfor fixed ratio m/k the period of 0.51s and amplitude 4 cm is not enough to determine the mass (and therefore acceleration).
805281
OK, first we need a formula for the motion, which, if it's simple harmonic motion is\nf(x) = Asin(kx)\nwhere A is 1.8*10^-4\nand k = 2000*2*pi\nSo f(x) = 1.8*10^-4*sin(4000*pi*x)\n\nTo compute the maximum force, we need the accelleration f''(x)\nf'(x) = 1.8*10^-4*(cos(4000*pi*x))*4000*pi, or \n1.8*4*pi*10^-1*cos(4000*pi*x)\n\nf''(x) = 1.8*4*pi/10*(-sin(4000*pi*x))*4000*pi\nor\n1.8*16*pi^2*100*(-sin(4000*pi*x))\nthe MAXIMUM acceleration is then 1.8*16*pi^2*100.\nThis is because you can find a value for x where -sin(4000*pi*x) is 1.\n\nUsing F=ma, the maximum force is 50 *1.8*16*pi^2*100 KG m/s^2, or approximately 1.42*10^6 KG m/s^2
312619
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.
716223
The change in velocity of an object undergoing uniform acceleration, a, for a given time, t, is given by:\nv = at\nwhere v is the change in velocity of the object.\n\nIf the object is accelerating at a constant rate of 10 m/s^2 for 5 seconds, the change in velocity the object experiences is 50 m/s.\nSince (we assume) the object started from rest, the objects final velocity after 5 seconds is 50 m/s.\n\nTo find the average velocity, divide the total displacement of the object by the total time.\nIn this case, the object has traveled 125 meters in the 5 seconds it has been accelerating (d = 1/2 at^2), so dividing this by 5 seconds and we get an average velocity of 25 m/s.
1001506
Dude! all you have to do is increase the velocity!\nOr Dude! I got it! E= MC2
483925
You may want to try Centripedal Force i.e. Turnings and slopes..\n\nBut it has some stuff to do with Physics as well. \n\nAlso you may want to find the maximum and minumum speeds at turnings. Include Frictional Force and try to minimize it using Calculus. \n\nTake Simple Circuits like circles, oval, parabola etc. Then take their equations and use calculus to find area under the curve etc. etc.\n\nWell thatz all comes to my mind for now!! :) BEST OF LUCK..
45821
are you sure that you want acceleration instead of velocity??\n\nwell, the acceleration is going to be 0.  since the speed is going to be the same the whole time, there won't be any acceleration.
91548
tried this problem, but when converting 2cm disance to meters, the acceleration was way out, too large.I used the equation V2 = U2 + 2aS to arrive at the acceleration (deceleration). This might have been the wrong approach.
396138
a=F/m, lim x->inf x/x = 1. The object will accelerate.
709444
Acceleration on any free falling object is always equal to the force of gravity, which is 9.8m/s^2.\n\nSince acceleration is a vector quantity, it can be denoted as positive or negative. In this case, since the object is moving downwar and not up, acceleration would be negative and so equal to -9.8m/s^2.\n\nGood luck.
952712
Since the angular acceleration is constant, you can take the average of 48 and 120 = 83 rads/sec.\n\nThus 7*83=581 revs.
1022160
I hope this will help. \nFor dynamics see reference 2