what is the ideal bound of vertical acceleration that translated to passengers in a car active suspension?
408573
Zero.  The whole point of active suspension is to anticipate bumps and dips at the incipient time that the wheels encounter them so that no disturbance is experienced by the passengers.  The "read ahead" of the road needs to take into account the vehicle speed and the response time of the system actuators so that the wheel is being pushed down at precisely the time it enters a pothole.  Otherwise, it's going to be like the hydraulic systems on some low-riders.  Similarly vehicle speed, actuator response time, and steering input need to be accounted for to preclude roll. (lateral rotation caused by centrifugal force.)
895549
suspension and floating
812368
I get the radial acceleration = ar = V²/r = (325*1000/3600)²/194 = 42.01 m/sec²\nI agree with your value for at = 6.69 m/sec²\n\nThe total acceleration = A = √(42.01² + 6.69²) = 42.539 m/sec²\n\nThe Cf required = A/g = 42.539/9.8 = 4.34\n\nTypical tire/asphalt Cf is around 1.0\n\nAmazing what a little bit of bank angle can do for you!
811400
If you ignore wind resistance the calculation is quite straightforward.\n\nAcceleration due to gravity is 9.8 metres per second per second.\nEvery second the car will accelerate by an additional 9.8 m/s\n\nYou convert 100 km/hr into m/s first:\n\n100 x 1000 (metres in a Km) / 3600 (seconds in an hour) = 27.78 m/s\n\nDivide that speed by the acceleration:\n\n27.78 / 9.8 = 2.83 seconds
856976
Extreme position! \nBecause acceleration is directly propotional to the displacement from the mean position and directed opposite to it.\n\na=-(ω^2)x
857622
1) I think your 2nd c.f. should say c. acceleration; that's the units you're using------ the c force of a mass m with acceleration a is f = ma, so youhave 1.5 kg*18 m/s² = 27 N\n\n2) First find the velocity: v = d/t = 2πr/t = 2π*.75m/.8sec = 5.89 m/s\nThen, acceleration: a = v²/r = 5.89²/.75 = 46.256 m/s²\nThen, the tension: T = ma = .1kg*26.256m/s² = 2.626 N\n\n3)  v = 2πr/t = 2π*90/12.3sec = 45.974m/s\na = v²/r = 45.97²/90 = 23.485 m/s²\nF = ma = 900*23.485 = 21,136.5 N\nThe lateral friction of the tires on the track
288399
How many feet does it take the car to reach 10 mph? What is the acceleration curve?
123977
Are you asking for the time it takes the car to accelerate 13.89 m/s....well then the answer is in the question.\n\nPerhaps you are asking for the distance  traveled while accelerating?\nchange in distance = 1/2at^2 + v_i*t\nwhere a is the constant acceleration, t is the time, and v_i is the initial velocity.\nthe car's acceleration = change in velocity / change in time
853743
(B) At the point of a dip between two hills the force of inertia and gravity would tend to push the car down on the pavement with more force. \n(A) At the top of the hill the opposite forces would tend to make the car lighter. \n\nOf coarse several other factors would come into play like the size, speed and weight of the car, and the severity of the hills. These factors would lessen or increase the amount of force at a given point.
461030
For the first problem, you need a law of motion. (final velocity)^2=(initial velocity)^2+2*acceleration*displacement. Substituting, you get 0=30^2+2*a*20m and a=22.5 m/s^2. Now, we need to write a force equation: ma=Tension force-weight. To find the mass of the elevator and its load, divide its weight by gravity (9.8 m/s^2) to get about 16.3 kg. And you already calculated the net acceleration. Our force equation can now be rewritten as (16.3 kg)*(22.5 m/s^2)=Tension force-160N. The tension force is about 207N.\n\nFor each of the weight scenarios in the second problem, you know the net acceleration to be 30 m/s^2. You can also find the mass of the passenger by dividing his/her weight by gravity. With the force equation ma=(force of feet up)-(weight) with respect to the passenger, you can find the force of his/her feet. Here is the work for the first scenario.\n\n16N/(9.8 m/s^2)=1.63 kg\n1.63 kg*30 m/s^2=(force of feet)-16N\nForce of feet=64.9N\n\nAnswers for the second and third scenario:\nb) 2600N\nc) 260N
974269
To stay on the road with no slip the vector of total force on a car F must be strictly normal to the bank with b=2.5 degrees. Vector F=Fx+Fy, Fy=m*g being the weight of the car, Fx=m*a=m*r*w^2 is centripetal force on the curve of r=120m, w=v/r is angular speed, v is linear speed, a=r*w^2 is centripetal acceleration. On the other hand as Fx=F*sin(b) & Fy=F*cos(b) or Fx/Fy=tg(b) or [m*r*(v/r)^2]/[m*g]=tg(b), hence v=sqrt(g*tg(b)*r)=7.17m/s=25.81 km/h. Hi to SS!
486362
Minimum weight allowed is 600kg, but some weigh as little as 440kg to allow the teams to add ballast they can move around in the car to help fine tune the handling / weight distribution, etc.
295799
k ill try to help u  even though i dont condone cheating. i just know how it is to not understand something and its due in a few hours and it sucks. just remember, nobody will be able to help u on the exam.\n\nso 1. \na. acceleration is change in velocity divided by time. \nsince it started from rest, initial velocity is 0 and then final velocity is given as 25.3 m/s. and the time is given as 0.59s. so to get acceleration   you would do  \na =  Vf-Vi / t\n  =  25.3 - 0 / 0.59\n  = 42.88......\n\nb. the force is equal to mass x acceleration. since you know the mass and acceleration, you just multiply them and you get F = 38421.69... N. \n\nc. you have to find the normal force in this case, which is the force that opposes gravity. like upward force. so in this case, gravity is acting on the driver and force of gravity is equal to mass x 9.8 = 666.4 N. and that means the normal force is 666.4N (since he is not moving up or down the total vertical force = 0)\n\n2. ok well in this case, you know the initial velocity is 0, the distance is 404m and the time is 4.944s. if its constant acceleration, you could assume that the average velocity is just Vi+Vf/2. so if thatz average velocity, then distance = average velocity x time\n     404 = 4.944s  x  (0+Vf)/2\n 404 x 2 / 4.944 = Vf\nVf = 163.43........m/s\n\nthe acceleration would just be Vf-Vi/t. and since you know Vf now  you just do   a = 163.43-0/4.944\na = 33.056....m/s^2\n\n3. ok well acceleration is 9.8m/s^2 (gravity)     distance is 9.7m. initial velocity is 0m/s. so this time, you have to solve 2 equations. the first one is the acceleration equation. a = Vf-Vi / t = Vf / t = 9.8\n9.8 = Vf/t\nVf = 9.8t\nand the other one is the distance = t  x  Vi+Vf/2   average velocity one. \n9.7 = t x Vf / 2\n19.4 = Vf x t\n19.4 = (9.8t) x t\n19.4 = 9.8t^2\nt^2 = 1.979...\nt = 1.406....s\n\nnow that you know the time, you find the Vf by putting it in any of those 2 equations. Vf = 9.8t = 9.8 (1.406)\nVf = 13.788... m/s\n\npart b. so in this case, you use the initial velocity as the one u just found (the velocity of the swimmer at the surface of the water)   so Vi = 13.788m/s\nand Vf = 0    since they come to a stop. and d = 2m\nso you have to find the acceleration. first you need to find the time by using the average velocity equation.\nd = t x Vi+Vf/2\n2m = t x (0+13.788)/2\n4 = 13.788t\nt = 0.29s\nnow that you know the time, you have to find acceleration\na = Vf-Vi/t\n  = 0-13.788 / 0.29\na = -47.5286m/s^2    its negative because its downwards and up is positive in our case. so the force = mass x acceleration and you just got acceleration and her mass is 65kg so you just multiply by 65kg to get F = 3089.36N\n\n\n4. well F = ma and you have the force. when they say it weighs this much newtons, then it means  use gravitational acceleration which is 9.8. so 2350N = m x 9.8m/s^2\nm = 2350/9.8\nm = 239.79kg\n\ni hope you actually understand this stuff. try to do your homework by yourself   dont take advantage of this site. youll just end up screwing yourself over. and if u dont get anythin, let me know and ill try to help
850174
You can start out with two equations:\n\nF = k*M*g   :  Where F (Newtons) is the static friction force of the car, k (unitless) is the coefficient of static friction which we will solve for, g (m/s2) is acceleration due to earth's gravity, and M (kg) is the mass of the car.\n\nFc = M*V^2/R  :  Where Fc (Newtons) is the centripetal force due to circular motion, M (kg) is the mass of the car, V (m/s) is the velocity of the car during the turn, and R (m) is the turn radius.\n\nThe car must maintain its grip on the road to meet the conditions of the problem, so we have to set F = Fc and solve for k, which will give the minimum value to maintain traction.\n\nSo, setting them equal and solving for k gives:\n\nk = V^2/(r*g) = 0.83513...\n\nYou should notice that the car's mass cancels out so we never needed it.  To get the answer, remember to convert 95 kph to meters/second and use 9.81 m/s2 for g.\n\nA value larger than k means that the tires have extra grip and could handle a tighter turn.  Any less than k, and the car's centripetal force would overpower the force of friction of the tires, and the car would slide.\n\nI hope this helped.
786122
Acceleration is defined as the change in velocity (speed) with respect to time. In your problem statement, the phrase "slowing down uniformly" implies that the change in velocity is happening at a constant rate.\n\nKnowing this, you can compute the acceleleration as:\n\na = (v2 - v1) / t\n\na = (5 - 20 m/s) / (4 s)\n\na = -3.75 m/s²\n\nUNDERSTAND THIS: The negative sign means that the velocity is slowing down. It does not imply that the velocity is negative.\n\nFor part (a), force is defined as the product of mass and acceleration. In other words, a force of 1 N is required to acclerate a 1 kg mass at a rate of 1 m/s². Mathematically:\n\nF = m * a\n\nYou're given the mass of the car, and you now know its acceleration, so the average force acting on the car is simply:\n\nF = m * a\n\nF = (2000 kg) * (-3.75 m/s²)\n\nF = -7500 N\n\nThe negative sign means that the force is acting against the car's direction of travel.\n\nFor part (b), the distance the car travels in this time is calculated by the distance equation:\n\nS = (1/2) * a * t² + vi * t + Si\n\nWhere vi is the car's initial speed and Si is the car's initial displacement (distance). Initially, at t=0 s, we'll assume that the initial displacement is 0 (because we only want to know how far the car travels from when we start the clock at t=0). Additionally, vi is given as 20 m/s. Plugging everything into the distance equation provides:\n\nS = (1/2) * a * t² + vi * t + Si\n\nS = (1/2) * (-3.75 m/s²) * (4 s)² + (20 m/s) * (4 s) + 0\n\nS = -30 m + 80 m\n\nS = 50 m\n\nThe car travels 50 m during this period of slowing down.\n\nHope that makes it all clear! :)
841492
a)The acceleration a = Force / mass = -6070 / 2600 = -2.33 m/s^2\nInitial velocity u = 34.9 m/s\nThe velocity at any time is given by v = u+a*t = 34.9 - (2.33*1.76) = 30.8 m/s\nb) v^2 - u^2 = 2*a*s, where s is the distance travelled in that time. \ns = ( v^2 -u^2)/2a = 57.8 m\nc) The car stops when u + a*t = 0.\nt = 34.9/2.33 = 15 s
739092
PE at release point = KE + PE at the top of the loop.\n\nmgh = 1/2 m v^2 + mg 2R   the 2R is he height of the loop.\n\nSolve for KE\n\nmgh-mg2R = KE = mg(h-2R)\n\nNow for the second one I believe you have to use Net F = ma\n\nAssume the normal force is 0, because the car is barly staying on the track. so Fgrav = m v^2/R\n\nso mg = mv^2/R\nso g = v^2/R\n\nNow going back to the energy equations, mgh = 1/2 mv^2 + mg2R\n\nm's cancel so gh= 1/2 v^2 + g2R\n\nsolve for h and you have h =    (.5v^2 + g2R)/ g  make that into 2 fractions and you have \n.5v^2/g  +  g2R/g  and v^2/g from the equation above = R\n\nso .5R + 2R = h  so 2.5R = h\n\nHope thats right!