a car move from calm state and within 12/0s with 1/80 m/s2acceleration move untill this speed reach to v(velocity).then 8/0s with this fixed speed v continue his way , then put on the brakes and after 10/0s stops .indicate speed v and indicate car movement acceleration when he put on the brakes.(suppose this acceleration is fixed).indicate whole distance that car went.
130528
first range of time:\n***************************************\ninitial values: v0 = 0 m/s; t - t0 = 12.0 s; s0 = 0 m\na = 1.80 m/s²\n\nfinal velocity v:\nv = v0 + a(t-t0)\nv = 0 m/s + (1.80 m/s²)(12.0 s)\nv = 21.6 m/s\n\nfinal position s:\ns = s0 + v0*(t-t0) + 0.5(a)(t-t0)²\ns = 0 m + (0 m/s)(12.0 s) + 0.5(1.80 m/s²)(12.0 s)²\ns = 129.6 m\n\nsecond range of time: (final values of first range are initial values of second range)\n***************************************\ninitial values: v0 = 21.6 m/s; t - t0 = 8.0 s; s0 = 129.6 m\na = 0 m/s²\n\nfinal velocity v:\nv = v0 + a(t-t0)\nv = 21.6 m/s + (0 m/s²)(8.0 s)\nv = 21.6 m/s\n\nfinal position s:\ns = s0 + v0*(t-t0) + 0.5(a)(t-t0)²\ns = 129.6 m + (21.6 m/s)(8.0 s) + 0.5(0 m/s²)(12.0 s)²\ns = 302.4 m\n\nthird range of time: (final values of second range are initial values of third range)\n***************************************\ninitial values: v0 = 21.6 m/s; t - t0 = 10.0 s; s0 = 302.4 m\nfinal values: v = 0 m/s\na = ? m/s²\n\nacceleration a:\nv = v0 + a(t - t0)\na = (v - v0)/(t - t0)\na = (0 m/s - 21.6 m/s)/(10.0 s)\na = -2.16 m/s²\n\nfinal position s:\ns = s0 + v0*(t-t0) + 0.5(a)(t-t0)²\ns = 302.4 m + (21.6 m/s)(10.0 s) + 0.5*(-2.16 m/s²)(10.0 s)²\ns = 410.4 m
477522
easy ;-)\n\nthe friction force, call it F, is equal to the normal force (the force perpendicular to a surface - in this case, the car's weight), times the friction coefficient, call it k. And the weight would be equal to the mass, m, times g.  So you'd have\n\nF = k * m * g\n\nnow the acceleration is the force divided by the mass, so:\n\na = F / m = k*g = (0.642)*(9.81)  (in metric)\n\nthe formula you suggest to use is the correct one ;-)\n\nnow i'm used to metric so i'll convert your mph to m/s, 40mph is 17.8 m/s.\n\nso you have\n\nd = (17.8)^2 / 2*(0.642)*(9.81)\nd = 25 metres\n\nif you want feet, and don't want metric in between, then you'd have to use 32 ft/s2 and not 9.81 m/s2 for g.  And you'd have to convert your 40mph in ft/sec (about 58.3 ft/s).\n\nhope this helps
376844
The force needed to stop the car is \n\nFs = M*a\n\nwhere M is the mass of the car and a is the acceleration (deceleration). For hydraulic braking, the pressure input (Pi) on the brake pedal side is the same as the pressure on the output (Po) brake pad side.\n\nPi = Fi/Ai = Fo/Ao = Po\n\nwhere Fi is the force applied to the brake pedal, Ai is the area over which the force is applied, Fo is the output force (braking force), and Ao is the area of over which the output force is applied. Since Fo = Fs, we can solve for the input force\n\nFi = Ai*Fs/Ao\n\nFi = (pi*Ri^2)*Fs/(pi*Ro^2)\n\nFi = (M*a)*(Ri/Ro)^2\n\nwhere Ri is the radius of the input piston and Ro is the radius of the output piston.
75670
look at the question and it provides the answer --- it is acceleration occurring at a uniform rate (ie. it is constant).  for example, from a stationary start, a uniform acceleration of 2m/s^2 will give speeds of 0s=0m/s, 1s=2m/s, 2s=4m/s, 3s=6m/s.
79918
The force of friction is dependent on two things, the Normal force between the "object" and the "surface" and the coefficient of friction between the two.\n\nThe normal force on a level surface is equal the weight of the object.  You are given the mass, from this, you can easily calculate the force of weight (and this F_n).\n\nYou are given the coefficient of friction, greek letter mu.\n\nThe force of friction can be found as,\nF_f = F_n * mu\n\nThe force of friction is always in the direction opposite that of the motion.  So if your traveling "forward", the frictional force is pointed "backward".\nWhat are the two components in the formula that affect the magnitude of the frictional force?  Just the normal force (which is dependent on mass and gravity) and the coefficient of friction.  For all intensive purposes, mu is constant, as is the mass and gravity of the object.  Therefore, the force of friction is constant.\n\nYou are given in initial velocity.  You are trying to find how far (change in distance) the car will travel while this constant force is being exerted on it before the car comes to a stop.\nThere are several ways this could be found.  I think the energy method would be the easiest, but I will cover the kinematic method as well....\nThe car has some initial kinetic energy when the problem starts out.  KE = 1/2 mv^2.  After the car comes to a stop, the car has zero KE...therefore some work had to be done on the system to change the energy (we assume a level road).  The change in KE equals the work done.  Work is also equal to the product of the force applied over the distancd applied.\nW = F * d\nSince the frictional force is what is causing this change in energy (which we know, it was calculated earlier), and we also know the change in KE, we can simply solve for the distance, d.\n\nThere is also another possible solution.\nWe know that the frictional force will cause an acceleration.\nF = m*a\nWe know that an acceleration will cause a change in velocity.  We know ,\nv_f^2 = v_i^2 + 2ad\nwhere v is velocity (i is initial, f is final), a is acceration, and d is distance.  Simple solve for d.
384036
Firstly, let me point out to you which answers are wrong and which answers are right. Otherwise you'll be just as confused as you were to begin with.\n\nThe first answer is confused. The second and third answers have no idea of what their talking about. The other answers are correct.\n\nHere's why the first answer is confused:\n\nSpeed and velocity use exactly the same units. \nSo it is perfectly acceptable to say "a car travels around a sharp curve a 60 km/hr". \n\nRestated, what we are saying is that the car moves around a curve at constant speed of 60 km/hr. The statement above does not include directional information. \n\n(However, suppose we said "a car travels around a sharp curve with constant velocity of 60 km/hr". Then, clearly this would be wrong. Velocity does change as it goes around the curve.)\n\nSecondly, let us recall some basic facts of 'Uniform Circular Motion' \nAn object that travels with Uniform Circular Motion has:\n- constant speed \n- changing velocity  \n- acceleration towards the centre\n\nNow, the car in question displays uniform circular motion. This is so even though we are considering a partial circular path.\n\nSo, we can use this import formula (you should memorise it) \n\n    ar = v ^2 /r\n\nThis relates the centripetal acceleration, ar, to the radius of the circle (or in the case of a partial circle, the radius of curvature), r.\n\nFrom the formula you can see that a sharp bend, i.e. a small r, will give a larger value of centripetal acceleration compared to a gentle bend, i.e a large r. \n\nSubstitute v=60 and different values of r into the formula above to verify this.
343170
Velocity:\nv=d/t\n\nd=distance\nt=time\n\nAcceleration:\na=(vf^2-vi^2)/2d\n\nvf=final velocity\nvi=initial velocity
559574
I assume it's starting from zero speed?  If so,\n\nd = v1t + 1/2at^2\nafter 2 seconds (i.e. 4 half-seconds), it's travelled\nd = 0(2) + 1/2(2)(2)^2\n   = 0 + 1/2 * 2 * 4\n   = 4m\n\nafter 2.5 seconds (i.e. the fifth half-second), it's travelled\nd = 0(2.5) + 1/2(2)(2.5)^2\n   = 0 + 1/2 * 2 * 6.25\n   = 6.25 m\n\nSo in the last half-second, it's travelled 6.25 - 4 = 2.25m.
341641
V=Vo - a*t\nD=Vo*t-a*t^2/2\nWhen braking without slipping,\nF=m*a\nm*g*0.642 = m*a\na=0.642*9.81=6.3 m/sec^2\nconvert mph to m/sec we get Vo= 17.8 m/sec\nThen,  17.8 = 6.3 * t  hence t =17.8/6.3=2.83 sec\nD=17.8*2.83 - 6.3*2.83^2 /2 =50.37-25.23=25.14 m
314934
Okay, I've thought about this and it's not possible to determine the rate of acceleration from the given criteria.  It is an "average" of 60 miles per second, BUT how many miles in the first second compared to the number of miles in the final, fifth second?  If that is known, you can project MPH from seconds 6 to 10 or the additional 1/4 mile.  How do you calculate the rate of acceleration in your theory?  How fast is the vehicle traveling at each point of the 1/4 mile distance?
305207
d = vt\ntP = 5tR\ndP = 300\ndR = 160\nvP + 50 = vR\n\n300 = vP * tP\n\n160 = vR * tR\n160 = (vP + 50) * tP/5\n160 = (300/tP + 50) * tP/5\n800 = 300 + 50tP\n500 = 50tP\n\n-----------------------------\ntP = 10 hours\ntR = 10/5 = 2 hours\nvP = 300/10 = 30 mph\nvR = 160/2 = 80 mph\n-----------------------------
441984
The friction is Ff = Uk*m*g = 0.71*m*g.\n\nThe car had original kinetic energy KEo of (1/2)*m*v^2. At 25 km/h, the car had KE25 of (1/2)*m*(25 km/h)^2. Convert to m/s before squaring. Find an expression for delta kinetic energy that the brakes scrubbed off before impact.\n\nThe friction had to do work equal to the delta kinetic energy to bring the car down to 25 km/h. Work is force*distance so\nW = Ff*d = 0.71*m*g*47 m\nand work is also is equal to delta KE\nSet them equal and solve for v.
511400
Since the vehicles lock together and travel at 45deg to the original motion of both vehicles, then the momentum of each before the collision must have been equal. We will call the mass of the Truck, Mt and the mass of the Car, Mc. The velocity of the Truck, Vt and the velocity of the car Vc. So:\n\nMtVt = McVc ...........(1)\n\nand \n\nThe total momentum immediately before the collision will be the same as the total momentum immediately after the collision:\n\nMtVt + McVc = (Mt+Mc)Vd ............(2)\n\nwhere Vd is the velocity of the Truck and Car combination immediately after the collision.\n\nAlso, we know that the kinetic energy of the Truck and Car combination immediately after the collision is:\n\n0.5(Mt+Mc)Vd^2................(3)\n\nNow all of this energy is consumed by the work done by the friction between the tires and the road over 18m. So:\n\n0.5(Mt+Mc)Vd^2 = W = Fx18....................(4)\n\nWhere F = (Mt+Mc)xa.....................(5)\n\nand a is the acceleration due to friction = 0.923x9.8 = 9.0454ms^-2\n\nNow, you have values for Mt, Mc and a, so you can calculate Vd from equation (4).\n\nFrom equations (1) and (2), and knowing Vd, you can calculate Vt and Vc.
520647
Yes you're on the right track, you know the momentum of the car along with the energy, both are conserved.  \n\nForce=Mass*Acceleration=Coefficient of Friction*Mass*Gravity\n\nYou can then solve for Acceleration=Coefficient of Friction*Gravity\nVelocity=Acceleration*t\nThen its simply a matter of filling t into this equation:\nDistance=1/2*Acceleration*t^2+Velocity*t
395489
hello\n\na)take the momentum of the sports car and divide by the total mass of  both cars. 25×920÷3220=7.14m/s\n\nb)using the same method as a), we can find the answer. but use a little logic here, if the total mass doubles then the final speed must be halved for conservation of momentum to be followed. v=12.5m/s\n\nc)KE=0.5mv². speed is halved. v² means kinetic energy should be quartered (0.5²). but mass was doubled due to there being 2 cars now. so overall kinetic energy has only been halved (0.5²x2). the other half is lost.
219373
Assume starts at Zero speed.\nMax speed is after 20 sec during which added 1.5m/s of speed each of these seconds.\nDistance during this phase d = Half a * (t squared)\nDistance during middle phase is  Max speed (as above) times 120 (NOT 2) seconds.\nDeceleration in third phase is Max speed (as above) divided by each of the 15 seconds it took to lose that speed. (If it does not work out as 4/3rds the acceleration, you have a problem!)\nDistance in the third phase is again d = Half a t^2  (Strictly speaking ut + 0.5at^2 with negative value of a, but no difference in this case as comes to a stop)\nDraw graph of speed v time (NB seconds - don't mix with minutes) find area by geometry.\nEdit - give you a clue - the distance is not 3900 and the deceleration is not 0.5 (nor even -0.5) as above.  Have you ever been tempted to give someone the wrong answer?
433053
wish I could solve em for you but ill try and help still.\n\n1) The acceleration should be the derivative of the equation for the car's velocity.  The original distance/time equation would be the anti-derivative of the velocity.  \n\n2 The equation to find out how long it will take to get there is h+ax-16x^2 in feet.  (a is the original speed and x is number of seconds and h can be the height of the building.)\n\nHope this helps