A car is braked so that it slows down with constant deceleration from 30ms to 15ms-1 over 75ms-1.?
919282
please make your question more understandable\n\nif it decreases from 30mph to 15mph over 75 metres then at this deceleration its logical to say it would take another 75metres to stop\nbut something tells me thats not the answer ! if you want me to do your homework at least have the decency to type in the full question.
589085
Most all car manufactures are installing Automatic Braking Systems (ABS) in the cars they sell.\n\nTerry
698048
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
757648
Friction force:\nF = m g c\n\nm = mass of car\ng = gravitational accel\nc = coeff. of friction\n\nCentripetal force:\nF = m v^2 / r\n\nm = mass of car\nv = velocity\nr = radius of curve\n\nSet the two forces equal to each other:\nm g c = m v^2 / r\n\nCancel m, plug in given values:\n9.8 m/s^2 * c = (12 m/s)^2 / 52 m\nc = 0.28257\n\nSo, c has to be at least 0.28257 to keep the car from slipping.
279308
There are two types of Traction Control. One system applies the brakes to a specific wheel when a sensor detects it slipping. The 2nd way is for the car to slow itself down by reducing the amount of fuel when it detects wheel slippage. Many cars with Traction Control use a both systems.
1170380
You have mass of car = 1500 kg.\nand speed = V = 25 meters per second.\n\nYou will recall the equation \nE = (1/2) * m * v^2\n\nKinetic Energy = 0.5 * mass * speed *speed.\n\nPunch these numbers into your calculator and you will get a numerical answer with units kilogram meters meters per second per second.\n\n\nThis is the kinetic energy of the car, which is removed by the brakes.\n\nI'm not going to do all your homework for you.  Just the first part to get you rolling.\nHave a great day.
557602
This is how you solve it. Use dimmensional canellation, but you will have to do it twice and subtract the difference\n\nA. (5 miles/1)  *   (1 hr/55 mi)  * (60 min/1 hr)\n   =5.45 minutes\n \n   \nB. (5 miles/1)*  (1 hr/65 miles) * (60 min/1 hr)\n   =4.6 minutes\n    \nYou saved 1.2 minutes. Not much at all. Which is what they say, speeding only saved 2 or so minutes in a 15-20 minute trip. BUt it tears the bejesus out of your engine
1007337
The KE of the moving vehicle is dissipated as heat in the brakes as the brakes use friction.\n\nThe answer to your second question is that no work has been done and therefore there has been no change in KE.\n\nIn terms of transfer of momentum, the really huge ball called the earth and the car did interact.  If the car started out going West and then ended up going East, the total momentum (angular and linear) between the car and the earth must be conserved.  This means that the earth was caused to change its rotation when the car turned around.  Obviously, the earth is so much bigger than the car that you don't even notice the change in momentum of the earth.
927736
Everyone above assumed a constant speed.  the more general formula is \n\nD = ∫v dt
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.
598306
nothing-\nthe system of atoms inside your car is stationary relative to the baloon.
378184
When the cylinder rolls, there is a force F=15N at the top and a static frictional force f preventing slipping.\neverything in usual notation,\nacceleration A=R*a (a=angular acceleration)\nNewton's second law implies'\nMA=F-f         (resultant force)\nalso for torques,\nIa=FR+fR         (for the axis passing through the centre)\nFRom the two equatins,\n2F=mA+Ia/R\ni.e.  2F=maR+Ia/r\n   a=2F/(mR+I/R)\nI=0.5MR^2=0.05\nplugging hte values,\na=16 \ntherefore, A=aR\n            =16*0.1\n           A=1.6
826419
Dead stop.  If it rebounds, it slows the following water, reducing spin.
792607
0 km/hr^2 .... because your speed doesn't change with time due to constant as you mentioned
185249
Don't listen to the other guy because it actually lowers!\nThe faster you go the pressure builds up and that airflow pushes down on the vehicle. MIKE
349995
It has 3 speeds, stop, go and crash.
822312
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