If acceleration is the change of speed then can we say that the speed of car is increasing instead............
828192
scientifically you can say it but with regard to saving energy, you can not:\nincrease speed ======= 2 words\naccelerate         ======= 1 word\n\nSave energy by using accelerate instead of increase speed
1174536
Maybe that means fast just got faster.
124905
There is no absolute reference frame, that is, there is no universal "stationary point" that we can use to say whether an object is moving.  Thus, every object is "at rest" in the frame of reference centered on itself.  In that frame of reference, an object's mass is always the same, its so-called rest mass.  That is the mass that you are likely thinking of when you say it doesn't increase.\n\nHowever, if the object is moving at a high speed relative to an observer, its mass is increased in the reference frame of that observer.  It will require more force to accelerate it a given amount, etc.
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! :)
180814
Good work, you answered your own question!\n\nWe can only feel changes in speed, or, perhaps more accurately, speed differentials.  \n\nWe don't feel like we're moving in a car because the AIR inside the car is moving right along with us.  If we were atop a train, for example, we would feel the movement of the air whizzing by our heads for sure!
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
777820
Well it is an increase of 33.25%, that's like comparing 40mph with 53mph in your car. BUT in more relevant terms, it depends on what you're doing with that, are you passing large quanities of data through it like video editing and such, or are you talking about mundane tasks such as word processing?  The difference is less noticeable if you're not working with large data streams.
534463
All kinds of things. All of a cars mechanisms are based on physical laws. I suppose this is for a school project so it might be enough if you talk about speed, acceleration etc...and maybe about the chemical energy that is being converted to heat energy which is being converted to kinetic energy (in school terms). When going through a curve you can talk about centripetal acceleration etc etc
51533
You are driving a car.  The car in front of you stops\nsuddenly.  You hit the brakes.  Tires squeal, but the car still moves. Inertia.
607432
While increasing flow of fuel u r increasing the speed of the driver unit (eg. increasing the speed of diesle engine which drives the alternator) bu thus more RPM and more power (however it will not increase more than the rating)
919955
Speed and Acceleration are related in that acceleration is the rate of change of your speed (should be velocity since you can take into account change in direction for accelerations).\n\nThe relationship can be mathematically expressed as:\ndv/dt = a.\n\nThat is, the change in velocity over the change in time.
298989
One must make a distinction between speed and velocity since they can have the same units, e.g., MPH for a car.  However, speed is a distance traveled  in any direction, even a curved path, during a unit of time while velocity is a vector meaning it is speed only in a given direction.  For a car on a curved road, the instantaneous velocity is identical to its speed.  If a car is traveling on a straight road exactly northeast, its velocity can be represented on paper by a vector arrow scaled to its speed pointing northeast which can be drawn as the diagonal within a box.  But the car is also manking progress north as well as east and smaller arrows drawn from the origin (original point) along the side and bottom of the box represent (to scale) the velocity north and east, respectively.\n\nAcceleration, which is an increase in velocity per unit time, is also a vector.  When a car is moving along a curved road, its speed may be held constant but its velocity and acceleration (or deceleration) will constantly change with respect to any particular compass direction.\n\nWhen a satellite spins around the earth say in a perfect circle, its speed may be constant but its velocity (speed in a given direction) is constantly changing, for example with respect to the moon.  Its instantaneous velocity always equals its speed (for example in a snap shot).  If the earth's gravity could be instantly eliminated (a mental exercise) the satellite would speed off into space in a straight line along the last tangent to its circular path.  In fact it is always falling due to earth's gravity just enough to compensate for its desire to fly off into space.  The sattelite is constantly accelerating towards earth and its velocity is constantly changing (due to the constant change of direction on a curved path) even though it travels at constant speed.\n\nTherefore, an object can not accelerate while it has a constant velocity but it can do so while it has a constant speed.  Hope that helps a little.
109202
Fnet= MA so you acceleration is 36km/h, so convert that to m/s, which is 10 m/s. To get acceleration, we know that the car started out at 18km/s, which is 5m/s. So the acceleartion = Change of velocity/Change of time. So that comes out 1 m/s/s. So there is 1000 Newtons on the car. The other guy's answer is wrong. I'm an engineering major. The standard Units in science is MKS, meters kilograms and seconds
212447
it's like the speedometer of a vehicle. the higher the reading the faster it is...
413558
Yes,but it's providing downforce to stick car to track better.The car is going to cut the air,and you sure don't want it to go under,then car would lift up and get airborn.
472797
The only time an object's speed will not be increasing or decreasing is when the acceleration equals zero.  When this is not true, the object is increasing or decreasing in speed.  As acceleration goes from positive to zero, the speed will begin to increase more slowly over time.\n\nAn example of this would be the speed of a car when approaching its maximum velocity.  The car will readily increase speed from 0 to 60, in say 8 seconds. As it approaches its top speed, say 115 MPH, it will do so more slowly as you get past 70 mph.  When the horsepower equals the power caused by wind drag, the acceleration will go to zero, the velocity will stop increasing, and the speed will be constant.
1160746
I am posting a three page answer in order for the response to make sense to you.  If you have a real interest in the question, it will give you new thoughts to consider.\n\n                        THE PROBLEM AND REPAIR OF RELATIVITY (C) 2004, Duane Ertle \n\n\nIn order to introduce the problem, I am going to quote a few sentences from a book "The Universe and Dr. Einstein," by Lincoln Barnett, published in 1960 by Mentor Books. The quotes are from pages 61 and 63 respectively. \n\n\nThe first concept involves that of mass.\n \n"In its popular sense, mass is just another word for weight. But used by the physicist, it denotes a rather different and more fundamental property of matter: namely, resistance to a change of motion. A greater force is necessary to move or stop a freight car than a velocipede; the freight car resists a change in its motion more stubbornly than the velocipede because it has greater mass. ... But Relativity asserts that the mass of a moving body is by no means constant, but increases with its velocity relative to an observer. In short, energy has mass!" \n\n\nNext, we need to look at the reason for this belief.\n \n"By further deduction from his principle of Relativity of mass, Einstein arrived at a conclusion of incalculable importance to the world. His train of reasoning ran somewhat as follows: since the mass of a moving body increases as its motion increases, and since motion is a form of energy (kinetic energy), then the increased mass of a moving body comes from its increased energy. ..." \n\n\nNow I am going to quote a paragraph from "The Bones of Time," copyrighted in 1978 by Duane Ertle, it is from the chapter "Flexible Time."\n \n"The reason a fast moving mass is so difficult to stop, or have a direction change in, is due to its becoming frozen in a linear manner so energy cannot pass through its total volume as happens when it is at rest. When a mass moves at one-hundred thousand miles a second, the energy existent and having potential movement at right angles to its direction of travel could be at most a distance of eighty-six thousand miles a second. When moving at the speed of light minus two feet per second, the potential energy existent at right angles to its forward motion would only be two feet in one second. At the speed of light the mass would convert from being a three-dimensional object and would become a single beam of high energy photons streaking through space, all existing in only one dimension. Obviously there would be zero energy potential at right angles to the direction of travel." \n\nMoving mass does not acquire additional mass as it moves. Atomic matter is composed of waveform. As electromagnetic energy (a wave) has differing energy values so, also, does mass at different speeds have an overall differing waveform. The waveform is derived from the moving mass itself, as energy at right angles to direction of travel decreases in line density with movement, and transfers a proportional amount of that frequency toward the direction of travel. In this sense greater mass is being added to the forward motion of the mass at the expense of the energy/mass at right angles to it. \n\nThere is an expression that deals with frequency density and an increase of energy as the lines of frequency increase. The expression in question says that E=hf. The energy of an electromagnetic wave is equal to that of a very small numerical constant times the frequency of the wave. The greater the frequency of the wave, the more energy the electromagnetic wave would have. Radio waves may have very long peak-to-peak distances of thousands of meters, while cosmic radiation is able to have its frequency so compressed that it acts as dimensional mass, and there may be trillions of frequencies, or waves within a single meter. Now consider how the frequency at right angles to direction of movement changes from having great potential in those direct