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Can acceleration overcome a constant force

  1. Dec 18, 2009 #1
    caveat: im probably using wrong terms.
    help me understand this as ive been told im wrong.

    it makes perfect sense that once a force is removed from a vehicle that it wont be able to sustain the same rate of acceleration when the force was applied. my argument is that i believe there is a calculable rate in acceleration that once power is removed there will be enough inertia to over come the forces acting against the vehicle; allowing the vehicle to continue gaining velocity to a point. i will add that i think this "calculable rate of acceleration" i mention is tangible, as i believe my vehicle has the ability to gain velocity during some shifts.

    heres me trying to explain again what makes sense to me with my very limited physics understanding.

    two identical cars, in identical conditions and are both traveling at 60mph when power is removed. one of the cars (car a) was using cruise control to maintain 60mph while the other car (car b) was under 2g's of acceleration when it reached 60mph.

    both these cars would travel the same distance after power was removed?

    its very difficult for me to except the explanation ive already received and perhaps with the saturation of intellects that frequent this site you can bring closure to this conundrum of mine=/.
     
    Last edited: Dec 18, 2009
  2. jcsd
  3. Dec 18, 2009 #2
    you should think about how ths situation would play out. what causes the cars to slow down?

    if you are comfortable with the idea that velocity is relative, try to imagine how this looks from the non accelerating car, if they are side by side when the power is removed.

    if not, then think about it like this: what you are suggesting requires that - all other things being equal - two cars side by side travelling at the same speeds, yet one of them has more energy, "accelerative inertia" based on the fact that in the past it has been accellerating, causeing it to deccelerate less.

    but decceleration is just the same as acceleration in the opposite direction, so the vehichle which slowed down more (car a) would have more "deccelerative inerta" after the power was removed.

    meaning that when that car (car a) finished slowing down, (came to a stop) it would then release its "deccelerative inertia" and roll backwards.

    make sense?
     
  4. Dec 18, 2009 #3
    i value your response and i will sleep on the explanation you provided. ill attempt to discern your sentiments tomorrow.
     
  5. Dec 19, 2009 #4
    There are two requirements for power:

    1) Power to accelerate; P = F dx/dt = F·v = m·a·v
    where a = acceleration and v = speed.

    2) Power to maintain speed, i.e., power to compensate primarily for air drag (which increases approximately as v3) and tire rolling resistance (which increases approximately as v). See
    http://en.wikipedia.org/wiki/Drag_(physics [Broken])

    Bob S
     
    Last edited by a moderator: May 4, 2017
  6. Dec 19, 2009 #5
    i find it hard to understand how the car would roll backwards so perhaps your trying to teach me my flawed reasoning =/

    its just hard for to comprehend how something that is accelerating so fast under power will stop accelerating completely once power is removed.
    i made this diagram to offer another perspective to allow others a potential different approach to help me with this.
    confused.jpg
    lets assume that just prior to removing power car b was about 80degrees on that graph what i tried to illustrate is that after power was removed it was able to still accelerate albeit only at 40 degree angle for short time before ultimately the wind and drive train resistance were able to start slowing it just like what happening with car a as soon as power was removed.

    im gonna spend more time on that link you provided and will return with some thoughts.
     
  7. Dec 19, 2009 #6

    Pythagorean

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    What if a lot of the power is being used to oppose a force that remains after you take away the power?
     
  8. Dec 19, 2009 #7
    ^ i can understand how that relates to the car a in my example but what im trying to understand if there is a calculable exception; that may look like car b on my graph above.
     
  9. Dec 19, 2009 #8

    Pythagorean

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    Well, the car isn't moving backward on your graph above. It's velocity is just decreasing, so it's still moving forward, just slower. It only looks like he's going backward from the perspective of the guy who is still accelerating.

    F = ma. The force applied is proportionate to acceleration, so there is no more acceleration when there is no more force.
     
  10. Dec 19, 2009 #9
    and this is the answer i got before as well=/
    can you try and relate to why i might have such a difficult time excepting this. it seems to me that the car b would have extra momentum that would account for it traveling a further distance than car a after power is removed. it just seems trivial to look only at the point in which the cars lose power when the events leading up to that point have little in common.
    is there any scenario in which the cars in my example with same isolated variables will result in a difference between the distance they decelerate?
     
  11. Dec 19, 2009 #10

    Pythagorean

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    The graph you gave was a velocity graph. It says nothing about distance. The car will in deed have momentum when the power is removed. . It's velocity doesn't go to 0, it's acceleration does. If there's opposing forces (like air resistance and rolling resistance) the velocity will drop to 0 until the car comes to a stop, but all that time that the velocity is not 0, it is still moving forward, just slower and slower.
     
  12. Dec 19, 2009 #11
    sorry, yes i should have made that clear, it was your reasoning that led to my obviously wrong statement about rolling backwards.

    perhaps lets try to define what it is you're having a problem with. stop me when you don't agree...

    acceleration is defined as the increase (or decrease) in speed, over time. you take the overall change in speed, and devide it by the time it takes and you get a value for acceleration.

    this applies to all initial speeds, including zero. ie, start at rest, accelerate to 60mph in 12seconds (if you've got an old banger like me) and your acceleration is
    (60 - 0)/12 = 5mph increase in speed every second, or 5mphps (miles per hour per second)

    that (60 - 0) is the change in speed, (final speed minus start speed)

    like wise, you can accelerate negatively, if you're going 10 mph, and you slow down to a stop in 5 seconds, your acceleration is found using the same formula:

    final speed = stopped (0mph)
    start speed = 10mph
    time = 5s

    hence, (0 - 10)/5 = -2mphps. the negative sign implies that you're slowing down not speeding up, but its still accelleration. (sometimes given as a posative and called retardation or decceleration)

    ok so far? good. now you can also have a negative speed. this is just going backwards.
    (maybe i should be using the word velocity here, but for our purposes it doesn't matter)

    important distinction to make here:
    the sign (+ or -) of speed is determined based on position, while the sign of acceleration is not. speed is a change of position, acceleration is not.

    given all of the above, do you see how you could accelerate posatively to a stop?
    if your starting speed was negative (going backwards), and you slowed to a stop, the value for acceleration is posative.

    thats a fair whack to cover if your not comfortable with it, so i'll leave it there for now untill you comment.
     
  13. Feb 17, 2011 #12
    I think I'll take a whack at this one too.

    Car (a) on cruise control is experiencing no acceleration before the engine power is removed. The sum of all the forces opposing the movement of the car are equal and opposite to the forces pushing the car forward. F=ma in this application means the SUM of all the forces (net force) is equal to the mass times its accelration. The sum of all the forces is zero so the acceleration is zero. At the instant this car has the force of the engine removed, the forces acting in the opposite direction (mostly friction) will cause the car to have negative acceleration and slow down.

    I think it also depends on how you worded the question. If at the instant the power of the engine is removed, car B has no positive force acting on it, then both cars will have the same stopping distance. Momentum is defined as mass times VELOCITY, acceleration has no effect on it. However, if car B still has a positive force acting on it at the instant the engine power is removed, it will move faster than 60 MPH by a little bit, because it has a positive force acting on it while car A doesn't. What is important here is what the acceleration and velocity of each vehicle is at the instant engine power is removed (t=0).
     
    Last edited: Feb 17, 2011
  14. Feb 17, 2011 #13
    Yes...after the power is removed the two cars have exactly the same KE and negative acceleration acting on them.
    The cars will travel exactly the same distance and will stop after exactly the same time.
     
  15. Feb 17, 2011 #14
    "car (car b) was under 2g's of acceleration when it reached 60mph."

    This part is confusing because it makes it seem like it might still be accelerating at the instant car (b) reaches 60mph.

    If all we care about is stopping distance, all we need to know is the velocity and accleration at the instant the engine power is removed. Car (a) had to accelerate to 60mph at some point, even if we don't know when. How the cars got to 60mph is irrelevant.
     
  16. Feb 17, 2011 #15
    thank you for the bump; last week i was going to resurrect this thread but decided against it.
    i agree the organization and wording of my questions that i have used throughout this thread have added to the ambiguity. at the time i did not realize in physics that words have a very defined meaning.
    having said that, i still think some people may be overlooking a concept ill attempt to describe; again my fault =/

    i have seen a couple responses say approximately; power removed they will be the same. i looked to my posts for the confusion and believe i found it. a large flaw with my original post is when i said, "...once power is removed there will be enough inertia...". i should have mentioned that the clutch pedal was going to be used to start the coasting. id imagine some responses would change now.

    when the clutch pedal is pushed in, will it remove all forces in both "car a" and "car b"?

    i had someone that agreed the cars would not behave similarly offer an explanation that in the "car b" scenario the forces are not removed instantly; said something along the lines of, time to get electricity to and from light bulbs.
    i think the second part of hologramandy's response suggests this may be possible.

    i have some things to do but ill check back in and attempt to describe it more if i need to.
    thanks for your time.
     
    Last edited: Feb 17, 2011
  17. Feb 18, 2011 #16
    Yes the car will cease to accelerate at the instant that power is removed. You might have a hard time grasping this because you have a concept of "inertia", and you naturally think it applies it to acceleration as well as velocity.

    It doesn't. The car ceases to accelerate immediately. I believe the only way to truly grasp this is to drive a reasonably powerful car, accelerate as hard as you can in 1st gear, and then remove power by pressing the clutch. You will feel an instant lack of acceleration "in the seat of your pants." For confirmation, look at the speedometer. You will see that it never goes past where it was at the instant you pressed the clutch.

    Another less intuitive approach which relies on the same principle is get on an elevator and have someone cut the cable. You won't hover in air for a few seconds before the "accelerative inertia" is overcome. Instead you will instantly begin to fall and you will instantly feel weightless. I don't recommend doing this in an elevator, but there are amusement park rides that simulate this.

    Perhaps an "accelerative inertia" would be a nice thing to have. People wouldn't die as often in car crashes or falls off of tall buildings, as it's not the fall that kills you but the sudden acceleration at the bottom. However, I'm sure you can see the paradox of "accelerative inertia" if two cars hit each other head on travelling at high speed. If the acceleration took it's time to catch up to the huge forces involved, the cars would end up passing through each other before they stopped.
     
  18. Jul 14, 2011 #17
    ltw97m3

    You are probably accurate in that your car appears to be accelerating from the speedometer, however there is a slight delay in the real-time display of this device.

    Try the opposite to accelerating - but braking from 100km/hr down to 60km/hr. Coast when the speedometer reaches 60km/hr. You will probably notice the car infact appears to continue to slow down to 55km/hr.

    I have wondered the same in the past noticing that as i switch gears the car never seems to break acceleration (according the speedometer) - which did not make sense..

    Cheers

    Chris
     
  19. Jul 14, 2011 #18

    sophiecentaur

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    What's needed is a return to the Basic Meanings of the quantities that are being discussed here and a visit to Newton's Laws of motion. They describe and predict, perfectly what will happen and why.
    Look up those three laws and see how they fit the situation. Don't try to re-invent or re-describe this basic stuff. Wikipedia will do the job for you.
     
  20. Jul 14, 2011 #19
    Time 0
    (A) 60km/hr
    (B) 0 km/hr
    Time 1
    (A) 30km/hr
    (B)60km/hr
    Time 2 (all power removed)
    (A)60km/hr
    (B)60km/hr

    Does it make sense that car A would now pass B? This is how car A is going backwards relative to car B. But at the moment when power is removed - both cars will be traveling at the same speed - and will coast the same distance (assuming the friction is the same acting on both cars)
     
  21. Jul 14, 2011 #20
    Sorry the above thread was suppose to show spaces - allowing the visualisation of car A relative to car B
    Time 0
    ----------(A) 0km/hr
    (B) 60km/hr
    Time 1
    -------------------(A)30km/hr
    ------------(B) 60km/hr
    Time 2
    --------------------------(A)60km/hr
    --------------------------(B)60km/hr
     
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