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Linear Momentum - Help Please

  1. Mar 14, 2006 #1
    I'm stuck with this.

    The force from your car's motor is 6,000 N. The mass of the car is 900 kg. What will be you final speed after driving a distance of 25 m
    a. if you start from rest
    b. if you have an initial speed of 20 m/s.

    I can get W=fd but I cant get how to find the speed (velocity)


    W=6000*25 = 150,000J


    I am at a loss where to pick up speed. I can get accleration by doing



    But accleration isnt the same as speed.

    I know there must be a way of working back from the info I have but I really just cant see it.

    I'd really appreciate some help.
    Last edited: Mar 14, 2006
  2. jcsd
  3. Mar 15, 2006 #2
    Do you know the Kinematic equations of constant acceleration. Read FAQ. It is given there. For your information - Why do you think acceleration isn't constant here?
  4. Mar 15, 2006 #3

    I'll take a llook at the FAQ then, but I'm still not sure what iit is I'm looking for.

    Its not that I dont think accleration cant be constant but as far I know (am told by our teacher) that acceleration and speed are not the same thing. This is my first time in the Physics arena and I know this may all seem pretty simple but for me I cant make the obvious connections because I don't have prior knowledge.

    Thanks for your reply. I'll see if I can get anything from the FAQ.
  5. Mar 15, 2006 #4


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    You really do need kinematic equations to do this question.
  6. Mar 15, 2006 #5
    Ok I did a search for Kinematic equations of constant acceleration and nothing I can see that helps.

    Clearly I need an equation - a pointer would be nice.

    My situation is a good example of how students get turned of from the sciences, its not because they do not want to learn (I certainly do) but rather because when it comes to the simple stuff everyone expects you to say "oh yeah thats easy".

    I think real world physics are important if not amazing but it seems to be a trial of errors and frustration when you start out because there seems to be an expectancy that we are born knowing the forumulas or something.

    I dont mind doing the work but I need to know what the work is. Otherwise I could be wasting my time and setting myself up to fail.

    So if anyoen could actually put my wheels on the right tracks and give me some links or info to follow I'd appreciate it. Telling me I should have an equation or understanding physics concepts is just resating the obvious and further killing any joy I might actually get from such an amazing subject.

    My 2 cents
  7. Mar 15, 2006 #6


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    Okay, firstly, you will need to use Newton's law:
    [tex]F = ma[/tex]
    to obtain the acceleration of the vehicle. Then you will need to use the kinematic equation:
    [tex]v^2 = u^2 + 2as[/tex]
    to obtain the final velocity. In the above equations [itex]F[/itex] is force, [itex]m[/itex] is mass, [itex]a[/itex] is acceleration, [itex]v[/itex] is final velocity, [itex]u[/itex] is initial velocity and [itex]s[/itex] is displacement.

    Post your working if you want help or it checking :smile:

    I would just like to say that it is better to learn new principles from a textbook, I would therefore recommend that you get yourself a good elementry physics textbook
    Last edited: Mar 15, 2006
  8. Mar 15, 2006 #7
    Ok thanks, I had the first part down but couldn't see the second formula and how to set it up (Which was really the problem) doing the work is easier. I have a class textbook but (I'll check) I dont recall seeing anything like the formula you have here. I was about to go off and try working with:

    P=w/t =f*d/t then work P=f*d/t

    Thus p=f*v and solve for v

    where P=power w=work f=force d=displacement t=time. :confused:

    Didnt get round to trying it out though.

    Anyway what you have given here just from what little I know so I will try it out and see if the numbers make sense.

    Oh and thanks for the help...sometimes having the answer helps understand the problem..it did here.
  9. Mar 15, 2006 #8
    NO, you don't, just use conservation of energy and it will work out easily. Where does all that energy from work go? Into kinetic energy therefore just set work equal to kinetic energy.

    In other words

    W = Fs

    W = (Delta)KE

    So then

    Fs = (Delta)KE
  10. Mar 16, 2006 #9


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    You are quite right, what I meant was it would be easier to answer both questions using kinematic equations.
  11. Mar 16, 2006 #10
    Ok for the record:

    @Hootenay I used your formula and the question was considered answered correctly.

    @D_Leet Your suggestion also worked and the instructor told me that this is what he was really looking for. The method you used would be true on all instances the first methos would only be tru if you were looking at a constant speed and were not concerned about any speed changes in between start and finish.

    So I guess your both right and you can fight amongst yourselves over what the instructor has told me.

    Whatever I would once again say thank you both for responding...I'll probably be back in the not too far distant future.
  12. Mar 17, 2006 #11


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    This statement is incorrect, the kinematic equation is concerned with constant acceleration, which implies a change in velocity between the 'start' and 'finish'. Therefore, you can use my method for any instances where a constant acceleration is used.

    You got the right answer, but make sure you know how :smile:
  13. Mar 18, 2006 #12
    Absolutely - so what do you think the instructor was getting at?
  14. Mar 18, 2006 #13


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    I think you tutor meant/said Variable Acceleration, if this was the case the kinematic equation would be useless, you would have to use conservation of energy.
  15. Mar 19, 2006 #14
    OK well thanks for your input. I appreciate it.
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