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Answear to magnitude of an object slowing to a halt? negative or positive?

  1. Nov 10, 2011 #1
    I have calculated magnitude of acceleration of an object slowing to a stop to be:

    -5.1m s^-1.

    I then had to find the magnitude of the unbalnced force applied to the object for it to stop which i found to be:

    So F= 1500 kg * 5.1 m s^-2

    Which equals 7650kg m s^-2

    Which equals 7650 N

    I have assumed this answear would have to be positive. so would this mean my answear in the first part should be positive and described as decelerating ?

    Any help would be most welcome!
    Last edited: Nov 10, 2011
  2. jcsd
  3. Nov 10, 2011 #2


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    The direction of the acceleration is the same as the direction of the force (that doesn't seem to be the case in your calculation). How is there any confusion if you get your arrows right in the first place and stick with any signs that come out of the calculation?

    It may help if you avoid using the word "deceleration". It really doesn't help except, perhaps, when you're driving. Just use negative acceleration when you are subtracting velocity (per second). That will end up with the velocity becoming negative (backwards) eventually.
  4. Nov 10, 2011 #3
    So as the magnitude of the aceleration is - 5.1m s^-1

    the force applied to the object woulb be -7650 N

    or am I getting mixed up still?
  5. Nov 10, 2011 #4
    The magnitudes of acceleration and anything else are always positive. The actual acceleration (and force) in this case is negative as your positive direction is the direction in which the mass is initially moving.
  6. Nov 10, 2011 #5
    I am still unsure if the answears should be positive or negative if someone could explain to me each case it would be a great help.
  7. Nov 10, 2011 #6
    Velocity, acceleration, force and others are vectors. Vectors have a magnitude (their numerical size) and direction. In your problem, the opposite directions are represented by positive or negative signs attached to the magnitude.

    If you say a force is 27 newtons, its direction is the same as if it were expressed as +27 newtons.
  8. Nov 11, 2011 #7


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    Correct. Mr Newton says this, too. The acceleration is in the same direction of the force so, whatever axes you choose, they will have the same sign.
  9. Nov 11, 2011 #8
    Sophiecentaur, you are correct that both the accleration and net force arenegative in this problem.
  10. Nov 11, 2011 #9


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    The essential thing with these problems is just to choose a direction as 'positive' and stick to the convention. I frequently have students who worry about whether to use UP or DOWN as positive for falling off cliffs and shooting rockets upwards etc. problems. They are very dubious when I tell them that it doesn't matter as long as they stick to their choice throughout. "Yebbut . . .?" I guess it's a confindence thing.
  11. Nov 11, 2011 #10
    thank you that has helped clear it up.
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