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Acceleration In One Dimension

  1. Sep 25, 2004 #1
    ****acceleration In One Dimension***

    Hi everyone, I need some major help right now, I dun understand some concepts, im a first time physics student. Can an object have a negative acceleration and be speeding up?
     
  2. jcsd
  3. Sep 26, 2004 #2
    Yeah, if it is going in the negative direction. E.g backwards
     
  4. Sep 26, 2004 #3
    it is all based on your frame of reference. that means that if you call the direction that the object is accelerating in the negative direction, then it will have a negative acceleration and be increasing its speed.

    frames of reference are important parts of setting up Physics problems. it is best to think about the problem before hand and set it up so that its final destination or the direction is is mostly moving in (the ground, point b, etc) is in a positive direction. it makes the problems much easier to do.
     
  5. Sep 26, 2004 #4
    I don't mean to hijack the thread, but I also have a similar question to ask: What is the difference between deceleration and negative acceleration? Are they the same thing?
     
  6. Sep 26, 2004 #5
    Deceleration means slowing down
    Acceleration means speeding up
    I think lol im a first time physics student i dun know :(
     
  7. Sep 26, 2004 #6
    I have never seen deceleration in my physics book. Always acceleration whether it is positive or negative.
     
  8. Sep 26, 2004 #7

    arildno

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    The pair acceleration/deceleration are "layman definitions" in that the pair is not used as a technical distinction, but as a fancy way of saying speeding up/slowing down
    (a pair I prefer to ac/dec).
     
  9. Sep 26, 2004 #8
    deceleration means you are in fact slowing down it is more of a broad term
    negitive acceleration just means you are speeding up in the direction opposite to the one defined as positive. But you could I guess say that you were decelerating from -25 m/s^2 to -20 m/s^2

    deceleration really can mean different things based upon how you use them. I generally think of it as something that means your speed is slowing down
    aka closer to 0 m/s
     
  10. Sep 26, 2004 #9
    Q40. A boy stands at the edge of a bridge 20.0 m above a river and throws a stone
    straight down with a speed of 12.0 m/s. He throws another pebble straight
    upward with the same speed so that it misses the edge of the bridge on the way
    back down and falls into the river. For each stone find (a) the velocity as it
    reaches the water and (b) the average velocity while it is in flight.
    Note: Ignore the affects of air resistance.

    (a) X= 20 m a=9.80m/s2 Vi=12.0 m/s Vf= ?????
    Vf2 = Vi2 + 2a (ΔX)
    Vf2 = (12.0 m/s)2 + 2(9.80 m/s2) (20 m)
    Vf2 = (144 m2/s2) + (19.6 m/s2) (20 m)
    Vf2 = (144 m2/s2) + (392 m2/s2)
    Vf2 = (536 m2/s2)
    Vf = 23.15167381 m/s
    Vf = 23 m/s

    Therefore the velocity of the stone as it reaches the water is approximately 23 m/s. Since the pebble was thrown straight upward with the same speed so that it misses the edge of the bridge on the way back down and falls into the river, I know that if the pebble is projected upwards in a vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity it has when it returns to the same height. That is, a pebble projected vertically with an upward velocity of +12 m/s will have a downward velocity of –12 m/s when it returns to that same height.

    (b) *AVERAGE VELOCITY*= ΔX/Δt

    STONE
    Xf = 20 m Vxf = 23 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????

    Vxf = Vxi + ax t Vavg= ΔX/Δt
    23 m/s = 12.0 m/s + (9.80 m/s2) t Vavg= 20m/1.12244898 s
    23 m/s – 12.o m/s = (9.80 m/s2) t Vavg= 17.81818181 m/s
    11 m/s = (9.80 m/s2) t Vavg= 17.81 m/s
    t = 11 m/s .
    9.80 m/s2

    t = 1.12244898 s
    t = 1.12 s



    (b) *AVERAGE VELOCITY*= ΔX/Δt

    PEBBLE
    Xf = 20 m Vxf = 12.0 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????
    Vxf = Vxi + ax t
    12.0 m/s = 12.0 m/s + (9.80 m/s2) t
    12.0 m/s – 12.0 m/s= (9.80 m/s2) t
    0 m/s = (9.80 m/s2) t
    0 m/s = t
    9.80 m/s2
    t = 0 m/s

    CHECK MY ASNWER???
     
  11. Sep 26, 2004 #10
    Acceleration vs. Deceleration

    How can you tell if the object is speeding up (acceleration) or slowing down (deceleration)? Speeding up means that the magnitude (the value) of the velocity is increasing. For instance, an object with a velocity changing from +3 m/s to + 9 m/s is speeding up. Similarly, an object with a velocity changing from -3 m/s to -9 m/s is also speeding up. In each case, the magnitude of the velocity (the number itself, not the sign or direction) is increasing; the speed is getting larger.

    Given this fact, an object is speeding up if the line on a velocity-time graph is changing from a location near the 0-velocity point to a location further away from the 0-velocity point. That is, if the line is moving away from the x-axis (the 0-velocity point), then the object is speeding up. Conversely, if the line is moving towards the x-axis, the object is slowing down.
     
  12. Sep 26, 2004 #11

    arildno

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    If
    [tex]\vec{v}\cdot\vec{a}\geq{0}[/tex]
    then the object "accelerates", otherwise it "decelerates".
    ([tex]\vec{v}[/tex] velocity, [tex]\vec{a}[/tex] rate of change of velocity (the proper meaning of acceleration)
     
  13. Sep 26, 2004 #12

    robphy

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    Some synonyms:
    deceleration, "slows you down" , "reduces your speed"
    symbolically (and without reference to coordinates): [tex]\vec a\cdot \vec v < 0[/tex].

    Proof:
    [tex]0>\vec a\cdot \vec v =\frac{d \vec v}{dt}\cdot \vec v=\frac{1}{2} \frac{d (\vec v \cdot \vec v)}{dt}=\frac{1}{2}\frac{d (v^2)}{dt}=
    \frac{1}{2}\frac{d}{dt}((\text{speed})^2)[/tex]
    So,
    [tex]0>\frac{d}{dt}(\text{speed})[/tex]
     
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