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Q in instantanous acceleration I want check my answer .

  1. May 29, 2010 #1
    Hi all

    1.JPG

    Q in instantanous acceleration ?



    View attachment 17036



    my answer is :



    a ) As we know Instanteouse acceleration :

    It is the limit of the average acceleration when the time interval to zero



    a = changing in v / changing t
    3.0-1.5 / 0.8 - 0.4 = 35225 m/s



    b )

    instantanous acceleration = Zero because it's straight line



    c ) here the gragh go down then the ( a ) will by ( - )



    a = changing in v / changing t



    3.0 - 1.5 / 1.6 - 1.4 = -0.4625



    this is my answer I want check
     
  2. jcsd
  3. May 29, 2010 #2
    Using your method part a) comes to 1.5/0.4=3.75m/s^2.Your method is right but something went wrong with your maths(used calculator perhaps and pressed wrong buttons?).Please note that you could have used the whole triangle and written a=3/0.8.

    You made a similar mistake in part c)

    You need to improve on your reasoning for part b).The acceleration is zero because it is a straight line parallel to the time axis,in other words there is no change of velocity.
     
  4. May 29, 2010 #3
    hi thanks

    but what are the mistake in part c ?
     
  5. May 29, 2010 #4
    You wrote a=3-1.5/1.6-1.4.It would be clearer if you wrote a=-(3-1.5)/(1.6-1.4)=1.5/0.2=-7.5m/s^2
     
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