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Finding acceleration

  1. Jan 23, 2015 #1
    1. Not sure if it's the right forum but I'll try. The velocity of an particle varies quadratic with a formula based on time

    a) Find the equation for acceleration and length as a function of time.

    b) a = 3 m/s^3 (I think it's alpha)
    Find the acceleration when t1=2s t2=5s and how big is the mean acceleration in the time intervall t2-t1.


    2. See picture


    3. a) I think you can use the formula aa18f61b4e7c9b1afebb4a98a91f0456.png saying u=0 (start velocity) and say a=(v-u)/t
    And do the same for cae80dc5147530bb7bddfa350e2e7b53.png where s= length. But is it that easy or are they looking for a different solution, as this is something I learned few years ago. Or do I have to antiderivate it, which would make it complicated because there is no known fact.


    b) If a) is right could I then say solve v(t) which gives me t=3 alpha= 3 --> v(t)=27 m/s, and then get acceleration by the 2nd formula aa18f61b4e7c9b1afebb4a98a91f0456.png v= 27 u=0 t=3, so acceleration is 9? Or am I on the wrong way?
     

    Attached Files:

  2. jcsd
  3. Jan 23, 2015 #2

    Doc Al

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    That's a bit vague. Is any additional information given? Such as velocity at time 0.

    Careful! Those formulas are for constant acceleration, which is not what you have here. (Note how the velocity is linear with time, not quadratic, in the formula.)

    Hint: Calculus made be involved.
     
  4. Jan 23, 2015 #3
    There is no additional information, but the formula (picture).

    Do you have any hint to how I get started on solving b)?
     
  5. Jan 23, 2015 #4

    Doc Al

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    Ah, I see it now. ##v(t) = 1/2 \alpha t^2##.

    First solve part a).
     
  6. Jan 23, 2015 #5
    Do I have to solve it as differintial equation? As I'm not allowed to use the formula for constant accerlation?

    I know I can derivate v'(t) to get a(t)= at where a= alpha, but how do get to the length formula?
     
    Last edited: Jan 23, 2015
  7. Jan 23, 2015 #6

    Doc Al

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    You'll have to use a bit of calculus. (How do distance, velocity, and acceleration relate to each other?)

    That's correct. You cannot use the standard constant acceleration formulas here.
     
  8. Jan 23, 2015 #7
    I know that the distance particle traveled is s(t)= antiderivative of v(t), but wouldn't I need additional information to find the constant?

    Edit: Already found the formula for acceleration a(t)= qt where q is a constant
     
  9. Jan 23, 2015 #8

    HallsofIvy

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    Not to find the distance travelled over a given time interval. That is the difference between two lengths

     
  10. Jan 23, 2015 #9

    Doc Al

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    To express the position, you'd need additional info. But for the distance traveled, you can just set s(t=0) = 0.

    Good. (That constant is given as ##\alpha##.)
     
  11. Jan 24, 2015 #10
    (I didn't know how to write alpha so I just called it q)
    So the formula for poisition as a function of time is S(t)=0 when t=0?

    and for b) since a(t)=qt where q=3 m/s^3 for t1= 2s and t2=5s
    a(2)=6 m/s^2 and a(5)=15 m/s^2
    mean accleration= 11/3 m/s^2
     
  12. Jan 24, 2015 #11

    Doc Al

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    Yes, that will give you the position from the starting point (where t = 0, presumably).

    Good.

    How did you calculate that?
     
  13. Jan 24, 2015 #12
    Sorry I meant 21/3 m/s^2 which is 7m/s^2
    a(2)+a(5) =6+15=21
    And t2-t1=5-2=3
     
  14. Jan 24, 2015 #13

    Doc Al

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    Careful! What's the definition of average acceleration?
     
  15. Jan 24, 2015 #14
    Since I already know what alpha is and time is I might be able to use the general formula for acceleration knowing the velocity at beginning and end.
    This gives me
    vo=3*(2)^2=12
    v1=3*(5)^2=75

    acceleration= (75-12)/(5-2)=21
     
  16. Jan 24, 2015 #15

    Doc Al

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    Good.

    You left out the factor of 1/2.
     
  17. Jan 24, 2015 #16
    Sorry this is stupid of me,
    (37,5-6)/3 =10,5
    v1-v0/delta(t)

    I have another question not sure if anyone is able to answer:

    Car with velocity 36 km/h colliding and stoping over a length of 1 meter (end velocity = 0). Find acceleration.
    I have solved it the simple way using v^2 = u^2 + 2as and got the right answer.
    Now my question is how do I set it up as an differential equation.

    My attempt:
    Since I know S(0)= 1 (saying start position is 1 and end is 0)
    velocity start = 36km/h and end is 0
    I also know that position is antiderivative of velocity
    S(t)=∫36
    S(t)= 36t+C

    Attempt on finding constant c.
    S(0)=36*0+C=1
    C= 1
    S(t)= 36t+1
     
  18. Jan 24, 2015 #17

    Doc Al

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    I would say that the start position is 0 and the end position is 1.

    You cannot assume that the velocity is constant. (After all, you are told that it is stopping.)

    All you'll do, by setting up a differential equation, is derive those kinematic formulas. Nothing wrong with that, as long as you start with the correct assumptions. Assume that the acceleration is constant, then you can integrate to get the position as a function of time.
     
  19. Jan 25, 2015 #18
    Do you mean derivating for example the formula
    v=v0+at?
     
  20. Jan 25, 2015 #19

    Doc Al

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    Yes. And, using a few tricks, all the others.
     
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