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Acceleration and speed .

  1. Oct 20, 2009 #1
    Hi , I'm new to to the forum . I have a very hard time with physic or maybe is undestanding the professor . I currently trying to solve this acceleration problems but I end up not understanding that I need to do next or how to solve them. If any you can help me I will really apreaciate it.


    1)The tires of a car begin to lose their grip on the road at an acceleration of 5 m/s2. At this accelaration , how long does the car need to reach speed of 25m/s starting from10 m/s?


    A=5 m/s2
    V1= 10 m/s
    V2=25 m/s




    T= 10m/s
    ------ = 2s
    5m/s2



    2) A car start from rest and reaches a speed of 40m/s in 10s. If its aceleration remain the same how fast will it be moving 5s later?

    V1=0
    V2=40m/s
    T=10s

    A= 40m/s -0 40 ms
    ------------ =--------
    10s 10s
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 20, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi nina09! Welcome to PF! :smile:
    No.

    Acceleration is rate of change of velocity …

    so how much is the velocity changing? and how long does it take to change? :wink:
    Yes … keep going! :smile:
     
  4. Oct 20, 2009 #3

    Delphi51

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    Hi Nina and welcome to PF!
    It is very natural to have trouble with any new skill at first; you just have to work hard at it and it may turn out that you have a lot of talent.
    is a good start. Look in your list of formulas for one that relates the velocities, acceleration and time. The one I'm thinking of is
    a = Δv/Δt. The Δ on the t doesn't make any difference since we start at time zero, but the Δ on the v does because you start at v=10. So you get Δt = Δv/a = (25-10)/5

    In the second problem, it is difficult to follow your calc because this forum removes spaces. I think you wrote
    a = Δv/Δt = 40/10 = 4 m/s^2.
    This is correct and knowing the acceleration you should be able to use the same formula again over 15 seconds to find the velocity.
     
  5. Oct 20, 2009 #4
    I wrote each step under the othe so is easier to see the results

    This is my solution to the second one , Did I did it correctly?
    1) 40m/s -0
    -------------=
    15s

    2)40m/s
    -------- =
    15s

    3) 2.6m/s^2
    ------------
    15 s


    I'm still having dificulty's with the first one since I can't find the formula.
     
  6. Oct 20, 2009 #5

    Delphi51

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    For the second one, it should be
    a = Δv/Δt = 40/10 = 4 m/s^2
    The car goes from 0 to 40 m/s in 10 seconds, not 15.
    In a second step you use the "same acceleration" of 4 with a time of 15 to get the speed after 15 seconds.

    In the first one, the formula is a = Δv/Δt. Your only mistake was to say Δv = 10 when it was actually 25-10.
     
  7. Oct 20, 2009 #6
    This my second try at the first one, is it wrong?

    25m/s-10ms
    5m/s^2= -------------=
    T

    15m/s x 5m/s^2
    ------------ =
    T

    =0.6m/s^2
     
  8. Oct 20, 2009 #7

    Delphi51

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    First one should be Δt = Δv/a = (25-10)/5 = 3 m/s^2.
     
  9. Oct 20, 2009 #8
    This is what I understood from the explanation for the second one:

    40m/s- 0
    ---------=
    10s

    40m/s
    ------- = 4m/s^2
    10s

    2 step

    4m/s^2
    --------
    15s

    =1m/s^2
     
  10. Oct 20, 2009 #9
    First one

    25m/s-10m/s
    -------------=
    5m/s^2


    25m/s-10m/s
    -------------= 3m/s^2
    5m/s^2
     
  11. Oct 20, 2009 #10

    Delphi51

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    First one looks good!
    Second one, second step should be
    a = Δv/Δt
    a*Δt = Δv after multiplying both sides by Δt
    Δv = a*Δt = . . .
     
  12. Oct 20, 2009 #11
    Does this * means multiplication?
     
  13. Oct 20, 2009 #12

    Delphi51

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    Yes, * means multiply. If you don't use some *s, things can get unreadable.
     
  14. Oct 20, 2009 #13
    This the second one, I know I'm doing something wrong but I just can't pinpoint where exactly I'm making the mistake.


    40m/s- 0
    ---------=
    10s

    40m/s
    ------- = 4m/s^2
    10s


    Second step

    The acceleration is 4m/s^2 and the time is 15


    A*t
    -----=
    V


    4m/s^2 *15s = 60m/s
    ------------
    V

    60m/s
    ------- =4
    15s

    ???
     
  15. Oct 20, 2009 #14

    Delphi51

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    I said a = Δv/Δt
    Use a=4, Δt = 15 and this is
    4 = Δv/15
    Multiply both sides by 15 to get the Δv by itself:
    15*4 = Δv
    60 = Δv

    Will try writing this in your notation. Have to use dots to make it line up.
    ......Δv
    4 = ----
    ......15

    ............Δv * 15
    15*4 = ----
    ............15

    ............Δv
    15*4 = ----
    ............1
    60 = Δv
     
  16. Oct 20, 2009 #15
    I think I got it , thank you very much.


    40m/s- 0
    ---------=
    --10s

    40m/s
    ------- = 4m/s^2
    --10s
    ------V
    4 = ----
    ------15

    -------Vx15
    15*4 = ----
    -------15

    ----------V
    15*4 = ----
    --------- 1

    60 = V
     
  17. Oct 20, 2009 #16

    Delphi51

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    I love that trick with the white dashes! Thank you!
    Another good trick is, if you need a symbol like Δ or θ, you just go to
    https://www.physicsforums.com/blog.php?b=347 [Broken]
    and copy it.
     
    Last edited by a moderator: May 4, 2017
  18. Oct 21, 2009 #17
    Thank you .
     
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