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Acceleration question HELP?

  1. Oct 13, 2009 #1
    A car is travelling at 19.0 m/s [W] when it enters a curved portion of the track and experiences an average acceleration of 0.270m/s^2 [N] for 62.0 seconds. Determine the velocity of the car after this acceleration. Include a vector diagram.


    -v1= 19.0m/s [E]
    a= 0.270 m/s^2 [N]
    delta t= 62.0 s
    v2= ?

    so i thought that u can solve this by just plugging into the formula

    a= v2-(-v1)/ delta T
    which gives me 36 m/s as v2, but it also says to use a vector diagram, which i get another answer of 25.3 m/s? HELP?
     
  2. jcsd
  3. Oct 13, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi bobsagget ! Welcome to PF! :smile:

    (try using the X2 and X2 tags just above the Reply box :wink:)
    Yes, you can, but remember velocities are vectors, and so obey the law of vector addition (and subtraction) …

    you can't just add the magnitudes. :wink:
     
  4. Oct 13, 2009 #3
    hey thanks for the info

    so do you have to swap the v1 to a -v1 so u can add the v2 and v1 together? like switch the the heading it has to its opposite
     
  5. Oct 14, 2009 #4

    tiny-tim

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    No, v1 is west and (v2 - v1) is north,

    so to do it the vector way, either add components, or use a vector triangle :smile:
     
  6. Oct 14, 2009 #5
    so if i were to put the v1 and deltav(v2-v1) into a vector triangle it would be like

    HYP= v2 lDelta v
    l
    ________ l
    v1

    then to solve its v2= delta v^2-v1^2
    v2=(16.7)^2 -(19)^2
    which would give me 25.3? is that correct?
     
  7. Oct 14, 2009 #6
    tried to make a traingle, but didnt work, but delta v is noorth and your v1 is west, and u have to solve the hypotenuse which is v2?
     
  8. Oct 14, 2009 #7

    tiny-tim

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    Yes, 25.3 is correct (and you meant +, not - :wink:)

    How did your triangle not work? :confused:

    the original velocity, 19, is west. The "added" velocity, 16.7, is north, and so the final velocity is the hypotenuse.

    Why is that worrying you? :smile:
     
  9. Oct 14, 2009 #8
    yes it was a + your right, I just made a mistake with the whole diagram thing but I got it :smile: thanks for your help, and then the heading of this is NW
     
  10. Oct 14, 2009 #9

    tiny-tim

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    You need to be more accurate about the direction …

    it's not exactly NW …

    what angle west of north is it? :smile:
     
  11. Oct 14, 2009 #10
    off the north line I got the angle to be 49 degrees, or w41n
     
  12. Oct 15, 2009 #11

    tiny-tim

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    (i've never seen "w41n", though i have seen "n49w", but if your professor says it's ok then of course it's ok)

    Yup! :biggrin:

    So the full answer is 25.3 m/s 49º West of North. :smile:
     
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