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Kinematics Acceleration Equation help please

  1. Jun 30, 2012 #1
    A car accelerates at 3 meters squared from rest for 10s. How far does it travel?

    a = 3 meters squared
    v1 = 0?
    v2 = ?
    Δd = ?
    Δt = 10s

    Why is velocity 1, 0? Why is delta t 10 seconds?
     
  2. jcsd
  3. Jun 30, 2012 #2

    tiny-tim

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    welcome to pf!

    hi jbeannie05! welcome to pf! :wink:
    v1 is the initial velocity (the starting velocity), and since the question says it starts "from rest", that means v1 = 0

    ∆t is the time from start to finish: since it says it accelerates "for 10s", that means ∆t = 10

    now apply the usual constant acceleration equations …

    what do you get? :smile:

    (btw, it's not 3 metres squared, it 3 metres per second-squared, 3 ms-2)
     
  4. Jun 30, 2012 #3
    v1 is given to you as 0 because the car starts from rest
    Δt is also given to you: the car accelerates during 10s
    With this you must have seen the formulas to deduce v2 (the velocity after 10s) and the total distance travelled.
     
  5. Jun 30, 2012 #4
    Δd=v1Δt+aΔt^2

    This should answer your question.
     
  6. Jun 30, 2012 #5

    tiny-tim

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    erm … 5 out of 10 :redface:
     
  7. Jun 30, 2012 #6
    Whoops, what I meant was;

    d=v1t+.5at^2
     
  8. Jun 30, 2012 #7
    A car accelerates at 3 metres per second-squared from rest for 10s. How far does it travel?

    a = 3 metres per second-squared
    v1 = 0?
    v2 = ?
    Δd = ?
    Δt = 10s

    So, a is the acceleration, which is at 3 metres per second-squared, the initial velocity is 0, because the question says it starts "from rest", Δt, this will determine the interval of time used in determining the velocity, I think, not sure, and v2 and Δd is unknown. Am I right so far?
     
  9. Jun 30, 2012 #8
    Start with distance travel=average velocity × time
    Here in the problem you have to find average velocity.
    You are given value of acceleration which is assumed to be constant.
    From this value you can find average velocity.
     
  10. Jun 30, 2012 #9
    you are right on so far
     
  11. Jun 30, 2012 #10
    A car accelerates at 3 metres per second-squared from rest for 10s. How far does it travel?

    a = 3 metres per second-squared (acceleration)
    v1 = 0? (initial velocity)
    v2 = ? (final velocity)
    Δd = ? (displacement) how far does something travel?
    Δt = 10s (interval) time

    So, a is the acceleration, which is at 3 metres per second-squared, the initial velocity is 0, because the question says it starts "from rest", Δt, this will determine the interval of time used in determining the velocity. v2 and Δd is unknown.
    Am I right so far?
     
    Last edited: Jun 30, 2012
  12. Jun 30, 2012 #11
    I'm not sure what you know about Kinematics, for a start one dimension.
     
  13. Jun 30, 2012 #12
    one dimension?
     
  14. Jul 1, 2012 #13

    tiny-tim

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    hi jbeannie05! :smile:

    (just got up :zzz:)
    completely! :smile:

    now look up the standard constant acceleration equations, and apply one of them :wink:
     
  15. Jul 1, 2012 #14
    A car accelerates at 3 metres per second-squared from rest for 10s. How far does it travel?

    a = 3 metres per second-squared (acceleration)
    v1 = 0? (initial velocity)
    v2 = ? (final velocity)
    Δd = ? (displacement) how far does something travel?
    Δt = 10s (interval) time

    So, a is the acceleration, which is at 3 metres per second-squared, the initial velocity is 0, because the question says it starts "from rest", Δt, this will determine the interval of time used in determining the velocity. v2 and Δd is unknown.

    Δd = v1Δt + one half 2 squared
    Δd = 0Δt + one half 3 metres per second-squared 10s 2 squared

    Is that the right equation to use?
     
  16. Jul 1, 2012 #15
    Δd is how far(vector) it is from the origin.
     
  17. Jul 1, 2012 #16
    Umm..what does this mean? Did you follow the link posted by tiny-tim?
     
  18. Jul 1, 2012 #17
    A car accelerates at 3 metres per second-squared from rest for 10s. How far does it travel?

    a = 3 metres per second-squared (acceleration)
    v1 = 0? (initial velocity)
    v2 = ? (final velocity)
    Δd = ? is how far(vector) it is from the origin.
    Δt = 10s (interval) time

    So, a is the acceleration, which is at 3 metres per second-squared, the initial velocity is 0, because the question says it starts "from rest", Δt, this will determine the interval of time used in determining the velocity. v2 and Δd is unknown.

    Δd = v1Δt + ½at^​2
    Δd = (0)(Δt) + ½(3m/s^​2)(10s)^​2

    Is that the right equation to use?
     
  19. Jul 2, 2012 #18

    tiny-tim

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    hi jbeannie05! :smile:

    (try using the X2 and X2 buttons just above the Reply box :wink:)
    yup! :biggrin:
     
  20. Jul 16, 2012 #19
    Δd = (0)(Δt) + ½(3m/s2)(10s)2​

    How would I calculate this on a calculator?
     
  21. Jul 16, 2012 #20

    tiny-tim

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    wot? a half times 3 times 10-squared??

    i did that in my head, and got 149.9999 :wink:
     
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