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Kinematics: Distance and time with limited acceleration and deceleration

  1. Jan 25, 2012 #1
    a motorcycle starts from rest point A and travels 300m along a straight horizontal track to point B where it comes to a stop. if the acceleration is limited to 0.7g and deceleration to 0.6g calculate the least possible time to cover the distance and maximum velocity reached.

    Can anyone point me in the right direction? i dont know where to start with this as the distance travelled at acceleration and deceleration is not defined and as they are not the same i assume you cant divide the distanc by 2

    Thanks
     
  2. jcsd
  3. Jan 25, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi MMCS! Welcome to PF! :smile:

    Call the two distances s1 and s2

    you know they have to add to 300 …

    then get two standard constant acceleration equations (one with a = 0.7 and one with a = -0.6), combine them with s1 and s2 = 300, and solve :wink:
     
  4. Jan 26, 2012 #3
    Thanks!

    Could you possible reccomend a formula to use, i cant find one that includes distance, time and acceleration without velocity and i cant use velocity because S1 has an U value of 0 but an unknown V and visa versa for S2, leaving 2 unknows in each equation t and u or v
     
    Last edited: Jan 26, 2012
  5. Jan 26, 2012 #4

    tiny-tim

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  6. Jan 26, 2012 #5
    Sorry im new to this:

    if i was to use this formula v2 = u2 + 2as for say S1 what whould i use as the distance as 300 is the distance for S1 and S2

    also i have tried to use this one s = ut + 1/2 at^2

    S1 = 0t + 1/2 x 6.9 x t^2

    S2 = ut + 1/2 x 5.9 x t^2

    S1 + S2 = 300

    3.45 x t^2 + -2.95 x t^2 + ut = 300

    0.5 x t^2 + ut = 300

    300/0.5 = t^2 + ut

    600 = t^2 + ut

    Because i dont know U for s2 i have 2 unknowns so i cant solve it
     
  7. Jan 26, 2012 #6

    tiny-tim

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    ah, but you know V for s2 instead (and a s and t) :wink:
     
  8. Jan 26, 2012 #7
    Where can i use V?
     
  9. Jan 26, 2012 #8

    tiny-tim

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    in s = ut + 1/2 at2

    you can either use negative t (so your "start time" is 0, and your "end time" is -t),

    or if you prefer you can memorise the alternative formula:

    s = vt - 1/2 at2 :smile:
     
  10. Jan 27, 2012 #9
    So we have

    S1 = 0t + 1/2 x 6.9 x t^2

    S2 = 0t - 1/2 x - 5.9 x t^2

    S1 + S2 = 300

    3.45 x t^2 + 2.95 x t^2 + ut = 300

    6.4 x t^2 = 300

    300/6.4 = t^2

    46 = t^2

    sqrt46 = t

    6.85 = t

    However i have 13.76s as the answer in my booklet? Have a gone wrong anywhere?

    Thanks for your help, hopefully iv nearly got it!
     
  11. Jan 27, 2012 #10

    gneill

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    One small problem -- you've assumed that the time intervals for acceleration and deceleration are equal. That is, you've used the same variable, t, for both the acceleration time and the deceleration time. This is unlikely to be true as the accelerations are not equal in magnitude.

    attachment.php?attachmentid=43182&stc=1&d=1327678980.gif
     

    Attached Files:

  12. Jan 27, 2012 #11
    So there would have to be seperate functions for t1 and t2? How would i get them?
     
  13. Jan 27, 2012 #12

    gneill

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    Study the diagram that I posted above. What are the slopes of the line segments making up the velocity vs time function? Can you determine a relationship between t1 and t2?
     
  14. Jan 27, 2012 #13

    tiny-tim

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    you have to minimise t1 + t2, given that they satisfy a equation of the form At12 - Bt22 = C

    there are various ways of doing this: one would be to write the thing you have to minimise (t1 + t2) as P (and maybe chuck in (t1 - t2) as Q) :wink:
     
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