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Acceleration and Algebra, 2 ppl going towards each other

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Sean and Steven are 1500 m apart from each other. Sean starts biking towards Steven accelerating from rest at a rate of 2 m/s2 until he reaches a cruising speed of 18 m/s. Steven begins rollerblading towards Sean, accelerating from rest at 1 m/s2 until he reaches a cruising speed of 12 m/s.
    When and where do they meet?


    2. Relevant equations

    Vf = Vi + at

    3. The attempt at a solution

    I did as my teacher said, used that to get the times at which they started to cruise at the constant velocity, then graphed it. And the area of the velocity graph = distance travelled, so I made them equal each other..

    Sorry if this isn't the right format or anything, I've just been working on this one question for like 2 hours and I just can't get it, and it's probably something so simple that I'm missing.
     
    Last edited: Feb 22, 2010
  2. jcsd
  3. Feb 21, 2010 #2

    rl.bhat

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    Homework Helper

    Hi AsianTheory, welcome to PF.
    Show your calculations. So that we can check where you have made mistakes.
     
  4. Feb 22, 2010 #3
    vf=vi+at
    18=0+2t
    9=t

    vf=vi+at
    12=0+1t
    12=t

    Okay, I can't really draw the graph, but here are my calculations.

    (18x9/2)+(18x-9) = (12x12/2)+(12x-12)
    6x = -12
    x = -2

    That already makes no sense, so I have no idea what to do.
     
    Last edited: Feb 22, 2010
  5. Feb 22, 2010 #4
    You did come quite far already.
    It would help if you explained a bit more what you are doing and why.

    (18x9/2) = 81m is the distance the cyclist travels when he's at cruising speed after 9 seconds.
    (12x12/2) = 72 m is the distance the rollerblader travels when he's at cruising speed after 12 seconds.

    81 + 18 (t-9) is the distance the cyclist is at time t (valid if t>=9)
    ........ is the distance the rollerblader is at time t (valid if t>=12)

    The only mistake you made was setting these equal to each other.
    Note that the distance of the rollerblader is measured as the distance from his starting point
    and in the opposite direction from the cyclist. He starts at 1500m and than goes
    towards the origin.
    The position of the rollerblader using the same coordinates as the cyclist is
    ............

    There's no way they they can meet before 12 seconds. (the cyclist travels less than 18*12 = 216m and the rollerblader less than 12*12 = 144m, so they can only meet after they
    stopped accelerating and the equations for their position that we just derived are valid.

    if you set the position of the cyclist equal to the position of the rollerblader, you get an
    equation for the time t, that they meet. The positon where they meet can then be found by
    substituting it in the equation for the position of the cyclist.
     
  6. Feb 22, 2010 #5
    Would I be able to use Vf2=(Vi2)+(1/2)(a)(t2)
     
  7. Feb 22, 2010 #6
    Nevermind, that doesn't work.. I have no idea what you said and I don't know what to do..
     
  8. Feb 22, 2010 #7
    I have a test on this tomorrow and it's pissing me off because it's making no sense..
     
  9. Feb 22, 2010 #8
    That's not useful. Sean & Stephen meet when they are done accelerating and move at constant speed.
     
  10. Feb 22, 2010 #9
    I would suggest using a system of equations
     
  11. Feb 22, 2010 #10
    I know, but I don't know what to do after that...
     
  12. Feb 22, 2010 #11
    What you had really seemed to contain most of the answer already.

    Sean the cyclist accelerated at 2 m/s^2 until he reached a speed of 18m/s.
    since [itex] v = v_0 + a t [/itex] and he starts at rest if follows that this happens when t = v_final/a = 9 s.
    The distance covered in 9 seconds is (1/2)at^2 = (1/2) * 2 * 81 = 81 m.
    After 9 seconds Sean moves with a constant speed of 18 m/s. The distance from his starting
    point at a time t is 81 + 18(t-9)

    Since you had (18x9/2)+(18x-9) I tought you had this already. (you should have used some parenthesis in 18x-9 and used t instead of x)

    You can compute the position at a time t for Stephen the rollerblader in exactly the same way,
    (you should do this yourself)
    but now this distance is measured backwards from a point 1500m up the road.

    If we want to compare the position of Stephen to that of Sean, we have to make sure
    they are both given in the same way, as a distance from the starting point of Sean the biker.
    If Stephen the rollerblader is a distance d from his starting point 1500 m away from the
    starting point of Sean, he is ............ m away from the starting point of Stephen

    Finally you can set these two distances from the starting point of Sean the biker equal to
    each other and then solve for t.
     
  13. Feb 22, 2010 #12
    Okay, I got that

    81 + 18(t-9) for Sean
    (12·12/2) + 12(t-12) for Steven which is also 72 + 12(t-12)

    This is where I am stuck, I have no idea what to do after this and I have a feeling I have to put 1500 somewhere..

    81 + 18(t-9) = D
    (12·12/2) + 12(t-12 = D

    Do I make these two equal to each other or not yet?
     
  14. Feb 22, 2010 #13
    Can anyone help me please? I have a test tomorrow on this...
     
  15. Feb 22, 2010 #14

    rl.bhat

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    Homework Helper

    During the accelerating period, both Stephen and Sean have traveled (81 + 72) m.
    Find the remaining distance between them when they stop accelerating. Let it be d.
    Now they are moving with uniform velocity. Start measuring the time from this point. When they meat they must have traveled same interval of time. Suppose they meat at a distance x from Sean. Then x/(velocity of Sean) = (d-x)/(velocity of Stephen).
    Substitute the values and find x. From that you can find t.
     
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