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Motion problem

  1. Oct 23, 2007 #1
    A pedestrian is running at his maximum speed of 6.0 m/s to catch a bus stopped by a traffic light. When he is 25 meters from the bus, the light changes and the bus accelerates uniformly at 1.0 m/s2. Find either {A} how far he has to run to catch the bus or {B} his frustration distance (closest approach).

    I have used d=Vf^2-Vi^2/2a and gotten 18 meters but im not sure if im correct..
     
  2. jcsd
  3. Oct 23, 2007 #2
    I dont think this is correct. You can see that the pedestrian has to run more than 25 meters to catch up to the bus so 18 cannot be the answer.

    I suggest you seperately write down your data for the pedestrian and the bus.
     
  4. Oct 23, 2007 #3
    ok so
    BUS
    V=1.0m/s^2

    PED
    a=6.0m/s
    d=25m

    could i possibly use d=Vave(t) to find time and use 25 meters as distance or is that wrong?..
     
  5. Oct 23, 2007 #4
    Bus has an acceleration of 1m/s^2, not velocity. It also has an initial velocity of 0 since it was at rest.

    Pedestrian has a constant velocity of 6m/s not acceleration. The acceleration is 0 because V is constant. And if V is constant then Vo and Vf are the same.

    you do not know the displacement, you only know that when the bus begins to acceleration the pedestrian is 25 metres behind. So it is a catch up game.
     
  6. Oct 23, 2007 #5
    right right my bad on the A and V mess ups sorry.
    and then would i use the d=Vf^2-Vi^2/2a to find a of the pedestrian?
    putting 25m into d?
     
  7. Oct 23, 2007 #6
    Well you can see if you use that formula you will get d=0/2a. Not what you want. Secondly 'a' of the pedestrian is 0 because he is running at a constant velocity.

    you dont have displacement, so you cant use 25m.

    I'll give you a headstart.
    displacement of the pedestrian is
    d=d+25
    and displacement of the bus is d=d.
    This is because the pedestrian is 25 meters behind the bus. If the bus displaced 'd' metres then the pedestrian must displace 'd +25m' at their meeting point if there is one.

    Think about equating certain equations to find the time they will be at the SAME distance(d).
     
  8. Oct 23, 2007 #7
    o alright.
    thank you
    im sorry im just way lost and i guess not thinking straight
    the only equations i have are the kinematics equations and i cant think of which one to apply that uses what i have/dont have
     
  9. Oct 23, 2007 #8
    Analyze each equation, try to equate the d's. So effectively you will have to move the 25 over. Don't be afraid! I am here to help.
    PED
    d=d+25
    Vo=6
    Vf=6
    a=0
    t=?

    BUS
    d=d
    Vo=0
    Vf=?
    a=1
    t=?

    More than enough information here
     
  10. Oct 23, 2007 #9
    Thank you for bearing with me [=!!
    so just to get to the 25 meter mark from 0 the pedestrian would have to run 150 seconds....
    wait

    so i did d+25=Vf^2-Vi^2/2a?
    so d+25=18m
    d=-7 so he is 7m away from the bus? so 25 plus the extra 7 the bus travles so he is 32m away from the bus?
     
  11. Oct 23, 2007 #10
    The pedestrian is running at 6 metres/second. To get to 25 metres, it will take him about 4.something seconds right? But during those 4.something seconds the bus is travelling forwards.

    Try a different equation. You will need two equations for 'd'. One for the pedestrian and one for the bus.

    Since 'd' is the same, you can equation the two equations and solve for something else.

    example:
    Bus: d=vot+0.5at^2
    Ped: d+25=vot+0.5at^2

    We dont have d, but we know they have to be the same.
    vot+0.5at^2 = vot+0.5at^2 - 25

    effectively we have eliminated d, and can solve for t.
    Go back to the starting of 'example:' and see if you can simplify those equations with the data known.
     
  12. Oct 23, 2007 #11
    ooo rigght..geez
    hmm well i am very horrible at this
    i dont know how to get t out of the equation effectively to solve for it..
    would i just divide both sides by t^2 then...no see i am just digging a deeper hole with this problem
     
  13. Oct 23, 2007 #12
    look back at the known data from post 8.
     
  14. Oct 23, 2007 #13
    well
    0m/s(t)+0.5(1.0m/s^2)(t^2)=6m/s(t)+0.5(0m/s^2)(t^2)-25
    then by taking out the ones that would equal 0 out i reduced it to..
    .5m/s^2(t^2)=6m/s(t)-25
    hmm
     
  15. Oct 23, 2007 #14
    cool looks like you have a quadratic!

    What is t(the time the boy would catch up to the bus)?
     
  16. Oct 23, 2007 #15
    i got t=-.064 and t=.3054
    ?
     
  17. Oct 23, 2007 #16
    Equation is:

    0.5t^2-6t+25 = 0

    How did you get those values? Did you use the quadratic formula?
     
  18. Oct 23, 2007 #17
    wait i think i switched my values
    so i got
    3.27 and -15.27
     
  19. Oct 23, 2007 #18
    If I plug these values into the equation for t, I dont get zero.

    Use the quadratic formula.
     
  20. Oct 23, 2007 #19
    so
    a=.5
    b=-6
    c=25

    you cant have a - square root right
    wouldnt those values give you a negative under the root sign?
     
  21. Oct 23, 2007 #20
    Yes they would! So what happened here?
     
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