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Not sure which formula to use for this problem!

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A women is traveling at 50km/h, she sees a stop sign and takes 1.4 s to react, she starts decelrating at 2.3m/s and there is a broken bridge 150 m ahead of the sign.
    Is she going to stop in time?

    2. Relevant equations
    I am not sure but i think one of these equations can be used:

    DeltaD = v1 * delta T + 1/2 + a (delta t)^2

    V2 = v1^2 +2 * a * delta d

    delta D = (v1+v2) / 2 * Delta T


    Delta D = V2 * Delta T - 1/2 * a (delta T)^2

    3. The attempt at a solution
     
  2. jcsd
  3. Oct 8, 2009 #2
    the reason they tell u that theres a broken bridge 150m ahead, is that they want u to use ure equations of kinematics to work out what her displacement wil be if she decellerates from 50 km/h to 0 , at 2,3m/s. remember to change it into m/s before u start.
     
  4. Oct 8, 2009 #3
    also what do u think is the importance of the 1,4s reaction time?
     
  5. Oct 8, 2009 #4
    I remember changing it to m/s and i got 13.8 m/s converting it from 50 km/h. I think i should use the Delta D = (v1-v2)/2 * Delta T. And i dont no the improtance of the 1.4 s reaction time
     
  6. Oct 8, 2009 #5
    they tell you that she has 150m to stop in. But she took 1,4secs to react. in other words she still drove at the initial speed for 1,4 secs before she startd to decelerate. Work out how far she drove in that time and deduct that from the 150m to know what distance she has to stop.
     
  7. Oct 8, 2009 #6
    Ok, so she travelled 31.74 m after deducting it from 150 i got 118.26. So she has to stop within 118.26 m. Now what formula do i use to figure out if she stoped in time?
     
  8. Oct 8, 2009 #7
    how did u get 31,74?

    D = V.t
     
  9. Oct 8, 2009 #8
    OH right right!!! i accidently times it by 2.3!! ok so its 19.32 m then she started to decelerate, then after subtracting it by 150 it is 130.68. So she has 130.68 m to stop
     
  10. Oct 8, 2009 #9
    use equation: V2 = V02 + 2aD

    with D being displacement
    remember the value of a is negative during deceleration


    BTW this is during 2nd part,....when she starts decelerating.
     
  11. Oct 8, 2009 #10
    OK i did it but i dont think its right:

    = 13.8^2 + 2 (-2.3) (130.68)
    = 190.44 + (-0.3) (130.68)
    = 190.44 + (-39.204)
    = 151.236

    Do i square root the answer? if this is right...
     
  12. Oct 8, 2009 #11
    D in the equation is the distance it takes her to go from 13,8 m/s to 0 m/s. in other words to stop. which is the answer u need.
    Do u understand why u can not substitute 130,68 into it?
     
  13. Oct 8, 2009 #12
    ok.. i understand, 130.68 can go into it because that is the distance she has to stop not the distance it takes her to stop.. so what do i put in instead of 130.68? 19.32?
     
  14. Oct 8, 2009 #13
    no u dont put anything into it. U want to find the value of D. U cant find the value of a variable if u substitute another value into it.

    put all values into equation, leaving D as it is. Then calculate the value of D
     
  15. Oct 8, 2009 #14
    ok, so the value of d is 190.14?
     
  16. Oct 8, 2009 #15
    nope


    V2 = V02 + 2aD


    What are the values of each variable? and wich one do u want to calculate?
     
  17. Oct 8, 2009 #16
    V^2 = 190.44 + 2 * (-2.3) D

    i need to calculate D
     
  18. Oct 8, 2009 #17
    good . whats the value of V^2 ?
     
  19. Oct 8, 2009 #18
    185.84? sorry it took so long... i didnt no the reply was in the second page.. i was waiting for it in the first page! :=)
     
  20. Oct 8, 2009 #19
    no worries

    V2 is the final speed, right? so what should the car's final speed be if he stopped? surely not 185
     
  21. Oct 8, 2009 #20
    If she stopped then it would be 0.
     
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