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Figuring Displacement from acceleration

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data

    A car starts from rest at a stop sign. It accelerates at 2.0 m/s^2 for 6.0 seconds, coasts for 2.0 s, and then slows down at a rate of 1.5 m/s^2 for the next stop sign. How far apart are the stop signs?

    2. Relevant equations

    not sure.

    3. The attempt at a solution

    I tried to split up this problem into 3 parts: from the first stop sign to after 6s when it begins to coast, when its coasting, and when it starts decelerating at the second stop sign. I think i figured out the first few parts, but how could i figure the displacement from the 3rd part? Thanks.
     
    Last edited: Sep 9, 2007
  2. jcsd
  3. Sep 9, 2007 #2

    mjsd

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    this is a constant acceleration problem

    relevant equations:

    [tex]s = ut +0.5 a t^2[/tex]
    [tex]s = 0.5(u+v)t[/tex]
     
  4. Sep 9, 2007 #3
    Can you show the work you've done so far?

    If you know the velocities at time t=6, t=8, t to get to stop sign. You can determine the distance. you may or may not need the final t.

    edit: ok you will need the 3 times, velocities at time t=6,8. You know starting/ending velocities.

    so you have 3 distances:
    accelating coasting decelarating
    x1 **** x2 *** x3

    to get the total distance just add those 3 up.

    When you show your work we can help out more.
     
    Last edited: Sep 9, 2007
  5. Sep 9, 2007 #4

    Kurdt

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    You will have to find the time it takes to come to a stop for the third part.
     
  6. Sep 9, 2007 #5
    thanks but what is "u" a variable for? plus i don't have the time it takes for it to decelerate. i only have the deceleration speed. so pretty much its just the initial velocity and acceleration that i have to work with. am i missing something?
     
  7. Sep 9, 2007 #6
    you have all you need. u stands for initial velocity in that equation.

    To start off with the first part. t=0 to t=6. acceleration = 2.0m/s^2

    so find the distance the car travels in that part. since the car starts from rest it's initial velocity =0.

    To get to the second part you need to know the initial velocity (final velocity for the first part). So use a kinematic equation to determine that velocity and you go from there.
     
  8. Sep 9, 2007 #7

    Kurdt

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    u is the standard (or most widely used) symbol for initial speed. v is commonly used for final speed.
     
  9. Sep 9, 2007 #8
    i love these types of problems :-]
     
  10. Sep 9, 2007 #9

    Kurdt

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    An alternative method is to draw a speed-time graph and find the area underneath the line.
     
  11. Sep 9, 2007 #10
    well for the first part of the problem, a=v/t, so i found the velocity by multiplying 2 m/s^2 by 6s and got 12 m/s. from this i got a displacement of 72m after 6 seconds.

    for the second part, i used the initial velocity of 12 m/s * 2seconds and got a displacement of 24m. This now gives me 96m of displacement for the first 8 seconds.

    Now, for the third part, i have acceleration as -1.5 m/s^2 and an initial velocity of 12 m/s. for the above equations, x=Vi+.5at^2, i have Vi and a, but i don't have t nor x. what am i missing. thanks again.
     
  12. Sep 9, 2007 #11

    Kurdt

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    You will have to use another of the kinematic equations to find the time. Specifically in case you don't know: [itex] v = u + at[/itex]
     
  13. Sep 9, 2007 #12
    you first part is wrong.

    12m/s is the final velocity at time 6s.

    in order to find the distance between t=0 to t=6 just do:
    x=x0+ut+1/2at^2

    and you know starting position/velocity.

    for the 3rd part you can find t if you use v=u+at. since you know the starting velocity, deceleration, and what will the final velocity when the car stops be?

    using that t you can solve for the 3rd distance and add them all up.
     
  14. Sep 9, 2007 #13
    ahh got it. thanks a lot guys!
     
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