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Calculate displacement based on velocity and acceleration

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data

    Basically, a car accelerates at -5.2 m/s² and is traveling at 45 km/h.

    2. Relevant equations

    How many meters will it take for the car to stop.

    3. The attempt at a solution

    I was using some online material and found a few equations. I found the time it takes to for the car to stop, (8.6538 seconds) but I'm unsure how to find the displacement from this. I was thinking taking the 45 km/h and converting to m/s (12.5 m/s) and multiplying by time. This should provide the distance or so I thought but it comes back wrong. (I got 108.173 meters as the distance)

    I also tried putting my numbers into this equation Vf²=Vi²+2a(delta)X but I got 194.127 and that too came back as incorrect.

    Can someone please explain what is wrong with my method and help me figure it out?
     
  2. jcsd
  3. Oct 2, 2008 #2

    alphysicist

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    Homework Helper

    Hi Planefreak,

    You have to convert the 45km/h to m/s for all of these calculations. First find the initial speed in m/s, and then use the equations.
     
  4. Oct 2, 2008 #3

    Mentallic

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    Homework Helper

    You are on the right track. :tongue2:

    There is no need to find the time in this question (if there was, I would've asked you how you obtained that time).

    Using [tex]V_{f}^{2}=V_{i}^{2}+2a \Delta x[/tex]

    Where:

    [tex]V_{f}=Final-velocity[/tex] In this case, when the car has stopped. i.e. 0 ms-1
    [tex]V_{i}=Initial-velocity[/tex] When it began breaking. i.e. 12.5 ms-1
    [tex]a=acceleration[/tex] The negative acceleration applied by the brakes. -5.2 ms-2
    [tex]\Delta x = Displacement[/tex] The distance covered during the braking process. Unknown.

    As you can see, using this equation is best since you know all variables except the one you are trying to obtain.

    Make sure you keep the acceleration negative, since it is decceleration. Good luck!
     
  5. Oct 2, 2008 #4
    Thanks, got it. I missed the obvious.
     
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