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Equation Help.

  1. Sep 7, 2009 #1
    Hello!

    in Physics we just started acceleration

    We were givin 3 equations

    v=V(not) + at

    x=V(not)t + 1/2at(Squared)

    V(Squared) =V(not) + 2ax


    Can someone please explain the variables and how to figure them out.

    With those equations I need to figure problems like this

    The speed limit in a school zone is 44 km/h. A driver traveling at this speed sees a child run onto the road 11 m ahead of his car. He applies the brakes, and the car decelerates at a uniform rate of 10.5 m/s2. If the driver's reaction time is 0.65 s, will the car stop before hitting the child?

    YES OR NO
    What is the stopping distance of the vehicle in m?


    And


    A boy throws a stone straight upward with an initial speed of 18 m/s. What maximum height will the stone reach before falling back down



    I really don't know where to start. I normally start listing the varables like
    V=
    a=
    t=
    x=

    But I do not understand what numbers to put in which areas, :confused:


    Thank you very much
     
  2. jcsd
  3. Sep 7, 2009 #2
    Well, to help you out a bit.

    x is a distance in meters
    t is a time in seconds
    V is a velocity in meters/second
    a is an acceleration in meters/second squared.

    Also, V(not) for this problem is the initial velocity given.

    Think it through, this is a very simple problem.

    Thanks
    Matt
     
  4. Sep 7, 2009 #3
    Thank you for your help Matt.

    How do I figure out time, if I only get 3/5 or 2/4 varibles?
     
  5. Sep 7, 2009 #4
    You are given the time. t = 0.65 seconds. I think you are misreading the problem. You are not calculating a time, you are calculating a distance. You need to determine if the driver stops within 11 meters. Not the time required for the driver to stop.

    Make sense?

    Thanks
    Matt
     
  6. Sep 7, 2009 #5
    I put the following

    x=11
    T=0.65
    a=10.5 m/s(squared)
    v=44 km/h

    Now I would have to plug the above into one of these

    v=V(not) + at

    x=V(not)t + 1/2at(Squared)

    V(Squared) =V(not) + 2ax


    This is where I get confused.

    I do not know which equation to use. I am thinking I would use
    V(Squared) =V(not) + 2ax

    Because it is the only one with an X

    So if I plug in the variables I would get

    44=0 + 2(10.5x)

    44 = 21x

    x=2

    So the distance would be 2m?

    Gosh I am confused lol

    Thanks again !
     
  7. Sep 7, 2009 #6
    You are solving for a distance, x, so use the equation that produces the distance, x, as the result.

    Thanks
    Matt
     
  8. Sep 7, 2009 #7
    So that would be x= v(not)t+1/2at(squared)


    ?
     
  9. Sep 7, 2009 #8
    Yes, that is the one. Also, as a check, make sure all of your units cancel to produce meters.

    Thanks
    Matt
     
  10. Sep 7, 2009 #9

    drizzle

    User Avatar
    Gold Member

    the driver's reaction time is not the time taken till the velocity reaches zero, you can find the time using the first equation:
    v=vo + at

    [don’t forget to subtract the driver's reaction time from the calculated time]
     
  11. Sep 7, 2009 #10
    Yes. That is correct.

    Thanks
    Matt
     
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