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Car going round corner

  1. Jul 23, 2007 #1
    1. The problem statement, all variables and given/known data

    A car moving at a speed of 27 m/s enters a curve that describes a quarter turn of radius 126 m. The driver gently applies the brakes, giving a constant tangential deceleration of magnitude 1.2 m/s2.


    2. Relevant equations

    a) Just before emerging from the turn, what is the magnitude of the car's acceleration?

    3. The attempt at a solution

    heres what i know, but cant seem to put it all together for this problem.
    v-initial = 27m/s
    v-final (as it emerges from the turn) = ?
    a = 1.2 (the radial acceleration component has to be zero because it is going round an arc right? (pie/2 .. also s = 197.9m )
    (the tangetial acceleration is the one that changes, correct?)

    I need to find v-final but how can i do that without the time????

    Please help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 23, 2007 #2

    mgb_phys

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    Homework Helper

    You can re-arange s = u t + 0.5 a t^2 to give
    v^2 = u^2 + 2 a S (Sorry for not putting it into latex)

    Good start is to write down all the numbers you know, or can easily work out.
     
  4. Jul 23, 2007 #3

    Doc Al

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    Staff: Mentor

    Not correct--just the opposite! The tangential acceleration is given as constant--the tangential speed changes, of course. The radial acceleration is not zero! Hint: How do you calculate centripetal acceleration?

    Use the distance.
     
  5. Jul 23, 2007 #4

    mgb_phys

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    Isn't the corner a red herring?

    Distance = 0.2 * 2pi*126 m
    U = 27
    a = 1.2 m/s^2

    v =
     
    Last edited: Jul 23, 2007
  6. Jul 23, 2007 #5

    Doc Al

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    Staff: Mentor

    Not at all. Note that they take care to specify "Just before emerging from the turn..."

    (And the OP had already calculated the distance.)
     
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