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Tangential deceleration

  1. May 10, 2008 #1
    Im trying to figure out the maximum deceleration possible for a car coming into a corner at two points.....

    I found the answer to the max. deceleration for the car right before the corner, but I cant get the correct answer for when the car is actually in the corner.....

    im looking for the maximum Tangential decerlation after just entering the corner.....

    the max. for right before the car enters is -8m/s

    Just looking for a strategy......

    m = 1500kg
    v = 15m/s
    r = 30

    max friction = 12,000N
  2. jcsd
  3. May 10, 2008 #2
    i guess i stared long enough

    it was: a^2 = a_n^2 + a_t ^2

    where a_n = v^2/r

    a_n = 7.5 m/s

    so we know the total a = -8m/s

    find a_t when car is entering corner

    a_t = ( (a^2) - (a_n^2) )^.5

    a_t = -2.78

    hence you are able to decelerate much better on a straight than in a corner....
    Last edited: May 10, 2008
  4. May 11, 2008 #3
    can u put the equations into brackets so i know that its eg v^(2/r) or (v^2)/r. i havent done this sort of stuff but it looks like you could just use logarithm functions to work it out.

    [when a=m^b b=(log(base 10)a)/(log(base10)m), LnA=B and A=e^b]
    hope that helps!
  5. May 11, 2008 #4
    actually this makes it more clear

    [tex]Normal Acceleraton = A_{n} =\frac{v^2}{r}= [/tex]

    magnitude of total Acceleration

    [tex] A = \sqrt{A_{n}}-{A_{t}} [/tex]

    solve for [tex] A_{t}[/tex]
    Last edited: May 11, 2008
  6. May 11, 2008 #5
    Im still working on learning the syntax for using latex....

    Actually this makes it easier:

    [tex]Normal Acceleraton = A_{n} =\frac{v^2}{r}= [/tex]

    Total Magnitude of Acceleration

    [tex] A = \sqrt{A_{n}}-\sqrt{A_{t}} [/tex]

    Solve for [tex] A_{t}[/tex]
    Last edited: May 11, 2008
  7. May 12, 2008 #6
    what are the values for the known letters? have you tried using simultaneus equations with logarithm functions? sorry i cant be of much help.
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