Tangential deceleration

  • Thread starter Jason03
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  • #1
Jason03
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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 can't 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
 

Answers and Replies

  • #2
Jason03
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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:
  • #3
brandy
161
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can u put the equations into brackets so i know that its eg v^(2/r) or (v^2)/r. i haven't 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!
 
  • #4
Jason03
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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]
 
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  • #5
Jason03
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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:
  • #6
brandy
161
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what are the values for the known letters? have you tried using simultaneus equations with logarithm functions? sorry i can't be of much help.
 

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