Maximizing Tangential Deceleration for Cars in Corners

In summary, the conversation is about finding the maximum deceleration possible for a car entering a corner at two different points. The maximum deceleration for right before the corner was found to be -8m/s, but the correct answer for when the car is actually in the corner has not been determined. The strategy for finding the maximum tangential deceleration after entering the corner is discussed, which involves using the equations a^2 = a_n^2 + a_t^2 and a_t = (a^2 - a_n^2)^.5. The values for m, v, and r are given, and the use of logarithm functions is suggested to solve the equations. Finally, the equations are simplified and the
  • #1
Jason03
161
0
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
 
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  • #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...
 
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  • #3
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
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
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
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.
 

1. What is tangential deceleration?

Tangential deceleration is the rate at which an object's tangential velocity decreases over time. It is a measure of how much an object's speed is changing in a circular or curved path.

2. How is tangential deceleration different from linear deceleration?

Tangential deceleration measures the decrease in an object's tangential velocity, which is the speed in a circular or curved path. Linear deceleration, on the other hand, measures the decrease in an object's linear velocity, or speed in a straight line.

3. What factors affect tangential deceleration?

The main factors that affect tangential deceleration are the object's mass, the magnitude and direction of its initial velocity, and the amount of force acting on it. The shape and surface of the object, as well as any external factors such as air resistance, can also have an impact.

4. How is tangential deceleration calculated?

Tangential deceleration can be calculated using the formula a = (v2 - v02) / r, where a is the tangential deceleration, v is the final tangential velocity, v0 is the initial tangential velocity, and r is the radius of the circular path.

5. Why is tangential deceleration important?

Tangential deceleration is important in understanding and predicting the motion of objects in circular or curved paths. It is also crucial in many real-world applications, such as designing roller coasters, calculating the speed of a car around a curve, and predicting the path of a satellite in orbit.

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