Reducing Stopping Time with Antilock Brakes

[negation]

Hi, I'm an undergraduate commencing my first year in Astrophysics and Mathematics.
It's the summer break as of now and I'm doing some self-study on the relevant topics when school commences.

Homework Statement

On packed snow, computerized antilock brakes can reduce a car's stopping distance by 55%. By what percentage is the stopping time reduced?

Homework Equations

none

 

[ShayanJ]

Consider the time-independent equation(as we call it) $v_2^2-v_1^2=2A\Delta x$.
For the case of stopping motion,$v_2=0$ and $A<0$ so let's define $a=|A|$,so we have $v_1^2=2a\Delta x$. $v_1$ can be taken to be the same in cases of having antilock brakes($\Delta x^*,a^*$) and having ordinary brakes($\Delta x,a$).So we have:$\frac{\Delta x^*}{\Delta x}=\frac{a}{a^*}=0.55$ which gives us $a^*=\frac{a}{0.55}$
Now consider the equation $\Delta x=-\frac{1}{2}at^2+v_1 t$. Let's say the decrease in time is equal to $\Delta t$ so we have $\Delta x^*=-\frac{1}{2}a^* (t-\Delta t)^2+v_1(t-\Delta t)$ and $\Delta x=-\frac{1}{2}at^2 +v_1 t \Rightarrow v_1=\frac{\Delta x+\frac{1}{2}at^2}{t}$

$\Delta x^*=-\frac{1}{2}a^*(t-\Delta t)^2+\frac{\Delta x+\frac{1}{2}at^2}{t}(t-\Delta t) \Rightarrow \frac{a^*t}{2}(t-\Delta t)^2-(\Delta x+\frac{1}{2}at^2)(t-\Delta t)+\Delta x^*t=0$
Now we have an quadratic equation in $t-\Delta t$ which can be solved easily.Then you can divide the answer by t to get your answer.

 

[negation]

Consider the time-independent equation(as we call it) $v_2^2-v_1^2=2A\Delta x$.
For the case of stopping motion,$v_2=0$ and $A<0$ so let's define $a=|A|$,so we have $v_1^2=2a\Delta x$. $v_1$ can be taken to be the same in cases of having antilock brakes($\Delta x^*,a^*$) and having ordinary brakes($\Delta x,a$).So we have:$\frac{\Delta x^*}{\Delta x}=\frac{a}{a^*}=0.55$ which gives us $a^*=\frac{a}{0.55}$
Now consider the equation $\Delta x=-\frac{1}{2}at^2+v_1 t$. Let's say the decrease in time is equal to $\Delta t$ so we have $\Delta x^*=-\frac{1}{2}a^* (t-\Delta t)^2+v_1(t-\Delta t)$ and $\Delta x=-\frac{1}{2}at^2 +v_1 t \Rightarrow v_1=\frac{\Delta x+\frac{1}{2}at^2}{t}$

$\Delta x^*=-\frac{1}{2}a^*(t-\Delta t)^2+\frac{\Delta x+\frac{1}{2}at^2}{t}(t-\Delta t) \Rightarrow \frac{a^*t}{2}(t-\Delta t)^2-(\Delta x+\frac{1}{2}at^2)(t-\Delta t)+\Delta x^*t=0$
Now we have an quadratic equation in $t-\Delta t$ which can be solved easily.Then you can divide the answer by t to get your answer.

I want to get used to the notation first.

1) What does (Δx∗,a∗) implies?
2) And if final velocity = 0, then why isn't there a negative in front of initial velocity?

 

[ShayanJ]

I want to get used to the notation first.

1) What does (Δx∗,a∗) implies?
2) And if final velocity = 0, then why isn't there a negative in front of initial velocity?

1)stopping distance and acceleration when the car has antilock brakes.
2)$-v_1^2=2A\Delta x \Rightarrow -v_1^2=-2a\Delta x \Rightarrow v_1^2=2a\Delta x$
I thought I made these points clear!

 

[negation]

1)stopping distance and acceleration when the car has antilock brakes.
2)$-v_1^2=2A\Delta x \Rightarrow -v_1^2=-2a\Delta x \Rightarrow v_1^2=2a\Delta x$
I thought I made these points clear!

That explains. Sorry man, it helps if there's 2 para spacing in between each lines. But thanks