Basic kinematics motion

1. Dec 5, 2013

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.

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

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

2. Relevant equations

none

2. Dec 6, 2013

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 lets 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 $\Delta x=-\frac{1}{2}at^2+v_1 t$. Lets 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.

3. Dec 6, 2013

negation

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?

4. Dec 6, 2013

ShayanJ

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!

5. Dec 6, 2013

negation

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

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