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Basic kinematics motion

  1. Dec 5, 2013 #1
    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

  2. jcsd
  3. Dec 6, 2013 #2


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    Consider the time-independent equation(as we call it) [itex] v_2^2-v_1^2=2A\Delta x [/itex].
    For the case of stopping motion,[itex] v_2=0 [/itex] and [itex] A<0 [/itex] so lets define [itex] a=|A| [/itex],so we have [itex] v_1^2=2a\Delta x [/itex]. [itex] v_1 [/itex] can be taken to be the same in cases of having antilock brakes([itex] \Delta x^*,a^* [/itex]) and having ordinary brakes([itex] \Delta x,a [/itex]).So we have:[itex]
    \frac{\Delta x^*}{\Delta x}=\frac{a}{a^*}=0.55[/itex] which gives us [itex] a^*=\frac{a}{0.55} [/itex]
    Now consider the equation [itex] \Delta x=-\frac{1}{2}at^2+v_1 t [/itex]. Lets say the decrease in time is equal to [itex] \Delta t [/itex] so we have [itex] \Delta x^*=-\frac{1}{2}a^* (t-\Delta t)^2+v_1(t-\Delta t) [/itex] and [itex] \Delta x=-\frac{1}{2}at^2 +v_1 t \Rightarrow v_1=\frac{\Delta x+\frac{1}{2}at^2}{t}[/itex]

    [itex] \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 [itex] t-\Delta t [/itex] which can be solved easily.Then you can divide the answer by t to get your answer.
  4. Dec 6, 2013 #3

    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?
  5. Dec 6, 2013 #4


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    1)stopping distance and acceleration when the car has antilock brakes.
    2)[itex]-v_1^2=2A\Delta x \Rightarrow -v_1^2=-2a\Delta x \Rightarrow v_1^2=2a\Delta x [/itex]
    I thought I made these points clear!
  6. Dec 6, 2013 #5
    That explains. Sorry man, it helps if there's 2 para spacing in between each lines. But thanks
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