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Math Help - urgent help with acceleration please

  1. #1
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    urgent help with acceleration please

    in coming to a stop, a car leaves skid marks 80m long on the highway. assuming a deceleration of 7.0m/s^2, estimate the speed of the car just before breaking.

    determine the stopping distances for an automobile with an initial speed of 90km/h and human reaction time of 1.0s: a) for an accelerrtion a=-4.0m/s^2; b) for a=-8.0m/s^2
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  2. #2
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    Quote Originally Posted by asnxbbyx113 View Post
    in coming to a stop, a car leaves skid marks 80m long on the highway. assuming a deceleration of 7.0m/s^2, estimate the speed of the car just before breaking.
    From an initial speed v the time to come to a halt under constant decelleration a, is t=v/a., and the distance travelled is s=(v/2)t.

    So plugging in the values given:

    80=(v/2)(v/7) = v^2/14

    so:

    v = sqrt[(14)(80)]

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by asnxbbyx113 View Post
    determine the stopping distances for an automobile with an initial speed of 90km/h and human reaction time of 1.0s: a) for an accelerrtion a=-4.0m/s^2; b) for a=-8.0m/s^2
    Stoping distance is:

    s = v t1 - v^2/(2a)

    where t1 is the reaction time, and all the variables are in consistent units.

    So convert the initial speed from km/hr to m/s and then plug in the other values.

    RonL
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  4. #4
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    in coming to a stop, a car leaves skid marks 80m long on the highway. assuming a deceleration of 7.0m/s^2, estimate the speed of the car just before breaking.

    Given:
    distance, d = 80 m
    deceleration, a = -7 m/sec/sec
    Final velocity, Vf = 0
    Initial velocity, Vo = unknown

    Find: Vo

    d = (average velocity)*(time)
    80 = [(Vo +Vf) /2]*t
    80 = [Vo +0](t/2)
    80 = (Vo)t/2 ---------------(1)

    Vf = Vo +at
    0 = Vo -7t
    t = (Vo)/7 ------------(2)
    Substitute that in (1),
    80 = (Vo)[(Vo)/7]/2
    80 = [(Vo)^2 ]/14
    (Vo)^2 = 80(14)
    Vo = sqrt(80*14) = 33.47 m/sec ----the speed of the car just before breaking.

    ---------------------------------------

    determine the stopping distances for an automobile with an initial speed of 90km/h and human reaction time of 1.0s: a) for an accelerrtion a=-4.0m/s^2; b) for a=-8.0m/s^2

    Human reaction time?
    Oh, it means what is the stopping distance from the moment the driver sees the danger or needs to stop immediately.
    So two distances are involved:
    ---1) the distance the car travels while it is still running at 90 kph. This is for 1 second.
    ---2) the distance the car travels the moment the brakes are hit. This is fot t seconds.

    Given:
    Vf = 0
    Vo = 90kph = (90km/hr)(1000m/1km)(1hr/3600sec) = 25 m/sec
    a = -4 m/sec/sec
    d = ?

    d = [(Vf +Vo)/2]*t
    d = [0 +25](t/2)
    d = 12.5t -------------------(i)

    Vf = Vo +at
    0 = 25 -4t
    t = 25/4 = 6.25 sec
    So,
    d = 12.5(6.25) = 78.125 m ------------for the braking.

    For the reaction time of 1 sec,
    D = (speed)(time)
    D = (25)(1) = 25 m

    Therefore, the stopping distance is 78.125 +25 = 103.125 meters. ---answer.
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