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Math Help - Did I do this right. F=ma to find Vi for my question.

  1. #1
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    Ontario Canada
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    Did I do this right. F=ma to find Vi for my question.

    Need help on this.
    A car whose mass is 2600 kg can produce an unbalanced braking force of 2765N.
    Calculate the car's initial speed in order to have a stopping distance of 3036.3m.
    Vi=?
    So...
    m=2600kg
    F=2765N
    s=3036.3m
    Vf=0

    So I found acceleration of a=F/m = 2765/2600 = 1.06346m/s^2
    So I have to find time=Square root of 2(3036.33)/1.06346 = 75.56605s
    So I would use this formula to solve for Vi? right
    Vi=Vf -at
    Vi = 0 - (1.06346)(75.56605) = -80.3614 m/s
    Is the negative right?
    IS THIS CORRECT if not please show me where I went wrong
    Thanks
    Jo
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  2. #2
    Senior Member
    Joined
    Feb 2008
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    383
    Quote Originally Posted by bradycat View Post
    Need help on this.
    A car whose mass is 2600 kg can produce an unbalanced braking force of 2765N.
    Calculate the car's initial speed in order to have a stopping distance of 3036.3m.
    Vi=?
    So...
    m=2600kg
    F=2765N
    s=3036.3m
    Vf=0

    So I found acceleration of a=F/m = 2765/2600 = 1.06346m/s^2
    So I have to find time=Square root of 2(3036.33)/1.06346 = 75.56605s
    So I would use this formula to solve for Vi? right
    Vi=Vf -at
    Vi = 0 - (1.06346)(75.56605) = -80.3614 m/s
    Is the negative right?
    IS THIS CORRECT if not please show me where I went wrong
    Thanks
    Jo
    Code:
    So I found acceleration of a=F/m = 2765/2600 = 1.06346m/s^2
    that should be minus. -2765 and -1.06346m/s^2

    s=vt-0.5*a*t^2

    3036.3=0-0.5*-1.06346*t^2

    3036.5/(0.5*1.06346)=t^2

    t=75.56605s

    Vi = 0 - (-1.06346)(75.56605) = 80.3614 m/s
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  3. #3
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    Babymilo,

    Not trying to step on your toes here, i just want to show an alternative.

    \frac{1}{2}mv^2 = KE of vi

    \frac{1}{2}mv^2 = work done by brakes

    work done by brakes = braking force x distance.

    vi = 80.36 m/s like you stated.
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