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Deceleration of a car.

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data
    The braking distance is the distance the car travels while decelerating once the brakes have been applied. A car of mass 450kg is travelling at a speed of 70MPH (31.3ms-1) when the driver makes an emergency stop.
    Calculate the deceleration of the car (Assume a uniform deceleration)

    We're also told that at 70mph/31.3ms-1 that the breaking distance is 75 m.


    3. The attempt at a solution
    I did initially try to do this using the equations of motion but that gave me a huge answer, and didn't take into account the mass of the car so I didn't think it was right.

    Any help is hugely appreciated!
     
  2. jcsd
  3. Dec 8, 2011 #2

    gneill

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    Staff: Mentor

    You should show your attempt, even if it has an error or two. That way we will know how to help.
     
  4. Dec 8, 2011 #3
    Ok, this is what I did.

    S=75
    U=31.3
    V=0
    A=?
    T=?

    s=u²+2as
    0=31.3²+2*a*75
    0=979.69+150a
    -979.69/150=a
    a=-6.53 ms-2

    Which looking back doesn't look too bad, when I did it first time I think I accidentally must have typed the wrong numbers into my calculator because I got -1000ms-2. I'm just thinking I did it wrong because I didn't use the mass.
     
  5. Dec 8, 2011 #4

    gneill

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    Staff: Mentor

    Looks good.

    One small nit though, in your first equation you put "s" on the left where you meant to put "V2". Clearly a typo, since you correctly substituted the value of V2 in your next line. So overall, well done.
     
  6. Dec 8, 2011 #5
    There's just one other part:
    The car is then filled with passengers so that its mass is doubled. Calculate its stopping distance when travelling at 70MPH. Assume the force calculated in b(ii) remains constant.

    I calculated the force as:
    F=ma
    F=450*6.53= 2938.5N

    How do I go about doing this is it:

    2938.5 = 900 * a
    2938.5/900 = 3.625

    So:
    v²=u²+2as
    0=31.3²+2*3.625*s
    -979.69/7.25=s
    s=135 metres.

    Thanks again!
     
  7. Dec 8, 2011 #6

    gneill

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    Staff: Mentor

    Check your value for the new acceleration. Looks like a finger problem. Otherwise you're method works okay.

    You could skip the force calculation if you realize that since F = MA, then A = F/M. Therefore if you double M you must halve A.
     
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