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Homework Help: (Crash) Dummies Doing Physics

  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    a. A set of crash tests consists of running a test car moving at a speed of 11.2 m/s (25 mi/hr) into a solid wall. Strapped securely in an advanced seat belt system, a 57.0 kg (126lbs) dummy is found to move a distance of 0.750 m from the moment the car touches the wall to the time the car is stopped. Calculate the size of the average force which acts on the dummy during that time.

    b. Using the direction of motion as the positive direction, calculate the average acceleration of the dummy during that time (in g's, with 1g=9.81m/s2).

    c.In a different car, the distance the dummy moves while being stopped is reduced from 0.750 m to 0.250 m, calculate the average force on the dummy as that car stops.

    2. Relevant equations
    w=force x distance
    Impulse=Favg x t

    3. The attempt at a solution
    i tried solving the first part using momentum:
    pf-pi=Favg x t
    but that didnt work

    I also tried using teh kinetic energy equation
    f x d = 1/2mv^2
    But this also didnt work..

    I dont know hat im doing wrong and i can seem to find anyother way of solving this problem.
  2. jcsd
  3. Nov 13, 2014 #2

    Simon Bridge

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    Last edited by a moderator: May 7, 2017
  4. Nov 13, 2014 #3
    The only thing you need to calculate the force is time. I think you might have to use some kinematic equations for that part, assuming that the dummy underwent constant acceleration.
  5. Nov 14, 2014 #4


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    Soapbox: This is a flawed question. Average acceleration is defined as change in velocity divided by change in time: ##\Delta v/\Delta t##. Ergo, average force is change in momentum divided by change in time: ##\Delta p/\Delta t##. There is insufficient information to calculate that. You can calculate an "average over distance" from change in energy and displacement, but that's different from the standard meaning of average force.
    As sam400 writes, to answer the question as posed you need to make some assumption about the pattern of deceleration. This will depend on the crumple characteristics of the vehicle. If it is constant deceleration then the average over distance happens to give the same answer, but more likely the deceleration will increase from 0 steadily up to a maximum before becoming approximately constant.
  6. Nov 14, 2014 #5
    Okay so i had done is this:
    Try 1:
    1/2mv^2 = F x d
    1/2(57)(11.2)^2 = F x (0.75)
    This gave me F = 4766.72N and since it acts on the dummie it is -4766.72N but this was wrong

    Try 2:
    Pi=mv = 57 x 11.2 = 638.4kgm/s
    i found that it takes 0.0669seconds for the dummie to travel 0.75m using proportions with the velocity
    Impulse = Favg x t
    638.4 = Favg x 0.0669 Favg = 9533.44 and again i put negative so -9533.44N but this was also wrong.

    Try 3:
    a = dv/dt
    a= 0-11.2/0.0669-0
    F=-9542N This didnt work

    Try 4:
    vf^2=vi^2 +2a(xf-xi)
    0=(11.2)^2 + 2a(0.75) a = 86.63m/s^2
    F=(57)(86.63) = 4766N which i had already entered and was wrong..
  7. Nov 14, 2014 #6
    The impulse of the dummy will be a negative one, so I feel it might be a sign related problem. If it's something online, it could be significant figure related too, unless the system lets you know about how close the value is from it.
  8. Nov 14, 2014 #7
    It tells me the amount of significant figures i need and i always enter the negative answer. when i input my answer it tell me this:

    The force that acts on the dummy has to do (negative) work on the dummy. It must slow the dummy and stop it. Thus the amount of work equals the kinetic energy of the dummy. From the equation of work done by a force we can calculate that force.

    Maybe this will help..
  9. Nov 15, 2014 #8


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    How does the question specify the number of sig figs required of your answer?
  10. Nov 15, 2014 #9


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    In try 1 you entered -4766.72N, but now you have the positive version. Did you try both?
    I see no reason to enter a negative value here. You are not asked for the work done. The force could be positive or negative depending on which direction you take as positive, and the question does not specify that.
    If you take the direction of movement of the car as positive then the displacement is positive and the force negative, giving negative work.
    If you take the direction of movement of the car as negative then the displacement is negative and the force positive, giving negative work
    And it specifies 6? But your conversions from lbs and miles were not that accurate. Only the first two digits of your answer are correct.
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