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Help solving for average force?

  1. Jun 22, 2016 #1
    1. The problem statement, all variables and given/known data
    When it crashes into a bridge support that does not move, a car goes from 85 km/h to 0 in 1.23 m.
    A) What is the impulse delivered to the 70-kg driver by the seat belt, assuming the belt makes the driver's motion identical to the car's motion? Assume that the initial direction of motion of the car is the positive direction.
    B) What is the average force exerted by the belt on the driver?
    C) If the driver were not wearing the seat belt and flew forward until he hit the steering wheel, which stopped him in 0.0145 s, what would be the average force exerted by the steering wheel on him?

    2. Relevant equations
    F = m*a
    J = delta p
    p = m*v
    3. The attempt at a solution
    So I got part A, -1650 N*s, by just converting 85 km/h to 23.61 m/s and multiplying that by the mass (70 kg). Then for part B I found what I thought was the acceleration by dividing the 23.61 m/s velocity by 1.23 m (73.8 s). I got 0.319992 m/s^2 and multiplied that by the 70 kg mass to get the force. This gave me an answer of -22 N, which was wrong. I then checked to see if I interpreted the "1.23 m" wrong, checking to see if I should've used 83 s instead of 73.8 s, and got - 20 N but this was still wrong.

    What am I doing wrong?
     
  2. jcsd
  3. Jun 22, 2016 #2
    Do you really think that it takes 73 seconds for a seat belt to stop you in a sudden collision?
     
  4. Jun 22, 2016 #3
    I mean no, but could you help me think of another way to approach this problem? What would be another way to find the acceleration so I can use F = m * a ?
     
  5. Jun 22, 2016 #4
    You determined the acceleration incorrectly. Are you familiar with the suvat equations? If so, which one of these equation involves velocities, acceleration, and distance only?
     
  6. Jun 22, 2016 #5
    ohhhh meters instead of minutes; dumb mistake, thank you.
     
  7. Jun 22, 2016 #6
    So which suvat equation do you propose to use?
     
  8. Jun 22, 2016 #7

    SteamKing

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    Where does the 73 seconds come from?
     
  9. Jun 22, 2016 #8

    Vf^2 = Vi^2 + 2ad
     
  10. Jun 22, 2016 #9
    Excellent. So now what do you get for a?
     
  11. Jun 22, 2016 #10

    jbriggs444

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    Science Advisor

    Oh, I see it. 1.23 m times 60 seconds per minute = 73 seconds.

    However, the 1.23 m actually means 1.23 meters.

    Obtaining a average (over time) force is difficult without assuming constant deceleration. Obtaining an average (over distance) force is easier.
     
  12. Jun 22, 2016 #11
    - 226.598 m/s^2 and -15862 N. Thanks again!
     
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