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Deceleration of a Car Driver

  1. Sep 16, 2006 #1
    I'd appreciate some help with this physics problem:

    A car is travelling 10 km/hr and crashes into a tree. The driver is thrown 1 m forward. What is his average deceleration?

    Here's my work:

    Velocity of driver: 10 km/hr = 10,000 m/hr = 250/9 m/s

    Time: 1 m / 250/9 m/s = .036 s

    So it took him .036 seconds to travel 1 m.

    Acceleration = final velocity - initital velocity / time

    0 - 250/9 m/s / .036 s = -771.6 m/s

    So his average deceleration is 771 m/s ?
     
  2. jcsd
  3. Sep 16, 2006 #2

    Janus

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    Your answer here is off by a factor of 10.
    Error above carries through.[/quote]

    So it took him .036 seconds to travel 1 m.

    [/quote] Besides the fact that your answer is still off by a factor of 10, I'm assuming from the way the question is stated, that the driver decelerates during that 1 m, so he will not be traveling at a constant speed over this meter.
    compounding your factor of 10 error from above.
     
  4. Sep 16, 2006 #3
    Oh, of course, that was foolish. It's 25/9 m/s and thus .36 s.

    Does that mean that the average deceleration is 7.716 ms/s ? Is that correct?

    So his speed is changing? I have no idea how that affects my answer of 7.71 m/s.
     
  5. Sep 16, 2006 #4

    Janus

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    It means that because he is slowing as he crosses that meter, the time it takes to cross that meter will be longer than what you calculated by assuming he maintained his initial velocity.

    If the driver has the initial velocity of the car and a final velocity of zero, what is his average velocity?
     
  6. Sep 16, 2006 #5

    radou

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    Car crashes? Drivers flying through the air? AGAIN?! :grumpy:
     
  7. Sep 16, 2006 #6
    Well, his average velocity would be half of the inital velocity? So 5 km/hr.

    That means 5 km/hr = 5000 m/hr = 25/18 m/s

    Time: 1 m / 25/18 m/s = .72 s

    So it takes him twice as long.

    And -250 m/s / 72 s = -1.929 m/s

    A deceleration of 1.929 m/s.

    Thanks for your patience.
     
  8. Sep 16, 2006 #7

    Janus

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    So far, so good
    remember:

    Also, the acceleration will be in m/sec/sec or m/sec² not m/sec
     
  9. Sep 17, 2006 #8
    Oh...I used the average velocity for inital velocity (that was a typo by the way, -250 m/s should have been -25/18).

    But initial velocity is 10 km/hr, and thus 250/9 m/s ?

    Acceleration: 0 - 250/9 m/s / .72 s = 38.58 m/s²

    I think I may finally have got it.
     
  10. Sep 17, 2006 #9
    10 km per hour is quite slow. It's 10000 metres per hour, which is 10000/3600 metres per second, or 2.77m/s. His average velocity was half this and therefore 1.38m/s. He covered a metre, so he was decelerating for 1/1.38s or .72 seconds. And he decelerated from 2.77m/s in .72s so his deceleration was 2.77/.72 or 3.85m/s/s.
     
  11. Sep 17, 2006 #10

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    Oops, You used the the "too large by factor of ten" value for your initial velocity again.
     
  12. Sep 17, 2006 #11
    I can't believe I did that again...OK, it's -25/9 m/s / .72 s = -3.85 m/s² (Farsight's answer).

    Thanks so much for helping me.
     
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