How Does Velocity Change in a Car Crash Impact?

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johns123
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Homework Statement



A problem in the text shows a 1700kg car hitting a wall
and crushing the front of the car. The car is moving at
49 km/hour constant velocity just before the impact. After
the impact a camera frame shows that the car has moved
or crushed about 2/9 of 2 meters in 5 x 10^-3 seconds, but
is not fully crushed yet. There are more frames showing the
rest of the crush until the car stops. Here's my question:

Original velocity before contact with the wall is ..

49 km/hour, or 49 x 10^3 / 3.6 x 10^3 = 13.6 meters/sec

the velocity I calculate in that first frame is ..

(2/9 x 2 meters) / (5 x 10^3) sec = .089 x 10^3 meters/sec

OR ! 89 METERS/ SECOND ! Well ?

This makes no sense to me, but D = R x T , and that's given in the text ?
 
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If it went 2/9*2 meters in .005 seconds, that is an average speed of 89 m/s.
And impossible, as you say. Any chance of posting the diagram? You could take a photo with a camera. Maybe if you name the text and page someone who has one could scan it.
 
Delphi51 said:
If it went 2/9*2 meters in .005 seconds, that is an average speed of 89 m/s.
And impossible, as you say. Any chance of posting the diagram? You could take a photo with a camera. Maybe if you name the text and page someone who has one could scan it.

The book is Physics For Scientists and Engineers vol 1 by Ohanian. And the text example with photos is in ch 11 on page 340. I don't know if it is a typo or not because my experience with Ohanian is if you work at it long enough, he is always right. Then, at the end of the chapter .. problem #3, he asks for me to plot the F x t and calc the average velocity for each frame. I can only believe that I'm not reading the text problem right .. or .. there's a typo, and the time is wrong .. maybe 5 x 10^2 ? ... dropping Vi from 13.2 to 8.9m/s in that interval.
 
In my experience a mistake can be very easily made at any point in a question. You might be best off to forget about all analysis and just describe the diagram as an observant artist would.