How Does Velocity Change in a Car Crash Impact?

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Homework Help Overview

The discussion revolves around a physics problem involving a car crash scenario, specifically examining the change in velocity of a car (1700 kg) that impacts a wall while traveling at a constant speed of 49 km/h. The problem presents a situation where the car crushes a portion of its front over a brief time interval, raising questions about the calculated velocity during the impact.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of average speed based on the distance crushed in a given time, questioning the feasibility of the resulting velocity. There are inquiries about the accuracy of the problem statement and potential typographical errors in the text. Some participants suggest sharing diagrams for clarity.

Discussion Status

The conversation is ongoing, with participants exploring various interpretations of the problem and questioning the assumptions made regarding the time and distance measurements. There is no explicit consensus, but guidance is offered regarding the need for careful analysis of the problem setup.

Contextual Notes

Participants note the possibility of a typo in the problem statement regarding the time interval, suggesting that the time might be incorrectly stated as 5 x 10^-3 seconds instead of a potentially larger value. The original poster expresses uncertainty about the calculations and the interpretation of the problem.

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.
 

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