How Does Velocity Change in a Car Crash Impact?

[johns123]

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 ?

 

[Delphi51]

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.

 

[johns123]

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.

 

[Delphi51]

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.

 

[Nickie]

I would approach this problem by first analyzing the information given and making sure all units are consistent. In this problem, we are given the mass of the car (1700kg), its initial velocity (49 km/hour), and the distance it travels (2/9 of 2 meters) in a given time (5 x 10^-3 seconds).

Using the formula D = R x T, we can calculate the velocity in the first frame to be 89 meters/second, which does seem quite high. However, it is important to note that this velocity is only for a very short duration, and the car may have experienced a rapid deceleration due to the impact with the wall.

Furthermore, it is also important to consider the concept of impulse, which is the change in momentum of an object. In this case, the car experiences a large change in momentum due to the impact, which could explain the seemingly high velocity in the first frame.

In order to fully understand the velocity problem in this car crash, further analysis and calculations would be needed, taking into account factors such as the force of the impact, the deformation of the car, and the change in momentum. It is also worth noting that in real-life situations, car crashes are complex and can involve multiple impacts and forces, making it even more difficult to accurately calculate the velocity at any given moment. It is important to approach these types of problems with caution and to consider all factors and variables at play.