Trying to reconstruct an accident

  • Thread starter bilbozilla
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  • #1

Main Question or Discussion Point

Hello all,

Here's the data I have -

Initial Elevation : 0 feet
Final elevation: -25 feet
Distance from roadway: 55 feet

Assuming the tree that was hit didn't lessen the velocity, and the vehicle didn't bounce, what would the velocity have been at the time the vehicle left the roadway at initial elevation?

How would I calculate the velocity? This is not a homework problem. I've got the totaled car to prove it. (Nobody was hurt) The rear-engine layout caused the vehicle to spin and the car went off rear-first.
 

Answers and Replies

  • #2
Vanadium 50
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First, you need a professional accident reconstructor, not some guys you met on the internet.

Second, we knew this wasn't a homework problem. You provided no information on what happened - just three numbers.
 
  • #3
This is a mental exercise for back-of-the-napkin calculations, nothing more. It is certainly not going to be used for anything other than a brief sanity check.

What other information do you need? I have gravity, which is constant, the hight, and the distance. Here are some more numbers.

The weight of the vehicle is 3,100# with driver. I didn't think about that one. What other variables do you need to get a rough estimate (again, just for kicks) of the velocity.
 
  • #4
DaveC426913
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Weight of vehicle is not needed. Initial height, final height and distance should be enough to calc velocity. But there are many other factors, such as angle of exit from roadway. However, taking it at face-value, I get:

a = 9.8 m/s^2
d(y) = 7.625m
t=?

http://en.wikipedia.org/wiki/Equations_for_a_falling_body#Overview"
t = sqrt(2 x 7.625 / 9.8)
t = 1.247s
This is duration of fall.

To travel 55ft (total displacement, not perpendicluar to roadway) in 1.247s would indicate an initial velocity of 13.45m/s or 48.3kph.

Again, this is an ideal scenario. It is of little use unless the scenario was textbook.
 
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  • #5
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t = sqrt(2d/g)
t = sqrt(2 x 7.625 / 9.8)
t = 1.556s
This is duration of fall.
I think you forgot to take the square root

sqrt(2 x 7.625 / 9.8) = 1.247
 
  • #6
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What you really need is a description of what happened. Apparently a tree is involved?
 
  • #7
DaveC426913
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I think you forgot to take the square root

sqrt(2 x 7.625 / 9.8) = 1.247
Thanks. Post fixed.
 
  • #8
DaveC426913
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What you really need is a description of what happened. Apparently a tree is involved?
I'm assuming the tree has marks on it that are 25 vertical feet below the road surface and that the tree is 55 horizontal feet from the spot where the car left the ground.

My assumption is implicit in saying this is a textbook calculation, given nothing but the numbers. If the poster wants to provide more detail, he's welcome to do so.
 

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