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The DMV manual says that

The faster you go, the less time you have to avoid a hazard or collision. The force of a 60 mph crash is not just twice as great as a 30 mph crash; it’s four times as great!

My physics is quite rusty, so I could not figure it out. I guess the above statement is correct, but how do we prove it?

Edit

I figured this out myself, but alternative methods or new ways of understanding are still welcome.

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    $\begingroup$ I think this can be answered by the basic equations of linear motion. But I can't post my answer till 8 hours...so silly. en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations $\endgroup$ Jun 25, 2013 at 19:26
  • $\begingroup$ What is the negative vote for ? I thought about it again. I think this could be the solution: F = ma, v^2 = u^2 + 2aS . v = 0 and S = "S1" for both cases. so, a = -u^2/2s .let F1 be the force for car traveling at 60mph F1/F2 = (U1)^2/(U2)^2 = (60/30)^2 = 4, F1 = 4F2, Proved....makes sense ? $\endgroup$ Jun 25, 2013 at 19:27
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    $\begingroup$ Just so you know, inflammatory or offensive posts are not acceptable here $\endgroup$
    – Jim
    Jun 25, 2013 at 19:32
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/535/2451 $\endgroup$
    – Qmechanic
    Jun 25, 2013 at 19:33
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    $\begingroup$ so much condescension going down all over this question; what site am i on $\endgroup$
    – Justin L.
    Jun 26, 2013 at 6:49

1 Answer 1

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This is pretty basic physics:
We know the following formulae

$$F=ma$$ $$a={v_f^2-v_i^2\over2\Delta d}$$ In both cases, the final velocity is $0$. Assuming you have the same room, $\Delta d$, to decelerate in a crash,

$$F=m{v^2\over2\Delta d}$$

Due to the square of the velocity, if you increase the impact speed by a factor of 2, you increase the impact force by a factor of 4.

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  • $\begingroup$ Thanks, I already did it. Stupid SO wont let me post my own answer here. $\endgroup$ Jun 25, 2013 at 19:29
  • $\begingroup$ please give me an upvote. Some crazy ppl just downvoted me. $\endgroup$ Jun 25, 2013 at 19:29
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    $\begingroup$ @Useyourhead you're new, so here's some free advice. Asking for upvotes is the only sure way of not getting them. $\endgroup$
    – Jim
    Jun 25, 2013 at 19:53
  • $\begingroup$ This, of course, has the built in assumption that $\Delta d$ is the same in both cases which is generally not true. It is more correct to say that there is four time the mechanical energy and leave it at that. $\endgroup$ Jun 25, 2013 at 19:57
  • $\begingroup$ @dmckee I agree, but the quoted DMV book said 4 times the force, so I attempted to provide their reasoning. I also did state explicitly that I assumed $\Delta d$ was the same in both cases. $\endgroup$
    – Jim
    Jun 25, 2013 at 20:00

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