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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.

Jim
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Use your head
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    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. https://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations – Use your head Jun 25 '13 at 19:26
  • 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 ?

     – Use your head Jun 25 '13 at 19:27
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    Just so you know, inflammatory or offensive posts are not acceptable here – Jim Jun 25 '13 at 19:32
  • Possible duplicate: http://physics.stackexchange.com/q/535/2451 – Qmechanic Jun 25 '13 at 19:33
  • @Jim - ok. But, don't you think there is too much unnecessary down voting going on ? – Use your head Jun 25 '13 at 19:33
  • While I didn't downvote you, I cannot claim to be omniscient and therefore cannot comment on whether or not the downvoting was unnecessary. – Jim Jun 25 '13 at 19:35
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    so much condescension going down all over this question; what site am i on – Justin L. Jun 26 '13 at 06:49
  • This question (v4) should in principle be closed as too localized. However, I think, it is more helpful to close it as a duplicate, even if it is strictly speaking not the case. – Qmechanic Jun 26 '13 at 21:39

1 Answers1

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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.

Jim
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  • Thanks, I already did it. Stupid SO wont let me post my own answer here. – Use your head Jun 25 '13 at 19:29
  • please give me an upvote. Some crazy ppl just downvoted me. – Use your head Jun 25 '13 at 19:29
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    @Useyourhead you're new, so here's some free advice. Asking for upvotes is the only sure way of not getting them. – Jim Jun 25 '13 at 19:53
  • 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. – dmckee --- ex-moderator kitten Jun 25 '13 at 19:57
  • @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. – Jim Jun 25 '13 at 20:00