# How - The force of a 60 mph crash is not just twice as great as a 30 mph crash; it’s four times as great! [duplicate]

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.

• 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 Jun 25, 2013 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 ? Jun 25, 2013 at 19:27
• Just so you know, inflammatory or offensive posts are not acceptable here
– Jim
Jun 25, 2013 at 19:32
• Possible duplicate: physics.stackexchange.com/q/535/2451 Jun 25, 2013 at 19:33
• so much condescension going down all over this question; what site am i on Jun 26, 2013 at 6:49

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.

• Thanks, I already did it. Stupid SO wont let me post my own answer here. Jun 25, 2013 at 19:29
• please give me an upvote. Some crazy ppl just downvoted me. Jun 25, 2013 at 19:29
• @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, 2013 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. Jun 25, 2013 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, 2013 at 20:00