If you double the speed of a vehicle how much does the force in a crash increase by?

2 Answers
Jun 12, 2017

The force will be between twice as great and four times as great, depending on some assumptions we make.

Explanation:

The force exerted by a vehicle in a crash will arise from its impulse, a measure of the change in its momentum. Usually the momentum starts high before the collision and ends up at zero after the collision.

The impulse is given by I = FtI=Ft, so the force will be given by F = I/tF=It

The impulse will be the change in momentum, from p=mvp=mv to p=0p=0, so we can re-write this as:

F = (mv)/tF=mvt

If we double the velocity, we go from v to 2v, so the expression changes to:

F=(mxx2v)/tF=m×2vt

This looks as though the force just doubles, but note that the time taken for the collision is likely to be shorter if the vehicle's velocity is higher. This is not a simple 1:1 relationship because it has to do with how the vehicle crumples, but if it did the time would be half as long:

F=(mxx2v)/(1/2t)F=m×2v12t

If that were the case, the force would be 4 times as great.

Jun 13, 2017

Quadruple

Explanation:

We know that W=FdW=Fd and W=Delta KE. We can set Delta KE equal to Fd.

We also know that KE=(1/2)(m)(v)^2:

(1/2)(m)(v)^2=Fd

Solve for F:

F=((1/2)(m)(v)^2)/d

Speed is simply the absolute value of v. Doubling in will yield:

F_N=((1/2)(m)(2v)^2)/d

Square the 2 and pull it out of the fraction. Leave ((1/2)(m)(v)^2)/d as it is. You will see why:

F_N=4(((1/2)(m)(2v)^2)/d)

We can use F=((1/2)(m)(v)^2)/d to plug into the equation we just made:

F_N=4F

So doubling the speed will quadruple the force.