Question

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

Answer 1

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 = Ft`, so the force will be given by `F = I/t`

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

`F = (mv)/t`

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

`F=(mxx2v)/t`

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)`

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

Answer 2

Quadruple

Explanation:

We know that `W=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.