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

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olgalumyblog83d · Accepted Answer

In a collision, a vehicle&#039;s force is determined by its impulse, which is a gauge of its change in momentum. Ordinarily, the momentum has a high initial value before the collision and zero after it.The impulse is given by I=Ft, so the force will be given by F=ItWe can rewrite this as: &quot;The impulse will be the change in momentum, from p=mv to p=0.&quot;F=mvtIf we double the velocity, we go from v to 2v, so the expression changes to:F=m×2vtAlthough it appears that the force is simply doubled, keep in mind that a faster-moving vehicle will probably cause the collision to happen sooner. Because of how the car crumples, this is not a straightforward 1:1 relationship, but if it were, the time would be halved:F=m×2v12tIf so, there would be a fourfold increase in the force.

Jace Nelson · Accepted Answer

We know that W=Fd and W=ΔKE.  We can set ΔKE equal to Fd.We also know that KE=(12)(m)(v)2:(12)(m)(v)2=FdSolve for F:F=(12)(m)(v)2dSpeed is just v&#039;s absolute value. Doubling in will result in:FN=(12)(m)(2v)2dRemove the 2 from the fraction by squaring it. Leave (12)(m)(v)2d as it is. You will see why:FN=4((12)(m)(2v)2d)We can use F=(12)(m)(v)2d to plug into the equation we just made:FN=4FTherefore, doubling the speed will result in a fourfold increase in force.

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

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