Problem — Car into a Wall (Average Force)

Problem

A car of mass $1200\;\text{kg}$ travels at $36\;\text{km/h}$ and hits a wall, stopping in $0{,}12\;\text{s}$. Estimate the average force of interaction between wall and car. Discuss why airbags and the deformation of the vehicle increase safety.

Solution

Step 1 — Converting the speed. It is convenient to work in SI units:

$v = 36\;\text{km/h} = \frac{36}{3{,}6}\;\text{m/s} = 10\;\text{m/s}$

Step 2 — Average acceleration (deceleration). The car goes from $v = 10\;\text{m/s}$ to $0$ in the interval $\Delta t = 0{,}12\;\text{s}$. In magnitude

$a = \frac{\Delta v}{\Delta t} = \frac{10\;\text{m/s}}{0{,}12\;\text{s}} \approx 83\;\text{m/s}^2$

Step 3 — Average force. From Newton’s second law

$\bar F = m\,a = 1200\;\text{kg} \times 83\;\text{m/s}^2 \approx 1{,}0 \times 10^5\;\text{N}$

$\ev{\bar F \approx 1{,}0 \times 10^5\;\mathrm{N}}$

This is an enormous force, about ten times the weight of the car itself.

Discussion — airbags and deformation. The change in speed $\Delta v$ is imposed by the crash and cannot be reduced. Rewriting the second law as

$\bar F = \frac{m\,\Delta v}{\Delta t},$

we see that the force is inversely proportional to the time $\Delta t$ over which the stop occurs. Airbags, seatbelts with pre-tensioners and crumple zones (the panels that “concertina” on impact) exist precisely to lengthen $\Delta t$: by spreading the stop over a longer time, for the same $\Delta v$, the average force on the occupants drops proportionally. A rigid wall stops the body in a fraction of a second; an airbag can lengthen the stop just enough to bring the force below the injury threshold.

Links

Topics: Dynamics Concepts: Newton’s second law · Uniformly accelerated motion