Question

A car moving at 5.86 m/s with a mass of 1,103 kg runs into a brick wall. If the car comesto rest over 129.9x10^-3s during the collision, what is the deceleration of the car in g's?

Answer 1

The force exerted on the car due to collision is,

$F=\frac{m(v-u)}{t}$

Plug in the known values,

$\begin{gathered} F=\frac{(1103\text{ kg)(0 m/s-5.86 m/s)}}{129.9\times10^{-3}s} \\ =\frac{(1103\text{ kg)(-5.86 m/s)}}{129.9\times10^{-3}s}(\frac{1\text{ N}}{1kgm/s^2}) \\ =-49758.1\text{ N} \end{gathered}$

The deceleration of the car is,

$a=\frac{F}{m}$

Plug in the known values,

$\begin{gathered} a=-\frac{49758.1\text{ N}}{1103\text{ kg}}(\frac{1kgm/s^2}{1\text{ N}}) \\ =-(45.1m/s^2)(\frac{1g}{9.8m/s^2}) \\ =-4.60g \end{gathered}$

Therefore, the magnitude of deceleration of the car is 4.60g.