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Answer :
The force exerted on the car due to collision is,
[tex]F=\frac{m(v-u)}{t}[/tex]Plug in the known values,
[tex]\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}[/tex]The deceleration of the car is,
[tex]a=\frac{F}{m}[/tex]Plug in the known values,
[tex]\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}[/tex]Therefore, the magnitude of deceleration of the car is 4.60g.
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