Use the relationship between the change in momentum of a particular particle and elapsed time to calculate the average force applied to the car.

Given data:

The mass of the car is\(m = 1500\;{\rm{kg}}\).

The speed of the car is\(v = 45\;{\rm{km/h}}\).

The time is \(t = 0.15\;{\rm{s}}\).

The relation of average force can be written as follows:

\(\begin{array}{l}F = \frac{{m\Delta v}}{t}\\F = \frac{{m\left( {{v_0} - v} \right)}}{t}\end{array}\)

Here,\(\Delta v\)is the change in speed of the car, and\({v_0}\)is the initial speed whose value is zero.

Plugging the values in the above equation,

\(\begin{array}{l}F = \left[ {\frac{{1500\;{\rm{kg}}\left( {0 - 45\;{\rm{km/h}} \times \frac{{1\;{\rm{m/s}}}}{{3.6\;{\rm{km/h}}}}} \right)}}{{0.15\;{\rm{s}}}}} \right]\\F = - 1.25 \times {10^5}\;{\rm{N}}\end{array}\).

Thus, \(F = - 1.25 \times {10^5}\;{\rm{N}}\) is the required average force.

The relation to calculate deceleration can be written as follows:

\(F = ma\)

Plugging the values in the above equation,

\(\begin{array}{c} - 1.25 \times {10^5}\;{\rm{N}} = \left( {1500\;{\rm{kg}}} \right)a\\a = \left( {83.3\;{\rm{m/}}{{\rm{s}}^2} \times \frac{{1\;{\rm{g}}}}{{9.8\;{\rm{m/}}{{\rm{s}}^2}}}} \right)\\a = 8.49g\end{array}\)

Thus, \(a = 8.49g\) is the deceleration.