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According to the conservation of energy, the final potential energy stored in the spring equals the car's initial kinetic energy.

Given data:

The mass of the car is\(m = 1200\;{\rm{kg}}\).

The maximum acceleration of the car is\(a = 4.0g\).

The initial speed of the car is calculated as follows:

\(\begin{aligned}{v_{\rm{o}}} &= 95\;\frac{{{\rm{km}}}}{{\rm{h}}}\\ &= 26.39\;\frac{{\rm{m}}}{{\rm{s}}}\end{aligned}\)

The final speed of the car is zero.

Assumptions:

Let k be the spring constant of the spring.

Let a distance compresses the spring \(\Delta x\).

Now the average force on the car by the spring is \(F = K\Delta x\)鈥�(i)

The action is equal to the reaction, then \(F = ma\). 鈥� (ii)

Comparing equations (i) and (ii),

\(\begin{aligned}k\Delta x &= ma\\k\Delta x &= m \times 4.0g\\\Delta x &= \frac{{4.0mg}}{k}\end{aligned}\) 鈥� (iii)

From energy conservation, the car's kinetic energy loss is equal to the potential energy gain of the spring. Then,

\(\begin{aligned}\frac{1}{2}k{\left( {\Delta x} \right)^2} &= \frac{1}{2}m{\left( {{v_{\rm{o}}}} \right)^2}\\k{\left( {\Delta x} \right)^2} &= m{\left( {{v_{\rm{o}}}} \right)^2}\end{aligned}\)鈥�(iv)

Using equations (iii) and (iv),

\(\begin{aligned}k{\left( {\frac{{4.0mg}}{k}} \right)^2} &= m{\left( {{v_{\rm{o}}}} \right)^2}\\k \times \frac{{16{m^2}{g^2}}}{{{k^2}}} &= m{\left( {{v_{\rm{o}}}} \right)^2}\\k &= \frac{{16m{g^2}}}{{{{\left( {{v_{\rm{o}}}} \right)}^2}}}\end{aligned}\)

Now substituting the values in the above equation,

\(\begin{aligned}k &= \frac{{16 \times \left( {1200\;{\rm{kg}}} \right) \times {{\left( {9.80\;\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right)}^2}}}{{{{\left( {26.39\;\frac{{\rm{m}}}{{\rm{s}}}} \right)}^2}}}\\ &= 2647.47\;\frac{{\rm{N}}}{{\rm{m}}}\end{aligned}\)

Hence, the value of the spring constant is \(2647.47\;\frac{{\rm{N}}}{{\rm{m}}}\).