What should be the spring constant of a spring designed to bring a 1500 car to rest from a speed of 95 so that the occupants undergo a maximum acceleration of 4.5 ?

Shall we imagine a compressed spring that uncoils or a stretched string that contracts? I'll imagine a stretched spring that is stretched to length L beyon it's natural length and has spring constant k. Let x = 0 be the initial position of the moving end of the spring at time t= 0.

By Hook's law the force the spring exerts is $\displaystyle F_s = k (L - x) $ for $\displaystyle 0 \le x \le L $.
The max force occurs when $\displaystyle x = 0 $. Then $\displaystyle F_{max} = k L $.

We have a limit on acceleration. $\displaystyle a_{max} = 4.5 g $.
Newton says that $\displaystyle F_{max} = m a_{max} $ where $\displaystyle m $ is the mass of the car.. So must make $\displaystyle k L \le m a_{max} $

The work done by the spring when it snaps back to its natural length will be the kinetic energy gained by the car. Work is the integral of force as a function of distance.

$\displaystyle \int_0^L k(L-x) dx = \frac{1}{2} m v^2 $ where $\displaystyle v $ is the final velocity of the car.
$\displaystyle \Big |_0^L (kLx - \frac {kx^2}{2}) = \frac{1}{2} m v^2 $
$\displaystyle ( kLL - \frac{kL^2}{2}) - ( kL(0) - \frac{(k(0^2)}{2}) = \frac{1}{2} m v^2 $
$\displaystyle \frac {kL^2}{2} = \frac{1}{2} m v^2 $

ILet's make the spring constant large enough so that it produces the max acceleration. Then we have two equations in the unknowns $\displaystyle k $ and $\displaystyle L $:
eq 1. $\displaystyle \frac {kL^2}{2} = \frac{1}{2} m v^2 $
and
eq 2. $\displaystyle k L = m a_{max} $

Solve for $\displaystyle k $ in eq. 2 and put the result in eq 1.

$\displaystyle k = \frac {m a_{max}}{L} $

$\displaystyle \frac {m a_{max}}{L} \frac{L^2}{2} = \frac{1}{2} m v^2 $