#### Answer

$x_{max}=3.24ft$

#### Work Step by Step

We can determine the car's maximum penetration in the barrier as follows:
$T_1+\Sigma U_{1\rightarrow 2}=T_2$
$\frac{1}{2}mv_1^2-\int_0^{x_{max}}90(10^3)x^{\frac{1}{2}dx}=\frac{1}{2}mv_2^2$
We plug in the known values to obtain:
$\frac{1}{2}(\frac{4000}{32.2})(75)^2-90(10^3)x^{\frac{1}{2}}(\frac{x^{3/2}}{3/2})|_0^{x_{max}}=\frac{1}{2}(\frac{4000}{32.2})(0)^2$
This simplifies to:
$x_{max}=3.24ft$