Answer
d=16.9ft
Work Step by Step
$v=v_{o\:}+\:a_ct$
For car B
$v_{B=}\:60\:-12t$
$s\:=\:s_o\:+v_ot\:+\:1/2a_ct^2$
$s_B\:=\:d\:+60t\:+\:\frac{1}{2}\left(12\right)t^2$
For car A
$v=v_o\:+a_ct$
$v_A\:=60-15\left(t-0.75\right)$
t is greater than 0.75
$s=s_o\:+v_ot\:+\frac{1}{2}a_{\:c}t^2$
$s_A=\left(60\cdot 0.75\right)+60\left(t-0.75\right)-\frac{1}{2}\left(15.\left(t-0.75\right)^2\right)$
where t>0.74
At the moment of closest approach, v_A= v_B
Therefore:
$60-12t\:=60-15\left(t-0.75\right)$
and t = 3.75s
The worst-case devoid of a collision would happen if
$s_{A}=s_{B}$
At instance 3.75 and considering s_{B} and S_{A}
$\left(60\cdot 0.75\right)+60\left(3.75-0.75\right)-7.5\left(3.75-0.75\right)^2=\:d+60\left(3.75\right)-6\left(3.75\right)^2$
157.5=d+140.625
d= 157.5-140.625= 16.9ft
