Answer
$3\;\rm m/s$
Work Step by Step
We can use the conservation of momentum to find the speed of the 30-ton car.
$$p_i=p_f$$
The two cars initially were at rest, so $p_i=0$
$$p_f=m_1v_1+m_2v_2=0\tag 1$$
where $m_1$ is the 30-ton car and $m_2$ is the 90-ton car.
Since the speed of $m_1$ is 4 m/s relative to $m_2$, so
$$v_1=v_2+4$$
Solving for $v_2$ since we need to find $v_1$;
$$v_2=v_1-4$$
Plugging into (1);
$$m_1v_1+m_2(v_1-4)=m_1v_1+m_2 v_1-4m_2 =0$$
Solving for $v_1$;
$$v_1=\dfrac{4m_2}{m_1+m_2}$$
Plugging the known;
$$v_1=\dfrac{4\times 90\times 10^3}{(30+90)\times 10^3}=\color{red}{\bf 3}\;\rm m/s$$
This means that the speed of the 30-ton car relative to the ground is 3 m/s.
