#### Answer

Please see the work below.

#### Work Step by Step

We know that the tension in the rope is given as
$T=Kx$
We plug in the known values to obtain:
$T=(1300)(0.55)=715N$
Now we can find the acceleration of the cars as
$a=\frac{T}{m}$
$a=\frac{715}{1900}=0.3763\frac{m}{s^2}$
The distance covered by cars is
$d=(\frac{1}{2})at^2$
We plug in the known values to obtain:
$d=(\frac{1}{2})(0.3763)(60)^2=680m$