You can derive position/time equations for each, but you must make sure the origin and time are in the same frame:
The accelerating car (starts at x=0 at t=0): x - x0 = v0t + (1/2)at2 or x = (1/2)at2
The constant velocity car: x-x0 = vt or x = x0 + vt
Plugging in the numbers (all units are MKS, so they mesh)
xNB = (1/2)(2)t2 = t2 and xSB = 300 - 25t (Note that v is negative, x0 and a are positive, choosing N as +)
We want to know when the NB and SB cars are at the same position for the same time (equate the positions and solve for t)
t2 = 300 - 25t (You can now use quadratic formula to solve for t and sub into equations for x)
Please consider a tutor. Take care.