# Velocity/Acceleration Problem

My engineering physics teacher gave the class practice problems for the upcoming exam, but I can't figure out how this one problem is supposed to be solveable.

A red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at Xr=0m and the green car is at Xg = 220m. If the red car has a constant velocity of 20kph, the cars pass each other at 44.5m, and if it has a constant velocity of 40kph, they pass each other at 76.6m. What are the initial velocity and constant acceleration of the green car?

So far, I've converted kph to m/s, 5.56m/s and 11.11m/s respectively, and divided the distances by those velocities to find the time it should have taken each car to reach those points, 8.01secs and 6.89secs respectively. Using the computed distances for the green car (220m - the distances of the red car) divided by the respective times to find the velocity of the green car at those points, 21.91m/s and 20.8m/s.

What has me stumped though is that if I use those velocities and the times to find the acceleration of the green car, I get -.99m/s^2, yet the book says I should get -2.0m/s^2. I'm not sure what I'm missing with this problem. Also, I would assume that I have to find the constant acceleration before I can find the initial velocity?

Because the distance covered by the green car in the time $t$ is the sum of that depending on the initial velocity v (given by $vt$) and that depending on the acceleration $a$ (given by $\frac {a}{2}\, t^2$), we have to solve the system
$$8.01\, v +\frac{a}{2}\, 8.01^2 =220-44.5=175.5$$ $$6.89\, v +\frac{a}{2} \, 6.89^2 =220-76.6=143.4$$
whose solutions are $v \approx 14.1 \,m/s\,\,$ and $a \approx 2.0 \,\, m/s^2\,\,$.