I'm having trouble with the understanding on this last problem from my homework set.
Two cars playing demolition derby are moving towards each other with constant velocity, as measured by Marvin, who is a stationary observer. As measured by Marvin, a green car on the left is moving rightward at 0.80 c and an orange car on the right is moving leftward at 0.60c.
(b) At the moment the green car passes Marvin, the separation between the two cars is 900 m in Marvin’s reference frame. What is the time interval between the green car passing Marvin and the collision of the two cars, as measured by Marvin?
I tried converting the orange car on the right's speed to green car's reference frame which gave me
$$v = -0.946c,$$
and then converted the 900m to that reference frame, by dividing by
$$\gamma = \frac{1}{\sqrt{1 - 0.8^2}} = 1.67.$$
So the distance in the green car on the left's reference frame is
$$\frac{900m}{1.67} = 540 m.$$
Then,
$$v = \frac{\triangle{x}}{\triangle{t}}.$$
so
$$\triangle{t} =\frac{540m}{0.946c} = 1.90 \mu{s}.$$
Then we multiply by $\gamma$ to get back to Marvin's reference frame and we get
$$t = 3.17\mu{s}.$$
My professor gave the hint that the distance should be 514 m in Marvin's reference frame. Which means that
$$\frac{0.8c}{514m} = 2.14 \mu{s}.$$
What am I doing wrong? Any help is much appreciated.