Relativity Problem with Three Reference Frames

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.

My wild guess (no offense, though):

(0) it's not meant entirely as a trick question, but

(1) there's a small but nevertheless important mistake in how you understood and presented the statement of the problem; because it seems a lot more sensible if the specification "(b)" is instead:

(b) At the moment the green (left) car passes Marvin, the separation between the two cars is 900 m in the reference frame of the green car.

And (2) the professor's hint was (or should have been) instead that

the distance should be 540 m in Marvin's reference frame.

(i.e. the distance between Marvin and the participant "M_right_gate" who was and remained at rest wrt. Marvin and who indicated being passed by the orange car (entering from the right) simultaneous to Marvin indicating being passed by the green car).

So apparently your calculation of the value 540 m (as distance between Marvin and the M_right_gate) is already done and at least mathematically correct.

Now, not to give away the final result entirely quite yet, all that remains is indeed best considered simply in the frame of Marvin (and of M_right_gate, and of "M_collision"), namely

t := distance[ Marvin, M_collision ] / (0.8 c), and

t := distance[ M_right_gate, M_collision ] / (0.6 c), where

distance[ Marvin, M_collision ] + distance[ M_right_gate, M_collision ] ==
distance[ Marvin, M_right_gate ] ==
540 m,

because, surely, the race track is meant to be straight; so Marvin, M_collision and M_right_gate are situated straight to each other.

Now solve for "t".

• I agree - there must be a misrepresentation of the problem, since Marvin clearly can't measure a distance to be both 900 m and 514 m. – user10851 Jun 26 '13 at 0:03

This is a trick question.

All the measurements you are doing are in Marvin's frame, so no transformations to any other frames are required. The time required is just the distance in Marvin's frame divided by the relative speed in Marvin's frame.