## anonymous 3 years ago Special Relativity question: Anne and Bob are driving along a long straight road. While passing a tree, Bob sets it on fire. Anne is approaching the tree with speed 0.6c; in Anne's frame of reference, Bob is exactly halfway between her and the tree when the first light from the fire reaches Anne's car. How fast is Bob driving with respect to the ground? Anyone know how to start this question? I'm having a lot of trouble setting it up...

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1. anonymous

|dw:1349228986644:dw|

2. anonymous

tell me if im correct or not:)

3. anonymous

Thanks for your answer! I'll look over it now! Quick question though: you said that the speed of light with respect to Anne is 1.6c, but I thought in any reference form light travels at speed c. And your answer is 1.6c for Bob, but again, I thought that speeds can't be higher than c.

4. anonymous

umm see we do not consider the Relative speed of light to any reference WHENEVER THE SPEED of REFERENCE is not comparable to the spee d of light bu in this case speed of anne is comparable(0.6 times the light), hence can be considered....now about the speed of bob im also thinking the sanme thing thats y i asked u to check:)

5. anonymous

@HELLSGUARDIAN I'm pretty sure that it's stated in the postulates for SR that nothing can move faster than c in ANY INERTIAL ref. frame. There are no accelerations here, so the postulate should hold.

6. anonymous

@vf321 Yes I believe so. What I've got so far (not sure if it's right): Consider the following in Anne's reference frame. - Anne's speed is 0 in her frame - The tree is moving towards her at 0.6c - Bob is moving towards her at some speed v - The firelight is moving towards her at c - at time t = 0, Bob and the tree are at the same position When the firelight has reached Anne at time t = t, it has traveled x = c*t metres, to where Anne is now. The tree has moved 0.6c*t metres closer and Bob has moved v*t metres closer. Since Bob is halfway between the tree and Anne, then we can say that (ct - 0.6ct)/2 = vt - 0.6ct A little algebra later and we get that Bob is moving at v = 0.8c in Anne's reference frame (towards her).

7. anonymous

does that look right? XD

8. anonymous

What might help you is the velocity addition formula for 1D SR. For a given reference frame velocity $$v_f$$, an object traveling an additional $$v$$ in that reference fram will have an unprimed velocity given by $\frac{v+v_f}{1+vv_f/c^2}$

9. anonymous

thanks!

10. anonymous

If my answer is right earlier for Bob's speed in Anne's frame, then i'm getting 0.385c as his speed in the ground's frame

11. anonymous

And oops, I think I accidentally deleted your last post... sorry!

12. anonymous

One thing I don't understand about your equation above - why is it so complicated? When Anne sees the tree on fire, Bob is halfway between Anne and the tree. From Anne's frame of reference, half the distance traveled by the light in some time t, $$ct$$, is equal to the velocity of Bob relative to Anne $$v$$ in the same time $$t$$. This gives us $\frac{ct}{2}=v{t}$$c/2=v$Or am I missing something?

13. anonymous

|dw:1349231562008:dw| is roughly how I set up part 1 to find v in Anne's frame

14. anonymous

v = c/2... just a minute

15. anonymous

The v, when relative to Anne, already "includes" the .6c

16. anonymous

That's what my formula is used for.

17. anonymous

The tree is "moving" as well, though, so the point where Bob is halfway between Anne and the tree will have moved since the light left the tree though?

18. anonymous

Yes. Good point. Then you're right.

19. anonymous

So using your equation above, $\frac{ v+v_{f}}{ 1+vv_{f}/c^2 }$ I guess I'd get $0.8c = \frac{ v+v_{f} }{ 1+vv_{f}/c^2} = \frac{ v + 0.6c }{ 1+ 0.6v/c } = \frac{v+0.6c}{\frac{c + 0.6v}{c}}$ $0.8(c+0.6v) = v+0.6c$ $0.8c + 0.48v = v + 0.6c$ $0.2c=0.52v$ $v = 0.385c$ as my answer