Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Cas

  • 2 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...

  • This Question is Open
  1. HELLSGUARDIAN
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1349228986644:dw|

  2. HELLSGUARDIAN
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    tell me if im correct or not:)

  3. Cas
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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. HELLSGUARDIAN
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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. vf321
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @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. Cas
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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. Cas
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    does that look right? XD

  8. vf321
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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. Cas
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks!

  10. Cas
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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. Cas
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. vf321
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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. Cas
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. Cas
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    v = c/2... just a minute

  15. vf321
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  16. vf321
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    That's what my formula is used for.

  17. Cas
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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. vf321
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes. Good point. Then you're right.

  19. Cas
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  20. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.