# Homework Help: Relativity Problem

1. Dec 5, 2008

### Aeighme

1. The problem statement, all variables and given/known data

Holy Time Dilation! The Joker has captured Batman and Robin and has superglued them together to the roof of a train, which is moving at a steady velocity of magnitude v = 0.64c toward a low bridge. Confident that his diabolical plan will finish off the dynamic duo for good, the Joker returns to his seat in the train. Fortunately, Batman remembered to pack some super-bat-glue solvent in his utility belt, and is able to apply the solvent to the superglue. At the instant they are freed, the superheroes simultaneously set their stop watches to zero, and then by pre-arrangement they start running along the train in opposite directions at speed u = 0.74c, measured in the frame in which the train is at rest. They each run for a time Δt' = 19.3 ns, as measured on their own stop watches, and then they each jump from the train, just before the train passes under the bridge. They hit the ground with a sharp thud, and come instantly to a stop. Miraculously, neither member of the Dynamic Duo is hurt.
According to the Joker, who is at rest with respect to the train, how far apart are Batman and Robin when they jump?

2. Relevant equations
t=t1*gamma
d=vt

3. The attempt at a solution

I tried finding the Joker's measured time and then calculated the distance..but it didn't work..nor did it seem right.

2. Dec 5, 2008

### Staff: Mentor

Sounds right to me. Show exactly what you did.

3. Dec 5, 2008

### Aeighme

19.3e-9s * (1/(1-.12)1/2= t

Then...

d=vt with v of joker wrt batman and time from above.

4. Dec 5, 2008

### Staff: Mentor

Where did you get a speed of 0.1c?

5. Dec 5, 2008

.74c -.64c

6. Dec 5, 2008

### Staff: Mentor

But Batman and Robin are moving at 0.74c with respect to the train, which is the Joker's frame. The speed of the train is irrelevant.

7. Dec 5, 2008

### Aeighme

Ohhh. so if I use .74 instead, I should get the correct answer.

8. Dec 5, 2008

### Staff: Mentor

Give it a try.

9. Dec 5, 2008

### Aeighme

19.3e-9*(1-(.74)2)-.5=t
t=2.86943e-8 s
d=vt
d=.74*3e8*2.86943e-8=6.37m

?

10. Dec 5, 2008

### Aeighme

and then I double that answer to find the distance between batman and robin?
or do I find robin's distance a different way?

11. Dec 5, 2008

### Aeighme

oh, got it. Just double d..and it works.

Thank you!

12. Dec 5, 2008

### Aeighme

As measured on the ground, how far apart are the two points at which the pair hit the ground

For this next part, I now use the velocity wrt the ground..which would be .1c?

13. Dec 5, 2008

### Staff: Mentor

Yes. But you'll have to figure out the speeds of Batman and Robin with respect to the ground.
No, speeds don't simply add and subtract in special relativity the same as they do in Galilean/Newtonian relativity. You need to use the formula for relativistic addition of velocity.

Realize that since they move in opposite directions, they have different speeds with respect to the ground.

14. Dec 5, 2008

### Aeighme

u=(u1+v)/(1+(u1/c2))

with u being the velocity I'm attempting to find, v being that of train and u1 their speed wrt train?

15. Dec 5, 2008

### Aeighme

Alrighty. I got the right answer.
Thank you again. This stuff is mind bending >.<

16. Dec 5, 2008

Excellent!