Relativity Problem. I feel so stupid.

1. Jan 9, 2005

Eich

How fast must a pion be moving in order to travel 15 m before it decays?
The avg lifetime, at rest, is 2.6 x 10^-8 s.

The answer is supposed to be 0.89c

This sounds so easy. Like, just plugging in the formula but I'm not getting it. I haven't done relativity in so long. I think I must've forgotten more than I thought. I can't even find my notes now. Can someone please tell me how to get to the answer?

2. Jan 9, 2005

Yapper

We arent even touching on reletivity in my AP Physics BC class so I cant help you

3. Jan 10, 2005

Eich

Thanks yapper for looking at the problem, at least.

Anyone else?

4. Jan 10, 2005

Yapper

Try the college boards

5. Jan 10, 2005

vincentchan

the answer is not .89c, $$\gamma = 1/ \sqrt{1-.89^2}$$~$$2$$ it must travel at much higher speed, at leat .99c up (Sorry, i don't have a calculator handy and cant give you the exact answer), your answer is propabaly right, don't alway trust the textbook answer...

Last edited: Jan 10, 2005
6. Jan 10, 2005

learningphysics

Find a formula for the time in the stationary frame in terms of the velocity v. Then use that time in the time dilation formula. T'=T(1-v^2/c^2)^(1/2). Then solve for v.

Remember T' is the time in the moving frame (in the pion's frame).

I get 2.66*10^-8m/s which is approximately 0.89c

7. Jan 10, 2005

Curious3141

Let the pion's resting lifespan be $\tau$. The distance it is required to travel from the perspective of a resting observer is $d$, and the pion's velocity relative to the resting observer is $v$. $c$ is the speed of light.

From the frame of the pion, it will always "measure" its lifespan as $\tau$. This is the proper time interval. The resting observer will measure this time interval as $\gamma \tau$ where $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$.

The observer measures the velocity of the pion as $v$, where :

$$\frac{d}{\gamma \tau} = v$$

$$\frac{d}{\tau} = \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}$$

Put the relevant values in and solve for v and the answer is $v = 0.89c$

8. Jan 10, 2005

Eich

Awesome. Thanks!
I got the same answer as vincentchan. And other people did too so I just left it at that. I'm so grateful it's been proved otherwise.

V.v.v. thankful. :rofl:

9. Jan 10, 2005

vincentchan

just wanna point out...
2.66*10^-8m/s is 0.89*10^-16c

10. Jan 10, 2005

vincentchan

y don't you simply use $$t = \gamma \tau$$
and again, v is not .89c

Last edited: Jan 10, 2005
11. Jan 10, 2005