How Fast Must a Pion Travel to Cover 15 Meters Before Decaying?

[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?

 

[Yapper]

We arent even touching on relativity in my AP Physics BC class so I can't help you

 

[Eich]

Thanks yapper for looking at the problem, at least.

Anyone else?

 

[Yapper]

Try the college boards

 

[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 can't give you the exact answer), your answer is propabaly right, don't alway trust the textbook answer...

 

[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

 

[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 $v = 0.89c$

 

[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.

 

[vincentchan]

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

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

 

[vincentchan]

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 $v = 0.89c$

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

 

[Curious3141]

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

Show us your reasoning.

 

[vincentchan]

oh, sorry, I thought its say the particle travel 15 mins... I didn't realize it is 15 meters, you are right, and so do the answer