Simple special relativity problem

[jethomas3182]

Homework Statement

From: http://www.phas.ubc.ca/~mcmillan/rqpdfs/1_relativity.pdf

The average lifetime of a pi meson in its own frame of reference is 26.0 ns. (This is its proper lifetime.)
If the pi meson moves with speed 0.95c with respect to the Earth, what is its lifetime as measured by an
observer at rest on Earth?

What is the average distance it travels before decaying as measured by an observer at rest on Earth?

Homework Equations

T=26 ns
v=.95c

t^2=T^2 + x^2
x=vt

The Attempt at a Solution

t^2 = T^2 + (vt)^2
t^2(1-v^2) = T^2
t^2(1-.95^2) = 26^2
t*sqrt(1-.95^2)=26
t=26/sqrt(1-.95^2) = 83.267 ns rounded to 83.3 ns Books answer: 83.3 ns

distance: x = vt = 83.3ns * 3*10^8 m/s = 24.99 m rounded to 25.0 Book's answer: 24.0

My answer is about 4% off.

83.267*.3=24.98
83.267*.29979=24.8

I got the median speed, but with halflife the median is mean*ln(2).

83.267*.29979/ln(2)=35.786 rounded to 35.8. Even worse.

Did the book make a typo, or am I missing something important?

 

[mfb]

The pi meson does not fly with the speed of light.

 

[jethomas3182]

The pi meson does not fly with the speed of light.

Thank you! Somehow I looked right over that.

 

[jethomas3182]

Reviewing this problem, I got

83.3*10^-9 * .95 * 3.0*10^8 = 23.74 meters

A more precise answer for the time is lower, a more precise number for lightspeed is lower. But the book's answer was 24.0, about 1% higher than my answer and their answer is claimed good to one part in 240 or so.

I don't see where I went wrong this time. It looks simple, but I don't see how to get it right.

 

[mfb]

Don't take the rounded value for the time, use a more precise one. This makes the distance even lower, however. 23.7 meters is the correct answer, rounding it to 24 is fine as well, giving 24.0 is wrong.

 

[jethomas3182]

Thank you! I wasn't confident that the book had it wrong.