A tourist walking at a speed of 1.3 m/s along a 7.5 km path that follows an old canal. If the speed of light in a vacuum were 3.0 m/s, how long would the path be, according to the tourist?
2007-03-19 01:01:17 · 3 answers · asked by christian m 2 in Science & Mathematics ➔ Physics
L = Lo {sqrt [1 - v^2/c^2]}
i think you missed 10^8 in speed of light and speed of tourist otherwise relativity will not come into picture i have corrected
Lo = is proper length = 7.5 km
[1 - v^2/c^2] = [1 - (1.3*10^8 / 3.0 *10^8)^2]
[1 - v^2/c^2] = [1 - (1.3/ 3.0)^2] = 0.81222
L = 7.5 km {sqrt [0.81222]}
L = 7.5 km {sqrt [0.81222]} = 6.759 km
path that will appear to tourist
2007-03-19 01:25:48 · answer #1 · answered by anil bakshi 7 · 0⤊ 0⤋
It would be foreshortened by the 'tau' value, which is given by:
(1-v^2/c^2)^1/2.
So, if c is 3, and v is 1.3, then v^2/c^2 = 1.69/9 = 0.18
1- 0.18 = 0.81.
(0.81)^1/2 = 0.90
7.5 km x 0.9 = 6.8 km.
2007-03-19 01:15:08 · answer #2 · answered by Ian I 4 · 0⤊ 0⤋
i.e travelling at 0.43c
L = L0 * (1 - (v^2/c ^2))^0.5
L = 0.90 * 7.5
L = 6.8 km (to 2 sig figs)
2007-03-19 01:12:22 · answer #3 · answered by gebobs 6 · 0⤊ 1⤋