Observer a, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of v 0.9c and determines its lifetime to be ta-112 ns. observer a places markers in the laboratory at the locations where the particle is produced and where it decays. how far apart are those markers in the laboratory? 30.24 (b) observer b, who is traveling along with the particle, observes the particle to be at rest and measures its lifetime to be τ 280 ns. according to b, how far apart are the two markers in the laboratory?

a) 30.24 m

Explanation:

The particle s a lifetime of 112 ns and travelled at 0.9c from the point of view of observer A.

0.9c = 2.7e8 m/s

112 ns = 1.12e-7 s

Since we have the time and the speed from the point of view of observer A, we can use those to calculate the distance from the point of view of observer A without any transformations

From the point of view of observer B, the particle will be at rest and the lab will be moving at 0.9c. The travel between the marker will take 280 ns from the point of view of observer B.

Therefore the distance from the point of view of observer B will be:

d = 2.7e8 * 2.8e-7 = 75.6 m

Now, there is an error here, because observer A should see the particle as having a longer life than as observer B sees it and the distance observer B sees should be shorter, I suspect a typo in the question.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

i would say