An observer moving toward earth with a speed of .95c noted that it takes 5 min f
ID: 1521225 • Letter: A
Question
An observer moving toward earth with a speed of .95c noted that it takes 5 min for a person to fill her car with gas. Suppose instead that the observer had been moving away from earth with a speed of .8c. How much time would the observer have measured for the car to be filled in this case? An observer moving toward earth with a speed of .95c noted that it takes 5 min for a person to fill her car with gas. Suppose instead that the observer had been moving away from earth with a speed of .8c. How much time would the observer have measured for the car to be filled in this case?Explanation / Answer
here
t1 = first observed time (=5 min);
t2 = 2nd observed time (unknown; this is the quantity we want to find);
V1 = observer's initial speed (=0.95c);
V2 = observer's final speed (=0.80c);
Lorentz factors for V1 and V2:
y1 = 1/sqrt(1(V1/c)^2)
y2 = 1/sqrt(1(V2/c)^2)
The "proper time" (the time measured by the person filling her car) is:
t' = t1/y1
The proper time is stated to be the same for both observations, so we also have:
t' = t2/y2
Combine those two equations and solve for t2
t2 = t1(y2/y1)
t2 = t1sqrt((1(V1/c)^2)/(1(V2/c)^2))
t2 = 300 * sqrt((1(0.95 c/c)^2)/(1(0.8/c)^2))
t2 = 156.125 s
t2 = 2.6 min