Answer- 1.58*10^2 m B.What is the difference in arrival times? Answer-7.59*10^-7

Question

Answer- 1.58*10^2 m

B.What is the difference in arrival times?

Answer-7.59*10^-7 s

C.A passenger on train A measures the length of the Winnepeg platform by a similar method. How long does she take to travel from one end of the platform to the other?

NEED HELP WITH C

D. From the point of view of a passenger on train A, how fast is train B moving? (Give your answer as a fraction of the speed of light, e.g. if you get 0.952c, you enter 0.952.)

Answer-0.937

E. How long does it take train B to pass the passenger on train A?

Answer- 2.73*10^-7 s

NEED HELP WITH F

G. How long will the trip from Winnipeg to Vancouver take according to a clock on train A?

NEED HELP WITH G

I need help with parts C,F and G. i inculded the other parts in case there were needed for the other answers

Explanation / Answer

A. Length contraction: L = Lo(1-(v/c)^2)
Proper length, Lo = 220 m
L = 220*(1-(0.695c/c)^2) = 220*(1-0.695^2) = 220*0.719 = 158.2 m

B. t = d/v = 158.2/0.695c = 158.2/(0.695*3.00×10^8) = 7.59×10^-7 s
When the rear of the train arrives, the front has traveled 158.2 m in 7.59×10^-7 s

This can also be calculated with time dilation by finding the proper time.
Time dilation: T = To / (1-(v/c)^2)

Under "normal" conditions the time would be: T = d/v = 220/(0.695*3.00×10^8) = 1.055×10^-6 s
To = T*(1-(v/c)^2) = (1.055×10^-6)*(1-0.695c/c)^2) = (1.055×10^-6)*0.719 = 7.59×10^-7 s

---> Difference in time arrival = 7.59×10^-7 s

C. Platform is 50 m: t = d/v = 50/0.695c = 2.4×10^-7 s
Passenger on train A measures the time to be:
To = (2.4×10^-7)*(1-0.695c/c)^2) = (2.4×10^-7)*0.719 = 1.73×10^-7 s
This means that the passenger measures the platform to be:
d = vt = 0.695c*(1.73×10^-7) = 36.07 m
With length contraction: L = 50*(1-(0.695c/c)^2) = 50*0.719 = 35.95 m

---> Time to travel the platform = 1.73×10^-7 s

D. Relativistic velocities: Vab = (Va + Vb) / (1 + VaVb/c^2)
Vab = (0.695c + 0.695c) / (1 + 0.695c*0.695c/c^2) = 1.39c / (1+0.695^2) = 0.937c

For a passenger on Train A, it would appear that Train A is stationary and Train B is
moving at 0.937c. Same thing other way around. A passenger on Train B experiences
that her train is the stationary one and that Train A is moving at 0.937c.

E. Time dilation approach:
Time for train B to pass "normally": T = d/v = 220/0.937c = 7.83×10^-7 s
The passenger measures this time to be:
To = (7.83×10^-7)*(1-0.937c/c)^2) = (7.83×10^-7)*0.349 =2.73×10^-7 s

Length contraction approach: L = 220*0.349 = 76.78 m
t = d/v = 76.78/0.937c = 2.73×10^-7 s

---> Time for Train B to pass passenger = 2.73×10^-7 s

F. T = d/v = 1900000/0.695c = 9.11 ms

G. To = (9.11×10^-3)*0.719 = 6.55×10^-3 s = 6.55 ms
In the Train A frame, the 1900 km distance is contracted:
L = 1900*0.719 = 1366.1 km => t = 1366100/0.695c = 6.55 ms

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