Question

A space station sounds an alert signal at time intervals of 1.00 h . Spaceships A and B pass the station, both moving at 0.400c0 relative to the station but in opposite directions.

Part A
How long is the time interval between signals according to an observer on A?
Part B
How long is the time interval between signals according to an observer on B?
Part C
At what speed must A move relative to the station in order to measure a time interval of 2.00 hbetween signals?

Answer 1

Answer:

(A). The the time interval between signals according to an observer on A is 1.09 h.

(B). The time interval between signals according to an observer on B is 1.09 h.

(C). The speed is 0.866c.

Explanation:

Given that,

Time interval = 1.00 h

Speed = 0.400 c

(A). We need to calculate the the time interval between signals according to an observer on A

Using formula of time

$\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}$

Put the value into the formula

$\Delta t=\dfrac{1.00}{\sqrt{1-(\dfrac{0.400c}{c})^2}}$

$\Delta t=\dfrac{1.00}{\sqrt{1-(0.400)^2}}$

$\Delta t=1.09\ h$

(B). We need to calculate the time interval between signals according to an observer on B

Using formula of time

$\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}$

Put the value into the formula

$\Delta t=\dfrac{1.00}{\sqrt{1-(\dfrac{0.400c}{c})^2}}$

$\Delta t=\dfrac{1.00}{\sqrt{1-(0.400)^2}}$

$\Delta t=1.09\ h$

(C). Here, time interval of 2.00 h between signals.

We need to calculate the speed

Using formula of speed

$\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}$

Put the value into the formula

$2.00=\dfrac{1.00}{\sqrt{1-(\dfrac{v}{c})^2}}$

$\sqrt{1-(\dfrac{v}{c})^2}=\dfrac{1.00}{2.00}$

$1-(\dfrac{v}{c})^2=(\dfrac{1.00}{2.00})^2$

$(\dfrac{v}{c})^2=\dfrac{3}{4}$

$v=\dfrac{\sqrt{3}}{2}c$

$v=0.866c$

Hence, (A). The the time interval between signals according to an observer on A is 1.09 h.

(B). The time interval between signals according to an observer on B is 1.09 h.

(C). The speed is 0.866c.