Question

A spaceship in distress sends out two escape pods in opposite directions. One travels at a speed v1 = + 0.70 c in one direction, and the other travels at a speed v2 = − 0.76 c in the other direction, as observed from the spaceship. Part A What speed does the first escape pod measure for the second escape pod? Express your answer using two significant figures.

Answer 1

To develop this problem it is necessary to apply the related concepts at relative speed.

When an observer perceives the relative speed of a second observer, the function is described,

$v' = \frac{v_1-v_2}{1-\frac{v_1v_2}{c^2}}$

Where,

$v_1$ = The velocity of the first escape pod

$v_2$ = The velocity of the second escape pod

c = Speed of light

v' = Speed of the first escape pod relative to the second escape pod.

Our values are given as,

$v_1$= 0.7c

$v_2$= -0.76c

Replacing we have,

$v' = \frac{v_1-v_2}{1-\frac{v_1v_2}{c^2}}$

$v' = \frac{0.7c-(-0.76c)}{1-\frac{(0.7c)(-0.76c)}{(3*10^8)^2}}$

$v' = \frac{0.7c-(-0.76c)}{1-\frac{(0.7c)(-0.76c)}{(3*10^8)^2}}$

$v' = 2.85*10^8m/s$

Therefore the speed of the first escape pod measure for the second escape pod is $v' = 2.85*10^8m/s$