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I'm independently studying SR right now. A practice problem in Hogg's text here, Chapter 4, problem 4-7 says (slightly modified):

Just prior, Hogg gives the equation to find the velocity of a cantalope thrown by B who's moving with respect to rest frame A. The final measured velocity

[tex]w = \frac{u+v}{1+\frac{uv}{c^{2}}}[/tex]

I will state below how I solved the problem and would appreciate it if someone offers consent or denial, thanks!

Let [tex]X_R[/tex], [tex]Y_R[/tex], and [tex]Z_R[/tex] be the velocities of rockets X, Y, and Z with respect to the Rest frame finish line.

We are told at the start that [tex]X_R = 0.9c[/tex].

We must calculate Y's velocity with respect to the finish line.

[tex]Y_R=\frac{X_{R} + 0.9c}{1+\frac{(X_{R})(0.9c)}{c^{2}}}=0.99447514c[/tex]

We must first calculate Z's velocity with respect to Y, then finally use Y's percieved velocity of Z to find Z's velocity with respect to X.

[tex]Z_Y=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c[/tex]

therefore

[tex]Z_R=\frac{Z_{Y} + 0.9c}{1+\frac{(Z_{Y})(0.9c)}{c^{2}}}=0.99998466c[/tex]

[tex]X_R = 0.9c[/tex] (this required no calculation)

[tex]Y_R = 0.99447514c[/tex] (this required one calculation)

[tex]Z_R = 0.99998466c[/tex] (this required two calculations)

*In an interplanetary rocket race there are three rockets: X, Y, and Z. Slow team X is travelling in their old rocket at speed 0.9c relative to the finish line. They are passed by faster team Y, and X observes Y to pass at 0.9c relative to themselves. But team Y observes fastest team Z to pass Y's own rocket at 0.9c. What are the speeds of teams X, Y, and Z relative to the rest frame of the finish line?*

Hint: The answer is not 0.9c, 1.8c, and 2.7c!Hint: The answer is not 0.9c, 1.8c, and 2.7c!

Just prior, Hogg gives the equation to find the velocity of a cantalope thrown by B who's moving with respect to rest frame A. The final measured velocity

*w*of the cantalope (thrown by B at velocity v who is himself moving at velocity u) as measured by A is:[tex]w = \frac{u+v}{1+\frac{uv}{c^{2}}}[/tex]

I will state below how I solved the problem and would appreciate it if someone offers consent or denial, thanks!

**Severian's Solution**Let [tex]X_R[/tex], [tex]Y_R[/tex], and [tex]Z_R[/tex] be the velocities of rockets X, Y, and Z with respect to the Rest frame finish line.

**Rocket X's velocity with respect to the finish line**We are told at the start that [tex]X_R = 0.9c[/tex].

**Rocket Y's velocity with respect to the finish line**We must calculate Y's velocity with respect to the finish line.

[tex]Y_R=\frac{X_{R} + 0.9c}{1+\frac{(X_{R})(0.9c)}{c^{2}}}=0.99447514c[/tex]

**Rocket Z's velocity with respect to the finish line**We must first calculate Z's velocity with respect to Y, then finally use Y's percieved velocity of Z to find Z's velocity with respect to X.

[tex]Z_Y=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c[/tex]

therefore

[tex]Z_R=\frac{Z_{Y} + 0.9c}{1+\frac{(Z_{Y})(0.9c)}{c^{2}}}=0.99998466c[/tex]

**Final Results**[tex]X_R = 0.9c[/tex] (this required no calculation)

[tex]Y_R = 0.99447514c[/tex] (this required one calculation)

[tex]Z_R = 0.99998466c[/tex] (this required two calculations)

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