The Mathematics of Speed Problem Set

[Felipe Doria]

Imagine that you have a multistage rocket in which the first stage blasts off from the ground at v1, the second stage blasts off later with speed v2 relative to the first stage, and the third stage blasts off later still with speed v3 relative to the second stage.

A

From the perspective of someone stationary on the ground, what is the speed of the second stage? (For ease, assume you are working in units for which c=1.)

a) v2

b) v1+v2

c) (v1+v2)/(1+v1v2)

d) (v1−v2)/(1−v1v2)

B

From the perspective of someone stationary on the ground, what is the speed of the third stage? (For ease, assume you are working in units for which c=1).

a) (v1−v2−v3+v1v2v3)/(1−v1v2+v1v3+v2v3)

b) (v1+v2+v3+v1v2v3)/(1+v1v2+v1v3+v2v3)

c) (v1+v2+v3)/(1+v1v2+v1v3+v2v3)

d) (v1+v2+v3+v1v2v3)/(1+v1v2v3)

C

You are on a rocket traveling at a speed of 0.75c away from Earth. You wish to launch a probe that will travel at a speed of 0.9c away from Earth.

How fast does the probe need to be going in the rocket's reference frame?

a) .15c

b) .46c

c) .34c

d) .25c

 

I've written the answers I think are correct in bold letters, are they correct? I don't quite understand how to apply the formula (v+w) / (1 + (v.w)/c) to 3 variables, could someone please explain B to me? Thank you for your help.

[Felipe Doria]

Anyone?

[Gamatronics]

Problems A and C are indeed correct. On problem B what they´re asking you is basically the same as problem A but now with respect to the second stage. The V at the second stage you already have it, that is the answer for problem A.

 

So you need to apply the same formula, that is V+W/(1+VW/c²), where V=v₃ and W=(v₁+v₂)/(1+v₁v₂/c²).​ Substitute that W into the formula and you´ll get your answer, remember that in this case c=1 so that makes the algebra easier when solving for V (v₃ in this case).