Physics 62 Special Relativity (35 of 43) Relativistic Sample Problem - Time

In this video I will finds time=? for a space ship, pursued by an enemy, to make it back to the safety of Earth.

Next video in the Special Relativity series can be seen at:

[namesee]

So you just jump off your spaceship at the base with instantaneous deceleration from .999c? Talk about a warm welcome home with massive kinetic energy.

[chavab]

A point that is slowly dawning on me -- with a concrete example in this video -- is that everything along the path of trajectory gets compressed. This is very confusing to me. From spaceship B's perspective, it is not clear why the path to Basecamp (A) is shorter than from A's perspective. From A's perspective, doesn't it seem like the ship is shortened?

[TDR]

Shouldn’t you divide by v instead of c?

[ThaoTran-ykfj]

i think you make a mistake. shouldn't the person on the spaceship will see the distance further than the person on earth because length is contracted.

so it should be L sub a = L sub b divided by gamma. Hence, L sub b = L sub a multiplied by gamma

[telvis]

The scenario is very funny, this guy just said you will leave the evil empire in the dust, are they not gonna follow you though? Lol. Great video though.

[zakirhussain-jsku]

I think relativistic effects remain restricted to object moving at relativistic speed. Such an object can't change anything outside its boundary limits.

[bus]

Excellent. Certainly gets the brain ticking!

[ShubhamGuptaGgps]

Sir i have a small dought that in this question you took dB=dA/gama

But in previous question you took LA=LB/gama

Why is it so?

[manog]

Motion is supposed to be relative. So if the base sees the spaceship to be 2.11 light years away and moving towards the base, the spaceship can be assumed as stationary and the base to be in motion towards the spaceship, and hence should be 2.11 light years away! So what is the peoblem then?

[Powerify]

But you are finding the time it takes base A and origin of B to coincide on B's perspective, but then on B's perspective he is stationary and base A is moving towards it, so the distance as measured by observer at B will be greater as according to him his reference frame is stationary

[vgovind]

Sir, I am sorry if this is a dumb question, but shouldn't the length transformation equation be the reverse according to the derivation?
That is, shouldn't L(sub A)=L(sub B)√1-v^2/c^2 ??
Also, if the reverse is correct, can you please explain how?

[AbhijeetSingh-nmll]

Wo we cover a distance in less time then it would take light to cover that same distance, are we going faster than light?

Or does that mean 34 days at our speed would equate to 2 years on base?

[anistu]

thank you. i didnt understand why you divided the distance as seen by B by the speed of light. did i miss something?