A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.600c. The pursuit ship is traveling at a speed of 0.800c relative to Tatooine, in the same direction as the cruiser. (a) For the pursuit ship to catch the cruiser, should the velocity of the cruiser relative to the pursuit ship be directed toward or away from the pursuit ship? (b) What is the speed of the cruiser relative to the pursuit ship?

Video transcript

A little fellow physicists today, we're going to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A race of spaceships is conducted by scientists on earth spaceship. A takes off first at 0.400 C relative to the earth. Spaceship B takes off a while later at 0.720 C measured relative to the earth and the same and in the same direction as the spaceship A I, if spaceship B should catch up with spaceship, A, what is the direction toward or away from of spaceship? A velocity relative to spaceship B I I determine the velocity magnitude of spaceship A measured by a pilot in spaceship B? OK. So our end goal is to solve for two different answers. Our first answer we need to find is we need to find the direction of spaceship A velocity relative to spaceship B if spaceship B should catch up to spaceship A and then our second answer that we're trying to find is to determine the velocity magnitude of spaceship A as measured by a pilot in spaceship B OK. So we're given some multiple choice answers. We're given answers for I and for I, I, so let's read them off to see what our final answer pair might be. A is away from B and 0.449 CB is away from B 0.320 C and then C is toward B 0.449 CD is towards B 0.870 C and E is away from B 0. C and F I mean E is away from B 0.870 C and F is towards B 0.320 C. OK. So first off, let us note that the speed of both spaceships are close to the speed of light. Therefore, we can recall and use the equation for relativistic velocity addition. And let's call that equation one. So that states that V subscript XVX prime is equal to VX minus U divided by one minus U multiplied by VX divided by C squared. OK. So to solve, for part, I, we must consider that the distance between spaceship B and spaceship A must decrease in order for spaceship B to catch up with spaceship A. Therefore, the velocity of spaceship A is relative to spaceship B and must be directed toward spaceship B. So it has to be towards spaceship B following that logic, it has to be toward. So that is our answer for I hooray, we did it. So now let's start solving for I I so to solve for part I, I let us consider that the un primed frame represents earth and that the primed frame is spaceship B. Our goal is to determine the velocity VX prime of spaceship. A note that we know the velocity of the primed frame for spaceship B is U and the velocity of spaceship A is V and is relative to earth. So we can write the following that VX prime is equal to VX minus U divided by one minus U multiplied by VX divided by C squared where C is the speed of light. So we can then go on to plug in all of our known variables to solve for VX prime. So let's do that. VX prime is equal to VX, which VX is given to us as zero point 400 C minus U which U in the problem is 0.720 C divided by one minus you which is zero point 720 C multiplied by 0.400 C which is VX. So it's VX multiplied by U or U multiplied by VX potato, potato, same thing. So that's all divided by C squared. So we have VX prime is equal to VX minus U divided by one minus U multiplied by VX all divided by C squared. The speed of light squared So when we plug that into a calculator, we should get minus or negative negative 0.449 C. OK. Wonderful. So to find the velocity magnitude, we must take the absolute value of negative 0.449 C. So when we take the absolute value of so whenever we take the absolute value of anything that's negative, it becomes positive. So that means that the velocity magnitude is 0.449 C and that is our final answer. So the velocity magnitude of spaceship A measured by a pilot in spaceship B has to be 0.449 C. So looking at our multiple choice answers, the correct answer has to be the letter C I towards B and I I 0.449 C. Thank you so much for watching. Hopefully that helps and I can't wait to see you in the next video. Bye.