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The Starship Enterprise returns from warp drive to ordinary space with a forward speed of $50 \mathrm{km} / \mathrm{s}$. To the crew's great surprise, a Klingon ship is $100 \mathrm{km}$ directly ahead, traveling in the same direction at a mere $20 \mathrm{km} / \mathrm{s}$. Without evasive action, the Enterprise will overtake and collide with the Klingons in just slightly over $3.0 \mathrm{s}$. The Enterprise's computers react instantly to brake the ship. What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant. Hint: Draw a position-versus-time graph showing the motions of both the Enterprise and the Klingon ship. Let $x_{0}=0$ km be the location of the Enterprise as it returns from warp drive. How do you show graphically the situation in which the collision is "barely avoided"? Once you decide what it looks like graphically, express that situation mathematically.

In this problem we have an astronaut, The ship traveling at .60. We have outside observer I for inertial observer. And I've just put out here some astronomical objects or some type of star planet, yeah. Whatever galaxy and that's at rest relative to. I. Now this picture here is relative the perspective of the this second inertial observer. I there. Hi frank. That's right. In terms of perspective of the astronaut the ship. Now the ship was moving to the right with V. I mean he will say that the inertial observer is moving to left with V. And likewise, I told you that this object here was at rest relative to I. So it will also move be said to be moving to the left with the Yeah. Now what we need both those pictures. The first question is, who sees the proper time or who give can give you the proper time and the proper length for the ship? Let's talk about proper time. First proper time. The vent your timing must occur to in your frame of reference at the same position. So you can use a single clock now to I let's say the astronaut was. They were together at some point. So I says, OK, I saw you at at my origin then he watches the astronaut on the ship traveled to this distance. Astronomical body of some nature. It's not the same position, not eyes origin, is it? It is not. So I will not give the proper time. Now. From a perspective, he he was with I and then I moved away to left and then at some later time this astronomical object came also to his origin. He's a rest. This frame was just sitting there. Everything else is moving. So yes or not. The clock with the astronaut gives proper time. Now let's talk about the length of the ship. The proper length it gets given in the frame where that object is at rest. Well the object of the ship is at rest is in the astronauts frank. So the astronaut gives growth proper time and the proper length of the ship. Now part B. Is asking for if I sees his clock Meet the 24 hour point where will the clock B. O. Clock read? And the astronauts frank. Remember I've seen dilated time longer. So let's calculate this is what we want to calculate delta T. Nothing managed to Argentina. This is given by a frame. Got the T. Is given by the clock and ice frank. Now let us rewrite this delta T. is the 24 hours. So we've got to get that in terms of that, does he not? In terms of delta T minus delta T. Square root one must be squared overseas square. Remember the time dilation formula has doubted he. Not proper time over 1 -50 squared overseas script. And we can factor out the delta T when my square root, when my feet squared, she squared. And we can put in our numbers now and I can put it in any units. I want that. He says it in a day hours might seem to be reasonable enough. 24 hours, one minus the square root, one minus 0.6 C. Divided by C. Square that. And this comes out to be 4.8 hours. So that means where ice clock reached 24 24 hours, 8:00 would be 4.8 hours behind that. Now in part C, they've given us the contracted distance, what I sees. They want to know what the proper length the length that they would give us. So the formula for length contraction will not square root of one mice v squared, oversee square. And we can solve for l not very easily. Uh over square root on my feet squared, she squared And they gave us 110 m for the contracted length seen by I and 1 -0.6 C. Divided by C squared. And this equals 130 eight meters, Which is longer than 110 has to be. It's the proper length. Okay, Party asked from the perspective of the astronauts, that means a frame. What is what is the astronauts? Relativistic total energy? Well, Ian from a perspective where he's got going anywhere to rest in that framework in a frame that astronaut is just sitting there at the origin. So all that the restaurant has his rest energy. So it's a 70 kilograms at times three times 10 to the eight meters per second. Got a square that And this works out to be 6.30 Times 10 to the 18 source. So that is not only the rest energy of the astronaut, but it also is the total mechanical energy. They start in that astronauts frame of reference everybody else is coming and going away from him now to I not only will the asteroid have rest energy, but it also is moving. So he will say that E and the astronaut is the kinetic energy plus the rest energy. Now can we get to connect energy? So this is a nice simple farmer that we can and that formula, it's one over the square root. One must be squared C squared minus one times EMC squared plus EMC squared well minus EMC squared plus EMC squared. This is these are gone. And so this is just leaving you with M C squared or a square root of one modest V squared for c square. That's it. A simple relationship for the total toto energy when there is connect energy And you can see it actually very easy to get from you at one if it's not moving the zero EMC squared is left, it's moving and you got this. So really you just need this one formula. Everything follows from it, arrest energy, you know, being the total energy. All the it works no matter what. Okay. Um, so putting in our numbers now this will be, we actually actually have the numerator. I don't have to do this project. P complete. We know the top is 10 to the as the top is this and then we just have to put in the bottom one Mattis 0.6 C or C skirt And this works out to be seven 88 Times 10 to the 18th tools, which is larger than there are answers to D, which is good because this has and the rest energy but connect energy, so better be large and that is the whole product.

Looking at our graph, we confined our meaningful domain and range As far as our domain goes. We don't really care about the negatives here. So we see that we're positive for speed, cause we don't care about negative from zero, uh, from zero to infinity and similar with our range. We don't really care too much when were negative. So we get from 10 2 20 Remember, 20 is not going to be included in that graph because that's where are asking. Toad is

All right. If M two is equal to 20 our initial mass of the second object is equal to 20. We get the F is equal to 20 times 20 plus 50 Nevada by 20 plus five. Typing that into a calculator or simplifying it a little bit. I get 400 plus 50 wishes for 50 over 25 which of course, is equal to 18. So we have that final velocity of the two objects repelling equal to 18.

All right. Plugging in all the information they gave us. We end up with this. Let's simplify that a little bit. We get 1 15 plus 20 m, too minus 100 divided by. I'm gonna put the M two first. This is like my variable being first. Two plus five. This, of course, simplifies one step further into 20 M too. Plus 15 over em. Two plus five. That tells us we have a vertical ass in tow. At and two equals five and a horizontal. Ask himto. That's why. Or the F in this case equals 20. Using that information, we end up with this graph we see here. I was actually wrong. Uh, we should have had our vertical ascent. Tenant. Negative five. I missed that. And I apologize. Negative five. And now we see that's accurate. It's right here. Horizontal Assam towed it. 20


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