Far in space, where gravity is negligible, a 425 kg rocket traveling at 75 m/s in the +x-direction fires its engines. FIGURE EX11.10 shows the thrust force as a function of time. The mass lost by the rocket during these 30 s is negligible. (b) At what time does the rocket reach its maximum speed? What is the maximum speed?

Video transcript

Hey everyone. So this problem is dealing with the impulse momentum theory. Let's see what it's asking us. A spacecraft with a mass of 3000 kg is moving in the positive direction with a velocity of 200 m per second through the empty expanse of space where the gravitational forces can be neglected, the spacecraft ignites its engines and the thrust force varies with time. As shown in the figure below, it is assumed that the mass lost by the spacecraft during the burn time is negligible. We are asked to determine the fault, what is the maximum speed attained by the spacecraft during the burn period? And at what time during the burn does the spacecraft achieve its maximum speed? Our multiple choice answers here are a 205 m per second at 25 seconds. B 210 m per second at 21 seconds. C 147 m per second at 28 seconds or D 155 m per second at 32 seconds. So the key, the first thing that we can, we call for this problem is the impulse momentum theory which states that the final momentum is equal to the initial momentum plus the impulse where impulse, I'm sorry, where momentum P is defined as mass multiplied by velocity and impulse is defined as the integral oh FDT or the force with respect to time. We can recall here that the integral is the same thing as the area under the curve. So the fact that we have a force curve force with respect to time we will solve for J by using this graph. So our impulse Jade is going to be this total area under the curve, we can split this into two parts. So we'll have a rectangle here and then we also have a triangle on this part. So our first half, so we'll call that, we'll call this area one and this area two. And so area one is simply the area of a rectangle. So that's the length multiplied by the quite. So 800 newtons is the height we go up. We see we're at 800 newtons on the Y axis and then on the X axis, we're at 10 seconds. And then for the second area, that's the triangles. We can recall that that's one half base times height or sorry, one half base multiplied by height. So one half multiplied by the base, which is 15 seconds multiplied by the height, which is also 800 newtons. And we get our impulse of 1.4 times 10 to the four Newton seconds. From the graph, we can see the impulse is positive the whole time. So speed will continue to increase, which means that the maximum speed is at the end of this impulse period or at a time of 25 seconds. And so we can then use our impulse go back to our impulse in momentum equation and solve for that final speed. So the mass of our spaceship multiplied by the final velocity of our spaceship is going to be equal to the mass of our spaceship multiplied by the initial velocity of our spaceship plus our impulse. And we have everything we need now that we've solved for impulse to solve for that final velocity. So we'll divide both sides by the mass of the spaceship to isolate the velocity, the final velocity variable and then plug in our known values. So the mass of the spaceship from the problem was given as 300 kg multiplied by the initial velocity given in the problem 200 m per second plus J. Our impulse we found was 1.4 times 10 to the four Newton seconds. I able to buy all of that by the mass. Again, 3000 kg plug that into our calculator and we get 205 m per second. And so again, 205 m per second at a time of 25 seconds is our answer. And that aligns with answer choice. A so that's all we have for this one, we'll see you in the next video.