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# When we estimate distances from velocity data, it is sometimes necessary to use times $t_0, t_1, t_2, t_3, ...$ that are not equally spaced. We can still estimate distances using the time periods $\Delta t_i = t_i - t_{i-1}$. For example, on May 7, 1992, the space shuttle $Endeavour$ was launched on mission STS-49, the purpose of which was to install a new perigee kick motor in an Intelsat communications satellite. The table, provided by NASA, gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. Use these data to estimate the height above the earth's surface of the $Endeavour$, 62 seconds after liftoff.

## $(10-0) 185+(15-10) \cdot 319 +\cdots+(125-62) \cdot 4151$

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So in this problem we are given That the space shuttle endeavour was launched on May seven, and were given a set of data here as the time and velocity or different uh positions which is this data right here that I have now entered in a spreadsheet. Just took a little spreadsheet into that data. We were asked to figure out use this data to figure out the height above the Earth's surface of the space shuttle Endeavour, 62 seconds after lift off. Okay, so first of all, what we do is we determine the tim the time differentials for each of these sets of data. So the first one from 0 to 10. Well, that's 10 seconds right? Which is what we have here. Next 1, 10- 15, another five seconds. 15-20. Another five, 2032 is another 12 and so on. Okay, so we're going to do is we're going to take that the distance this time is time times velocity gives us distance, right? Yeah, so we have the under the curve, we have that, we can draw rectangles of this, this this delta time the DT with and then the height. We got to figure out what the height will be because we can either draw somewhere the left corner of those, those rectangles is on the curve or the right corner. The right upper corner is on those curves. Okay, so how do we do that? Well, what we'll do is we'll do the left corners, first to the left end points first, which means we take delta time times the previous time. Select on this first one to be zero times 10 is zero And then five times 1 85 Gives us 9 25 and so on. Down through here until we get to 62 seconds. Then we add those up and that gives us 31,893 ft. All right, so then we do the same thing with the right in point on those triangles. We would take the 10 seconds for the first interval. I was the 185. That would give us 1850. We do that for the rest of these time periods. Down through here until we get to 62 seconds again. And those up that now gives us 5 54,694 ft. Okay, this is feet mrs feet. So then we just do the average of these two. So I'm doing this in a spreadsheet. Well, we'll just do this, we'll take this one plus that one friend of mine too. So I get 43294 feet. And so that is the approximate height. Their estimation of the height 62 seconds after launch launching

DM
Oklahoma State University

Integrals

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