Consider the position function s(t) =−16t^2+100t representing ...

Question

Consider the position function s(t) =−16t^2+100t representing the position of an object moving vertically along a line. Sketch a graph of s with the secant line passing through (0.5, s(0.5)) and (2, s(2)). Determine the slope of the secant line and explain its relationship to the moving object.

Accepted Answer

Welcome back, everyone. In this problem, a ball is thrown vertically with its height in meters given by the equation H of T equals -10 T2 + 50T. Draw a graph of H with the Secant line passing through 11 and 2 H2. State the relationship of the ball by finding the slope of the scant line. For our answer choices, each choice shows the graph of H with the second line, and A says the slope is -20, which represents the average rate of change of the height of the ball that's decreasing at an average rate of 20 m per second in the time interval open bracket 12 closed bracket. B Says that it's 20 where the slope of the sea can land represents the average rate of change of the height of the ball, which is increasing at an average rate of 20 m per second in the time interval. C says that it's -20 where the slope of the sea can line represents the average rate of change of the height of the ball, which is decreasing at an average rate of 20 m per second in the time interval, and the D. Says that it's 20 where the slope of the second line represents the average rate of change of the height of the ball, which is increasing at an average rate of 20 m per second in the time in the in the time interval. Now, to help us solve this problem, we're supposed to draw the graph, but if we can at least do a sketch of what the graph should look like, then we can compare it to our answer choices to see which one is correct once we also have the correct slope of the second line. So first, let's figure out the two points, OK? That it our problem says that our graph of H goes through, OK, or passes through along with the CC line before we can do our plot, OK? Now remember Let me just make a note here, OK. Remember that our problem told us, OK. That H which is sorry, H of T. Which is equal to -10 T2 plus 50T. If we're going to find H1 and H of 2, essentially we're trying to find the heights at these times where T equals 1 and T equals 2. Now at T equals 1. Each of one Is going to be equal to -10 multiplied by 12 + 50. Which is going to be 50 multiplied by 1, apologies, which is going to be negative 10. Let me write it beside it. It's going to be -10. Plus 50 Which is going to be equal to 40, so each of 1 equals 40. At T equals 2. Then each of 2 is going to be equal to -10 multiplied by 2 squared. Plus 50 multiplied by 2, and now that's -10 multiplied by 4 which is -40, plus 50 multiplied by 2 which is 100, and the 40 plus 100 equals 60. So this tells us that the points 140. And 260 are on our graph. So now we need to plot these points representing the height of the ball over time, and then we can sketch our curve. Now, notice for our curve here, since the coefficient of our squared term is negative and it's a quadratic expression, then our graph is going to be a parable opening downwards. OK. So for our sketch here, right. So let's say that this is T and then we have H of T, OK. Then no, we're gonna have and let's say this is we're 1 and this is 2. Just try that a little bit here. And then let's say that we're going up, so we go up to 60, and that means halfway between it is gonna be 30. So that means just a little above that, it's going to be 40. Now we know that we have the points 140 and 260. Then a parabola opening downwards is going to be going through our graph, OK? So it'll probably looks something like this. And let me just try to draw a nice. Ive here And let me just make one more try just to show that it's a curve, OK? So a curve may look, a curve would look something like this. And now we know that we have our second line, so let me put that in blue here. That's going to be passing through these two points. OK. OK, as a matter of fact, let me just kind of fix this up a little bit. OK. Let me just carry the curve in a bit more just to show that it's a curve, but remember at the end of the day we'll be comparing this to uh our graph itself. So it's gonna probably look something like that. And then we're gonna have our second line, OK? 4 points, 140 and 260. Now that we have an idea of what our graph should look like, we can find the slope of the sea and light because now we know that the slope. For any graph, the slope m is equal to the difference between y coordinates divided by the difference between X coordinates. So no, that's going to be equal to H 2 minus H 1 divided by 2 minus 1. So that's going to be 60 minus 40 divided by 1 and 60 minus 40 equals 20. Therefore, our slope is 20. Now the slope of the C line represents the average rate of change of the height of the ball between 1 2nd T equals 1 and 2 seconds T equals 2. So in this case. If the slope is 20, that tells us the ball's height is increasing at an average rate of 20 m per second during this time interval. So in summary, the slope of the CA length through the points 140 and 260 on the graph of H of T is 20, which tells us the height of the ball increases at an average rate of 20 m per second between T equals 1 and the T equals 2 seconds. Now if we go back to our answer choices. OK, when we go back to our answer rices here and we take a look at what the graphs look like here we can tell that D. OK, D Is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Consider the position function s(t) =−16t^2+100t representing the position of an object moving vertically along a line. Sketch a graph of s with the secant line passing through (0.5, s(0.5)) and (2, s(2)). Determine the slope of the secant line and explain its relationship to the moving object.

Key Concepts

Position Function

Secant Line

Slope of the Secant Line

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