The accompanying figure shows the plot of distance fallen vers...

Question

The accompanying figure shows the plot of distance fallen versus time for an object that fell from the lunar landing module a distance 80 m to the surface of the moon.

a. Estimate the slopes of the secant lines PQ₁, PQ₂, PQ₃, and PQ₄, arranging them in a table like the one in Figure 2.6.

b. About how fast was the object going when it hit the surface?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says during a physics experiment, a ball is dropped from rest at a height of 120 m above the ground. The figure shows the plot of distance falling versus time for the ball as it descends due to Earth's gravity. Estimate the slopes of the second lines MN1, MN2, and MN3. Approximately approximately how fast was the ball moving when it reached the ground. Round your answers to one decimal place. Below the problem, we're given a graph that has distance and meters plotted on the y axis and time in seconds along the x-axis. We also have a curve plotted on this graph, and this curve begins at the origin and is constantly increasing, passing through the points of N1, and 2, N3, and M. Now the first part of this problem was to determine the slopes of the sequent lines MN1, MN 2, and MN 3. And so, in order to determine the slopes of those secret lines, we need to determine the coordinates of the points M, N1, N2, and in 3. So, if we look at our graph, we can read those coordinates directly from the graph. So for our point in one, it is located at the coordinate of 2.20. The point in 2 is located at the coordinates of 3.45. The point in 3 is located. At the coordinates of 4.80. And point M is located at the coordinates of 5120. And so now we're going to use these points in order to determine the slope of ours lines. So the firsts line that we're going to look at is ours line MN1. So, we're looking for a slope, so we'll label it as M sub MN 1. And recall that the slope of a second line can be given by the change in the Y coordinates divided by the change in the X coordinates. So, for our slope here, for the second line of MN1, we're gonna have the quantity of 120 minus 20 in quantity, divided by the quantity of 5 minus 2 in quantity. And when we evaluate this expression, this is going to be approximately equal to 33.3 m per second. So here we've rounded to just one decimal place, and this has units of meters per second, since our distance was given in meters and our time in seconds. Now, we need to find the slope of our next secret line. So, will that be M sub MN 2. So we're looking at the second line MN 2. And this is going to be equal to, again, we'll have our change in Y coordinates divided by our change in X coordinates. So we'll have the quantity of 120 minus 45 in quantity, divided by the quantity of 5 minus 3 in quantity. And evaluate this expression, this is going to be equal to. 37.5 m per second and again rounding to one decimal place. So that is the 2nd slope that we needed to find. The last slope that we need to find is going to be the slope of the second line of MN 3. And this is going to be equal to, again, our change in our Y coordinates divided by our change in X coordinates. So this will be the quantity of 120 minus 80, and quantity, divided by the quantity of 5 minus 4. In quantity. And when we evaluate this expression, this is going to be equal to 40 m per second. So here we've now calculated all 3 of the slopes for our secret lines. So now we just need to approximately determine how fast our ball is moving when it reaches the ground, and we'll label that speed as V. So here we're looking for the instantaneous velocity, which would be the slope of the tangent line to a curve at the point where the ball reaches ground, which is going to occur at point M on our graph. But if we look at a curve, we see that the curve between points M and point N3, that part of the curve is approximately linear. So the slope of the second line of MN 3 is going to give us approximately the slope of our tangent line at point M. So that means that the speed of the ball, approximately when it hits the ground, is going to be 40 m per second, which is the slope of the second line of MN3. That is the last part of this question answered. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Secant Lines

Slope of a Line

Instantaneous Velocity

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