The following table gives the position

Question

The following table gives the position $s\left(t\right)$ of an object moving along a line at time $t$ . Determine the average velocities over the time intervals $\left\lbrack1,1.01\right\rbrack$ , $\left\lbrack1,1.001\right\rbrack$ , and $\left\lbrack1,1.0001^{}\right.]$ . Then make a conjecture about the value of the instantaneous velocity at $t=1$ .

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with velocities. This problem says during a test run, a drone flies in a straight line and this position sft at different times T is noted, calculate the average velocities over the intervals of 5 to 5.01 from 5 to 5.001 and five from 5 to 5.0001. Make a presumption on the instantaneous velocity at T equal to five. Below the question. We're given a table that contains two rows. The first row is labeled T and it has values of 55.00015 0.001 and 5.01. The next row is labeled SFT and it has values of 32 32.00179985 32.017982 N 32.1782. We're given four possible choices as our answers. For choice A we have 14.81 14.985 14.9987. It appears that the instantaneous velocity at T equal to five is 15. For choice B we have 14.81 14.985 14.9987. It appears that the instantaneous velocity at T equal to five is 14. For choice C we have 17.82 17.982 17.9985. It appears that the instantaneous velocity at T equal to five is 18. And for choice D we have 17.82 17.982 17.9985. It appears that the instantaneous velocity at T equal to five is 32. Now, this question asks us to calculate the average velocities over the three different time intervals given in the problem. So we're gonna start off with the first time interval which is going from 5 to 5.01. And so we need to calculate the average velocity over this time interval. So if we call your definition for an average velocity, that is just gonna be our change in position divided by our change in time over the given time interval. So we'll just note the average velocity as the average and that is gonna be equal to our change in position, which is gonna be s of 5.01 minus s of five. And that quantity is gonna be divided by our change in time, which is going to be 5.01 minus five. So we can use the table that we were given in the problem to substitute in values for our two positions. So doing so will have the average is equal to for of 5.01. When I look at the table, uh T equal to 5.01. S of T has a value of 32 0.1782. I'll subtract from that our position at T equal to five. And again, when I look at the table, this has a value of 32. I'm gonna take the quantity and divide it by our time interval. So when I do the subtraction in the denominator from the life line above, I have 5.01 minus five, which gives me 0.01. So when I plug these values into my calculator, I'll get the average is equal to 17.82. So that is the average velocity for this time interval that I'm looking for, we're gonna repeat the same exact process for the other two time intervals. So the next time interval that we're going to look at is going from 5 to 5.001. And again, we'll use our same definition of average velocity. So we'll have the average is equal to s of 5.001 minus S of five. And that quantity is gonna be divided by the quantity of 5.001 minus five. Again, substituting in values for our positions using the table will get, the average is equal to, for s of 5.001. We're gonna have the value of 32.017982. Then we'll have minus 32. And that quantity is gonna be divided by our time man, which is 0.001. Plugging these values into our calculator will get, the average is equal to 17.982. And so that is the average velocity for this time interval. So our last time interval that we're going to look at is going from 5 to 5.0001. And again, we'll just use our formula for average velocity. So we'll have, the average is equal to s of 5.0001 minus s of five. That quantity is gonna be divided by the quantity of 5.0001 minus five. Again, substituting in values for our position using the table will have the average is equal to 32.00179985 minus 32. And that quantity is gonna be divided by our time interval which is 0.0001. So now we can plug those values into our calculator and we'll get the average is equal to 17.9985. And so that is the average velocity for this time interval. Now, the last part to this question asked us to make a presumption about the value of the instantaneous velocity at t equal to five. To recall the, that the instantaneous velocity is the velocity that you get when you um calculate your average velocity. But when the time interval is going towards zero. So if we look at the values that we've calculated for each of the values, the time interval is actually getting smaller and smaller, closer and closer to zero. And if we look at the values for our average velocities, they get closer and closer to the value of 18. So it appears that our instantaneous velocity, which we label as V it's gonna be equal to 18 at T equal to five. So that is gonna be the last part to this question. And if we take all of our values as well as our um assumption here and look at our answer choices, the correct answer choice actually corresponds to answer choice C So I hope this explanation has been helpful and I'll see you in the next video.

Key Concepts

Average Velocity

Instantaneous Velocity

Limits

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