The velocity in ft/s of an object moving along a line is given...

Question

The velocity in ft/s of an object moving along a line is given by v = ƒ(t) on the interval 0 ≤ t ≤ 8 (see figure), where t is measured in seconds.

a) Divide the interval [0,8] into n = 2 subintervals, [0,4] and [4,8]. On each subinterval, assume the object moves at a constant velocity equal to the value of v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0,8] (see part (a) of the figure)

Accepted Answer

Hello. In this video, we are told that a car traveling along a straight road has a velocity in meters per second, represented by the graph of V equal to FFT on the interval from 0 to 12 as shown. The time in seconds, divides 0 to 12 into N equals to 2 subintervals, 0 to 6 and 6 to 12. We want to estimate the displacement of the car on 0 to 12 by assuming that the car travels at a constant velocity equal to the value of V evaluated at a midpoint of each sub-interval. OK. So, first, to summarize what is given to us, we are working on an interval from 0 to 12, and we are told to split this interval into two sub-intervals. The first sub-interval is going to be defined from 0 to 6, and the second interval is going to be defined from 6 to 12. Now, what we want to do here is we want to estimate the displacement of the car on the overall interval. Now, in this case here, displacement. Is equal to the velocity located at the midpoint multiplied by the width. Of each subinterval. So, there's two things we need to solve for each of the intervals we are working with. We need to solve for the displacement on both intervals from 0 to 6 and 6 to 12, and we need to figure out the width of the intervals and the velocity at the midpoints. Let's go ahead and start by solving for the width of the sub-intervals. Now, the width of the sub-intervals is going to be defined as delta T, which is going to be the width of the interval in time. Now, this is going to be defined as our upper boundary minus our lower boundary, divided by the amount of sub-intervals that we are working with. In this case here, since we are working on an overall interval from 0 to 12, this will be defined as 12 minus 0, divided by the amount of sub-intervals we're working with, which is 2. This will simplify to 12 divided by 2, which is going to give us 6. So each of our sub-intervals is going to have a width of 6, and this is going to be the general width we will use to solve for the displacement on each of the sub-intervals. Now, starting with the interval from 0 to 6, the first thing we want to do is we want to solve for the velocity located at the midpoint. Now, the midpoint on the interval from 0 to 6 is going to be located at 3, and by using the graph, the value at this midpoint is going to be 12. So we know that the velocity at the midpoint on the interval from 0 to 6 is going to be 12. And so that means that the displacement for this interval is going to be 12 multiplied by 6, which is going to give us a value of 72 m per second. That is going to be the displacement for the first interval. Now we want to repeat this process, but for the interval from 0 to 12. Or from 6 to 12. So, we need to find the velocity at this sub-interval's midpoint. Now that midpoint is located at the value of 9, and by using the graph, the velocity located at this point is going to have a value of 18. So the velocity at the midpoint is 18, and that means that the displacement on the interval from 6 to 12 is going to be defined as 18, multiplied by 6, which is going to give us a value of 108. And so now what we want to do is you want to take the sum of both of these displacements in order to find the overall displacement from 0 to 12. So, for the displacement from 0 to 12, that is going to be the sum of 72 plus 108, and that is going to give us a total value of 180 m. So the total displacement for the car traveling from 0 to 12 is going to be 180 m, and with that being said, the overall solution to this problem is going to be C. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Key Concepts

Velocity Function

Midpoint Rule

Displacement

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